U.U.D.M. Project Report 2018:18

Homological Algebra for Quiver Representations

Mateusz Stroiński

Examensarbete i matematik, 15 hp Handledare: Martin Herschend Examinator: Veronica Crispin Quinonez Juni 2018

Department of Mathematics Uppsala University

HOMOLOGICAL ALGEBRA FOR QUIVER REPRESENTATIONS

MATEUSZ STROIŃSKI

Abstract. Quiver representations arise naturally in the study of representation theory for associative algebras. A par- ticularly simple case to consider is that of representations of finite acyclic quivers, which is the object of study of this thesis. In this thesis we give an explicit presentation of some elementary constructions in the category of quiver representations, which then allows us to formulate and discuss some basic homological algebra within that category. The tools of homological algebra can then be applied to study the canonical problem of represen- tation theory: classification of the indecomposable representations.

Contents 1. Introduction 1 2. First definitions 1 3. Projective and Injective Representations 11 4. Irreducible morphisms and almost split sequences 23 5. Auslander-Reiten Translation Revisited 32 References 38

i 1. Introduction Quivers, and in particular their representations, owe their significance to their role as an important tool in study of representations of finite-dimensional algebras (and more generally also Artinian algebras), especially due to the fundamental developments of that field made by Peter Gabriel (resulting in Gabriel’s famous theorem presented in [3]) and those made by Maurice Auslander and Idun Reiten, resulting in an entire field of study known as Auslander-Reiten theory (basic elements of which are introduced in the following text; for a more detailed and advanced text we refer to [2] ). The following text focuses on finite-dimensional representations of finite acyclic quivers, and hence treats them as an independent object of study rather than a tool used in study of other objects. The first part of the text introduces the category RepK Q of finite-dimensional representations of a given finite acyclic quiver together with its fundamental properties (in particular, RepK Q is a K-linear category), gives explicit descriptions of some essential categorical notions realized in quiver representations, such as direct sums and products, pullbacks and pushouts and also kernels and cokernels.

Second part introduces projective and injective representations, describes the fundamental family {P (i)}i∈Q0

of indecomposable projective representations, and similarly a family {I(i)}i∈Q0 of indecomposable injective representations, and then shows that those families generate the projective respectively injective represen- tations of given quiver Q, that is, given a projective representation P , P can be decomposed as a direct sum with summands in said family, P = L (Lni P (i)), and similarly for injective representations. The i∈Q0 j=1 extensive presentation of projective representations is followed up by a discussion of projective resolutions, concluding that every representation admits a projective resolution of length 1. Those facts then allow us to conclude that similar statements hold for injective representations of Q, when considered (due to a duality op between RepK Q and RepK Q shown in the same part) as the projective representations of the opposite quiver Qop. The second part concludes with a brief introduction to the Auslander-Reiten translation - a map τ : RepK Q → RepK Q whose significance is revealed in the fourth and final part of the text. The third part of the text gives a more concrete description of the relation between representation theory of quivers and that of finite-dimensional algebras, by presenting an equivalence of RepK Q and mod-KQ - the category of finite dimensional KQ-modules, where KQ denotes the path algebra of Q. It focuses more on homological algebra of those modules, introducing important notions such as almost split sequences and irreducible morphisms and their direct relation to the fundamental question of representation theory (here of finite-dimensional algebras, although for this text the path algebras of finite acyclic quivers, being a proper subset of finite-dimensional algebras, is the object of focus) - the classification of indecomposable representations. This question is even more relevant to the fourth part of the text, which reveals some important properties of the aforementioned Auslander-Reiten translation, most notably it being a bijection between the set of indecomposable non-injective modules and that of indecomposable non-projective modules. It also introduces the Auslander-Reiten quiver Γ(mod-A) of mod-A (where A denotes a finite-dimensional algebra), which, using τ gives a visual representation of the structure of indecomposable modules of A, hence providing crucial information about the category mod-A itself. Finally it should be noted that the first half of the text follows (roughly) chapters 1 and 2 of [7], whereas the latter half consists of parts of chapter IV of [1]. Hence the reader is referred to those texts for a more thorough exposition of the material presented here.

2. First definitions

Definition 2.1. A quiver Q = (Q0,Q1, s, t) consists of

• Q0 - a set of vertices • Q1 - a set of arrows • s : Q1 → Q0 - a map mapping an arrow to its starting point • t : Q1 → Q0 - a map mapping an arrow to its terminal point

Definition 2.2. M is a representation of Q over field K if and only if M = (Mi, ϕα)i∈Q0,α∈Q1 , where Mi is a K-vector space, and ϕα : Mi → Mj is a K-linear map, provided that s(α) = i, t(α) = j

Throughout this text we assume the following: 1 • All representations considered are finite-dimensional, i.e. each Mi is finite-dimensional. • All representations are over an algebraically closed field K (unless otherwise stated).

0 0 0 Definition 2.3. Let M = (Mi, ϕα)i∈Q0,α∈Q1 , M = (Mi , ϕα)i∈Q0,α∈Q1 be two representations of quiver Q. 0 A morphism of representations f : M → M is a collection of linear maps (fi)i∈Q0 such that for each α ∈ Q1, the following diagram commutes:

ϕα Mi Mj

fi fj 0 0 ϕα 0 Mi Mj

In particular, f is an isomorphism if each fi is an isomorphism of vector spaces. Finally, we denote the set of morphisms from M to M 0 by Hom(M,M 0)

Remark 2.4. Given representations M, N, P of quiver Q, and given morphisms g ∈ Hom(M,N), f ∈

Hom(N,P ), let f ◦ g = (fi ◦ gi)i∈Q0 . Clearly f ◦ g is in Hom(M,P ) and thus representations over K of Q together with their morphisms constitute a category, denoted by RepK Q .

0 0 Proposition 2.5. Given two representations of Q, M,M ∈ RepK Q, the set of morphism Hom(M,M ) 0 forms a vector space, i.e. we have Hom(M,M ) ∈ VectK .

0 0 Proof. Let f, g ∈ Hom(M,M ) We know that ∀i ∈ Q0 : fi, gi ∈ Hom(Mi,Mi ) ∈ V ectK and thus by setting f + g = (fi + gi)i∈Q0 , kg = (kgi)i∈Q0 (∀k ∈ K) we can satisfy all the vector space properties, provided that our morphisms are well defined, which is checked as follows:

0 0 0 ϕα ◦ (fi + gi) = ϕα ◦ fi + ϕα ◦ gi = fj ◦ ϕα + gj ◦ ϕα = (fj + gj) ◦ ϕα and 0 0 ϕα ◦ kfi = k(ϕα ◦ fi) = k(fj ◦ ϕα) = kfj ◦ ϕα 

Proposition 2.6. 0 = (0i, 0α) is a zero object of RepK Q i.e. it is both initial and terminal in RepK Q .

Proof. Let M = (Mi, ϕα)i∈Q0,α∈Q1 ∈ RepK Q 0 is initial: ∀i ∈ Q0, 0i is initial in VectK i.e. ∃!ψ ∈ Hom(0i,Mi) = 0 and 0 ◦ ϕα = 0 = 0 ◦ 0, so the morphism is unique and well-defined. 0 is terminal: ∀i ∈ Q0, 0i is terminal in VectK i.e ∃!λ ∈ Hom(Mi, 0i) = 0 and ϕα ◦ 0 = 0 = 0 ◦ 0, so the morphism is unique and well-defined. 

0 0 0 Definition 2.7. Let M = (Mi, ϕα)i∈Q0,α∈Q1 ∈ RepK Q and let M = (Mi , ϕα)i∈Q0,α∈Q1 ∈ RepK Q. We 0 0 0 h ϕα 0 i define the direct sum of M and M as M ⊕ M = (Mi ⊕ M , 0 ) i 0 ϕα

0 0 Proposition 2.8. Let M,M ∈ RepK Q then M ⊕ M is: Q 0 0 a) The product M M of M and M in RepK Q ` 0 0 b) The coproduct M M of M and M in RepK Q Proof. For both parts we use the fact that the notions of finite product and finite coproduct coincide in Q ` VectK , i.e for V,W ∈ VectK , we have V ⊕ W ' V W ' V W

First let N = (Ni, γα)i∈Q0,α∈Q1 ∈ RepK Q a) Let f ∈ Hom(N,M), f 0 ∈ Hom(N,M 0). 0 1 0 Let π = (πi)i∈Q0 : M ⊕ M → M, where πi = [ ]i, (i.e define π pointwise as projections from the direct h fi i 0 sum onto the left summand), g = (gi)i∈Q0 = 0 : N → M ⊕ M (i.e define g pointwise as morphisms fi 0 induced by product property of M ⊕ M ). The following diagram then commutes for all i ∈ Q0 due to product property of M ⊕ M 0: 2 Ni

f 0 i i f gi

0 0 Mi (M ⊕ M )i Mi πi 0 πi

Moreover, the choice of g is unique as gi is the unique morphism making said diagram commute. So all that needs to be checked is that π and g are well-defined, i.e for all α ∈ Q1 these diagrams commute:   ϕα 0   0 ϕ0 0 α 0 (M ⊕ M )i (M ⊕ M )j

πi πj

ϕα Mi Mj

γα Ni Nj   gi ϕα 0 gj   0 ϕ0 0 α 0 (M ⊕ M )i (M ⊕ M )j And indeed we have

h ϕα 0 i h ϕα 0 i ϕα ◦ πi = ϕα ◦ [ 1 0 ] = [ ϕα 0 ] = [ 1 0 ] 0 = πj ◦ 0 i j j 0 ϕα 0 ϕα and also h ϕα 0 i h ϕα 0 i h fi i h ϕα◦fi i h fj ◦γα i 0 ◦ gi = 0 ◦ 0 = 0 0 = 0 0 ϕα 0 ϕα fi ϕα◦fi fj ◦γα

0 h fj ◦γα i as both f and f are well-defined morphisms, and finally 0 = gj ◦ γα fj ◦γα 0 0 1 b) Let f ∈ Hom(M,N), f ∈ Hom(M ,N). Similarly to a), let ι = (ιi)i∈Q0 , where ιi = [ 0 ]i (i.e define 0 0 ι : M → M⊕M pointwise as inclusions of the left summand into the direct sum), g = (gi)i∈Q0 : M⊕M → N, 0 0 where gi ' [ fi fi ] (i.e define g pointwise as morphisms induced by coproduct property of M ⊕ M . Just like in the case of product, our desired diagram now commutes pointwise for all i ∈ Q0 and it also does so uniquely, all due to direct sum being the coproduct in VectK , and yet again all we need to check is that g and ι are well-defined. For ι we have: h i h i ϕα 0 ϕα 0 1 ϕα 1 0 ◦ ιi = 0 ◦ [ ] = [ ] = [ ] ◦ ϕα 0 ϕα 0 ϕα 0 i 0 0 j and for g we proceed as follows:

0 0 0 h ϕα 0 i h ϕα 0 i γα ◦ [ fi fi ] = [ γα◦fi γα◦fi ] = [ fj fj ] ◦ 0 = gj ◦ 0 0 ϕα 0 ϕα 

Definition 2.9. M ∈ RepK Q is indecomposable if and only if M 6= 0 and there are no non-zero represen- tations L, N such that M = L ⊕ N

Theorem 2.10 (Krull-Schmidt Theorem). Given a representation M ∈ RepK Q, M admits a unique (up to reordering of summands) direct sum decomposition into indecomposable summands, i.e M ' M1 ⊕ · · · ⊕ Mn such that M1,...,Mn are indecomposable.

Proof. Theorem 1.2 in [7]  3 Proposition 2.11. RepK Q has pullbacks, i.e., given the following diagram in RepK Q : A B

f g D

there is a representation X ∈ RepK Q together with morphisms ρA ∈ Hom(X,A), ρB ∈ Hom(X,B) consti- tuting a pullback of the diagram. Lemma 2.12. The category R -Mod of left R-modules (for some ring R) has pullbacks. Proof. Given a diagram as above in R -Mod, let X = {(x, y) ∈ A ⊕ B : f(x) = g(y)} = Ker(d), where d = (f, −g): A ⊕ B → D, and let ρA, ρB be the restrictions of canonical projections πA, πB to X. Clearly g ◦ ρB = f ◦ ρA, so X completes the pullback square, thus it remains to show that the square is universal, i.e. given Y ∈ R -Mod, qA ∈ Hom(Y,A), qB ∈ Hom(Y,B) such that f ◦ qA = g ◦ qB, ∃!h : Y → X such that qB = ρB ◦ h, qA = ρA ◦ h, which is illustrated by following diagram: Y qB h

ρB qA X B

ρA g f A D

qA We achieve that by letting h = (qA, qB) = [ qB ] (which is well-defined as f ◦ qA = g ◦ qB). Clearly we have

ρA ◦ h = ρA ◦ (qA, qB) = qA, ρB ◦ h = ρB ◦ (qA, qB) = qB 0 0 so the diagram commutes. Uniqueness of h can be proven as following: given h ∈ Hom(Y,X), ρA◦h = ρA◦h, 0 0 0 ρB ◦ h = ρA ◦ h and thus h = h, but both those conditions are necessary for h to satisfy the diagram above, 0 0 and thus if h does so, we get h = h. 

This gives us a foundation to prove the same property for RepK Q just like we did with earlier statements: by borrowing the pointwise construction from VectK (= K -Mod) and endowing it with some canonical choice of arrow-wise structure so that all the morphisms are well defined.

h i ϕαXi 0 Proof of proposition 2.11. Let X = (Xi, )i∈Q0,α∈Q1 (where A = (Ai, ϕα),B = (Bi, ψα)), where 0 ψαXi Xi is the pointwise pullback of

Ai Bi

fi gi Di

Let ρA : X → A = (ρA )i∈Q , ρB : X → B = (ρB )i∈Q . Then ρA ∈ Hom(X,A) as ϕα ◦ ρA = ϕα ◦ [ 1 0 ] = h i i 0 i 0 i i ϕαXi 0 [ 1 0 ] ◦ . Analogous statement holds for ρB. Thus, X completes the commutative square. i 0 ψαXi

We still need to prove that X does it in a universal manner. Let Y = (Yi, ωα)i∈Q0,α∈Q1 ∈ RepK Q, qA ∈

Hom(Y,A), qB ∈ Hom(Y,B) such that f ◦ qA = g ◦ qB Let h = (hi)i∈Q0 be the morphism defined pointwise by Xi being pullback pointwise (i.e. Xi’s universal property of pullback of diagram above in V ectK ). It is clear that h is the only possible candidate, as for all i ∈ Q0, hi is the unique morphism making the diagram commute at i, which is necessary for h to make the corresponding diagram in RepK Q commute. Thus we only qA need to show that h is well-defined, i.e. h ∈ Hom(Y,X). Recall from Lemma 2.12 that hi = (qA , qB ) = [ q ] h i i i B ϕαXi 0 We need to show that given α : i → j ∈ Q1, we have ◦ hi = hj ◦ ωα In particular, the following 0 ψαXi holds: h i h i h i ϕαXi 0 ϕαXi 0  qAi  ϕαXi ◦qAi 0 ◦ hi = ◦ qB = 0 ψαXi 0 ψαXi i 0 ψαXi ◦qBi 4 Now since qA ∈ Hom(Y,A) we have that ϕα ◦qAi = qAj ◦ωα and that equality implies ϕα Xi ◦qAi = qAj ◦ωα A similar equality holds for B and ψ. Thus

h i h i h qA ◦ωα i h qA i ϕαXi 0 ϕαXi ◦qAi 0 j j ◦ hi = = qB ◦ωα = qB ◦ ωα = hj ◦ ωα 0 ψαXi 0 ψαXi ◦qBi j j 

Proposition 2.13. RepK Q has pushouts, i.e. given the following diagram in RepK Q : D f g

A B

There is a representation X ∈ RepK Q together with morphisms ιA ∈ Hom(A, X), ιB ∈ Hom(B,X) consti- tuting a pushout of the diagram. Lemma 2.14. R -Mod has pushouts. Proof. Consider a diagram as above in R -Mod. Let X = (A ⊕ B)/K, where K = {(f(d), g(−d) | d ∈ D} = (f, −g)(D). Let ι : A → A ⊕ B, ι0 : B → A ⊕ B be the usual inclusions, π : A ⊕ B → X be the canonical 0 0 0 projection. Let γA = π ◦ ι, γB = π ◦ ι . Clearly γA ◦ f = γB ◦ g as π ◦ ι ◦ f − π ◦ ι ◦ g = π ◦ (ι ◦ f − ι ◦ g) = 0 as Im(ι◦f −ι0 ◦g) = Ker(π). So the pushout square commutes. Now we need to show that it indeed is a pushout square, i.e. has the following universal property: given Y ∈ R -Mod, qA ∈ Hom(A, Y ), qB ∈ Hom(B,Y ) such 0 that qA ◦ f = qB ◦ g, there is a unique morphism h ∈ Hom(X,Y ) such that β = h ◦ γA, β = h ◦ γB, which is illustrated by the diagram below: g D B

f γB qB γ A A X h

qA Y 0 Let ω = (qA, qB) i.e. the unique morphism such that ω ◦ ι = qA, ω ◦ ι = qB. We have 0 0 ω ◦ (ι ◦ f − ι ◦ g) = ω ◦ ιf − ω ◦ ι ◦ g = qA ◦ f − qB ◦ g = 0 so K ⊂ Ker(ω), and thus by factorization theorem there is a unique morphism η : X → Y such that η ◦π = ω Thus η ◦ γA = η ◦ π ◦ ι = ω ◦ ι = qA and similarly for B, so the diagram commutes. η is also unique such 0 0 morphism: if η ∈ Hom(X,Y ) makes the diagram commute, it must have η ◦ π ◦ ι = qA = η ◦ π ◦ ι, and 0 0 0 0 0 η ◦ π ◦ ι = qB = η ◦ π ◦ ι so by universal property of coproducts we have η ◦ π = η ◦ π, and so η = η as π is epi. 

Proof of proposition 2.13. Let X = (Xi, χα)i∈Q0,α∈Q1 , where Xi is the pullback of

Di fi gi

Ai Bi

and χα : Xi → Xj is defined by χα((a, b) + Ki) = (ϕα(a), ψα(b)) + Kj. (where A = (Ai, ϕα),B = (Bi, ψα)).

In analogy to the proof about pullbacks, we first show that γA = (γAi )i∈Q0 is a well-defined morphism: ¯ χα ◦ γAi (a) = χα ◦ πi ◦ ιi(a) = (ϕα(a), 0) + Kj = πj ◦ ιj ◦ ϕα = γAj ◦ ϕα

so γA ∈ Hom(A, X), and similarly for B, and thus X completes the square. Let Y = (Yi, κα)i∈Q0,α∈Q1 , qA =

(qAi )i∈Q0 , qB = (qBi )i∈Q0 such that qA ◦ f = qB ◦ g, i.e. Y completes the square. Let h : X → Y = (hi)i∈Q0 consist of morphisms induced pointwise by pushout property. Clearly, h is the only possible candidate for the induced pushout morphism. It remains to show that h is a well-defined morphism from X to Y , which it is if and only if for all arrows α ∈ Q1, α : i → j, the following equation is satisfied: κα ◦ hi = hj ◦ χα.

We obtain that result as following: for qA ∈ Hom(A, Y ) we have κα ◦ qAi = qAj ◦ ϕα Now by the pointwise 5 pushout property of h we know that qAi = hi ◦ γAi and that qAj = hj ◦ γAj . Now, by putting that to our

former equation, we have the following: κα ◦ hiγAi = hj ◦ γAj ◦ ϕα and finally, as γA ∈ Hom(A, X), we have

χα ◦ γAi = γAj ◦ ϕα we get

κα ◦ hi ◦ γAi = hj ◦ χα ◦ γAj ⇐⇒ κα ◦ hi ◦ πi ◦ ιi = hj ◦ χα ◦ pij ◦ ιj 0 0 and similarly for B we get κα ◦ hi ◦ πi ◦ ιi = hj ◦ χα ◦ pij ◦ ιj and thus by universal property of coproducts we get κα ◦ hi ◦ πi = hj ◦ χα ◦ πi so κα ◦ hi = hj ◦ χα as πi is epi.  We can use our construction to directly obtain the kernel and cokernel representations. Definition 2.15. Let C be a category with a zero object 0. Let A, B ∈ C and f ∈ Hom(A, B). A pair (Ker(f), ι) such that Ker(f) ∈ C, ι ∈ Hom(Ker(f),A) is said to be the kernel of f it it satisfies the following universal property: Given D ∈ C and h ∈ Hom(D,A) such that f ◦ h = 0 (i.e. the unique morphism that factors through 0, that is f ◦ h = 0¯ ◦ ϕ, ϕ ∈ Hom(D, 0) with 0¯ being the unique morphism in Hom(0,B)), there is a unique morphism φ : D → Ker(f) such that h = ι ◦ φ, i.e. the following diagram commutes: D ∃!φ h

f Ker(f) ι A B

Proposition 2.16. In a category C with a zero object 0 and pullbacks, given f ∈ Hom(A, B), f has a kernel (X, γ) and it is given by the following pullback:

X 0 0

γ 0 f A B Proof. The universal property of kernels defined above is properly contained (as a diagram) in the following diagram induced by the universal property of pullbacks: D 0 ∃!φ

0 h X 0 γ 0 f A B  This yields the following definition:

Definition 2.17. Given A = (Ai, ψα)i∈Q0,α∈Q1 ,B ∈ RepK Q, f ∈ Hom(A, B) we define the kernel of f as the

representation Ker(f) ∈ RepK Q given by Ker(f) = (Ker(fi), ψα Ker(fi))i∈Q0,α∈Q1 , ι ∈ Hom(Ker(f),A) =

(ιi)i∈Q0 , where ιi is the canonical inclusion of Ker(fi) into Ai in VectK We treat cokernels similarly: Definition 2.18. Let C be a category with a zero object 0. Let A, B ∈ C, f ∈ Hom(A, B). A pair (Coker(f), π) such that Coker(f) ∈ C, π ∈ Hom(B, Coker(f)) is said to be the cokernel of f if it satisfies the following universal property: Given D ∈ C and a morphism h ∈ Hom(B,D) such that h◦f = 0, the following commutative diagram exists:

f A B π Coker(f) h ∃!φ D 6 Proposition 2.19. In a category C with a zero object 0 and pushouts, given f ∈ Hom(A, B), f has a cokernel (X, γ) and it is given by the following pushout:

f A B

0 γ 0 0 X Proof. Once again, the universal property of cokernels is properly contained in the following diagram induced by the universal property of pushouts:

f A B

0 γ h 0 0 X ∃!φ

0 D  This motivates the following definition:

Definition 2.20. Given representations A, B = (Bi, ψα)i∈Q0,α∈Q1 , and a morphism f ∈ Hom(A, B) we define the cokernel of f as the representation Coker(f) ∈ RepK Q given by Coker(f) = (Coker(fi), χα), where χα(bi + fi(Bi)) = ψα(bi) + fj(Bj) and π ∈ Hom(B, Coker(f)) = (πi)i∈Q0 , where πi is the canonical projection from Bi onto Coker(fi) Definition 2.21. A morphism of quiver representations f is: a) mono if Ker(f) = 0 b) epi if Coker(f) = 0 c) an isomorphism if it is both epi and mono.

Definition 2.22. A representation L ∈ RepK Q is said to be a subrepresentation of M ∈ RepK Q if there is a monomorphism g : L → M.

Theorem 2.23 (First Isomorphism Theorem). Let A = (Ai, ϕα),B = (Bi, ψα ∈ RepK Q, f ∈ Hom(A, B). Then Coker(Ker(f) ' Ker(Coker(f)) Proof. Fortunately, the corresponding theorem for vector spaces equips us with an explicit isomorphism, ∼ i.e. Coker(Ker(fi)) = Ai/ Ker(fi) ' Im(fi) = Ker(Coker(fi)) With γi : Ai/ Ker(fi) −→ Im(fi) defined as follows: γi(xi + Ker(fi)) = f(xi). By applying the definitions in said order we get that Coker(Ker(f)) =

(Ai/ Ker(fi), χα) with χα(x + Ker(fi)) = ϕα(x) + Ker(fj), and Ker(Coker(f)) = (Im(fi), ψα Im(fi))

Now let γ = (γi)i∈Q0 . Then γ is a morphism as for all α : i → j ∈ Q1 and all x + Ker(fi) ∈ Ai/ Ker(fi) we have (ψα Im(fi) ◦γi)(x+Ker(fi) = (ψα Im(fi))(fi(x)) = (ψα ◦fi)(x) = (fj ◦ϕα)(x) = γj(ϕα(x)+Ker(fj)) =

(γj ◦ χα)(x + Ker(fi)) so (ψα Im(fi)) ◦ γi = γj ◦ χα. Moreover, since γi is an isomorphism for all i ∈ Q0, its corresponding kernel and cokernel representations assign the 0 vector space to every i, and thus (as 0 is the zero object), they both are the zero representation, so γ is mono and epi, and thus an isomorphism.  It’s about time now to formally introduce the notion of image.

f Definition 2.24. Given A −→ B in RepK Q , we let Im(f) ∈ RepK Q be the representation given by Im(f) := Ker(Coker(f)).

f0 f1 f2 Definition 2.25. A sequence ··· −→ M1 −→ M2 −→· · · in RepK Q is exact at Mi if Im(fi−1) = Ker(fi), and f g is an exact sequence if it is exact for all i. In particular, exact sequences of the form 0 → L −→ M −→ N → 0 are said to be short exact. 7 Definition 2.26. a) f : L → M in RepK Q is called a section if there is h ∈ Hom(M,L) such that h ◦ f = 1L b) g : M → N in RepK Q is called a retraction if there is h ∈ Hom(N,M) such that g ◦ h = 1N f g c) A short exact sequence 0 → L −→ M −→ N → 0 is said to split (equivalently, is said to be split exact) if f is a section.

f g Proposition 2.27. Let 0 → L −→ M −→ N → 0 be a short exact sequence in RepK Q . Then: a) f is a section if and only if g is a retraction b)If f is a section, then M ' L ⊕ N. Moreover, let η be the morphism induced by g being a retraction and let h be the morphism induced by f being a section. Let ιL, ιN be the canonical injections from L respectively M to L ⊕ N and let πL, πN be the canonical projections from L ⊕ N to L respectively N. Then the following diagrams are isomorphic: f g 0 L M N 0 h η

ιL πN 0 L L ⊕ N N 0 πL ιN

To prove that proposition we employ the strategy used earlier: we first prove the same result in category R -Mod to obtain candidate morphisms that satisfy our desired property pointwise, and then prove that those candidate morphisms indeed are morphisms.

f g Lemma 2.28. Let 0 → L −→ M −→ N → 0 be a short exact sequence in R -Mod. Then: a) f is a section if and only if g is a retraction b) If f is a section, then M ' L ⊕ N. Moreover, let η be the morphism induced by g being a retraction and let h be the morphism induced by f being a section. Let ιL, ιN be the canonical injections from L respectively M to L ⊕ N and let πL, πN be the canonical projections from L ⊕ N to L respectively N. Then the following diagrams are isomorphic: f g 0 L M N 0 h η

ιL πN 0 L L ⊕ N N 0 πL ιN

Proof. a) " =⇒ ": If g was mono (it is epi due to exactness of the sequence) we would just set η = g−1, but that need not be the case, and although no elements in N have an empty preimage, some preimages need not be singletons, making the inverse mapping not well-defined. We thus "fix" that problem with diagram chasing. Let n ∈ N, m, m0 ∈ g−1(n). We have m − m0 ∈ Ker(g), so m − m0 ∈ Im(f) and hence there is f −1(m − m0), which is unique as f is mono by exactness of the sequence. Now as f is a section, −1 0 −1 0 0 we have h ∈ Hom(M,L) such that h ◦ f = idL, and (h ◦ f)(f (m − m )) = f (m − m ) = h(m − m ), so (f ◦ h)(m − m0) = f(f −1(m − m0)) = m − m0 and thus m − (f ◦ h)(m) = m0 − (f ◦ h)(m0) and g(m − (f ◦ h)(m)) = g(m) − (g ◦ f)(h(m) = g(m) as g ◦ f = 0. So, given n ∈ N, we let η(n) = m − (f ◦ h)(m) −1 for any m ∈ g (n) which by above is well-defined and satisfies g ◦ η = idN And so g is a retraction. " ⇐= ": in analogy to the construction above, given m ∈ M, and letting η ∈ Hom(N,M) be the morphism induced by g being a retraction (i.e. g ◦ η = idN ), we have g(m − (η ◦ g)(m)) = g(m) − (g ◦ η ◦ g)(m) = g(m) − (g ◦ η)(g(m) = g(m) − g(m) = 0 so we have m − (η ◦ g)(m) ∈ Ker(g) and thus there is a unique element f −1(m − (η ◦ g)(m)) and we let h(m) = f −1(m − (η ◦ g)(m)) and given l ∈ L we have (h ◦ f)(l) = −1 −1 f (f(l) − (η ◦ g ◦ f)(l)) = f (f(l)) = l. So h ◦ f = idL and we’re done. Remark 2.29. The constructions given above commute, i.e. given a section f with h ∈ Hom(M,L) such that h ◦ f = idL, if we first construct a "retraction morphism" η from h as we did above, and then construct a "section morphism" h0 from η as above, we get h0 = h, as h0(m) = f −1(m − η(g(m))) = f −1(m − (g−1(g(m)) − (f ◦ h)(g−1(g(m)))) = f −1(m − m + (f ◦ h)(m)) = h(m) 8 b) We prove this statement in several steps: η i) 0 ← L ←−h M ←− N ← 0 is a short exact sequence Proof: idN = g ◦ η is mono, so η is mono, idL = h ◦ f is epi, hence h is epi. Now all that remains is to show Ker(h) = Im(η). Let m ∈ Im(η). Thus there is a unique element n such that m = η(n), so g(m) = (g ◦ η)(n) = n, and m ∈ g−1(n), so by earlier definition of η we have m = η(n) = m − (f ◦ h)(m), hence m = m − (f ◦ h)(m), so (f ◦ h)(m) = 0. Moreover, h(m) = 0 as f is mono, so Im(η) ⊆ Ker(h). Now let m ∈ Ker(h). Then h(m) = 0, so by the definition of h we get f −1(m − (η ◦ g)(m)) = 0, so m − (η ◦ g)(m) = 0 as f is mono, and thus m = η(g(m)). Hence m ∈ Im(η), so Ker(h) ⊆ Im(η), and thus Ker(h) = Im(η). Thus the sequence is exact. ii) f ◦ h + η ◦ g = idM Proof: Given m ∈ M we have m − (f ◦ h + η ◦ g)(m) = m − (f ◦ h)(m) − (η ◦ g)(m) = m − f(h(m) − (g−1(g(m)) − (f ◦ h)(g−1(g(m)))) = m − f(h(m)) − (m − f(h(m))) = 0 iii) M ' L ` N with injections f, η Claim: Given X ∈ R -Mod, ϕ ∈ Hom(L, X), ψ ∈ Hom(N,X), there is a unique γ ∈ Hom(M,X) such that the following diagram commutes: f η L M N γ ϕ ψ X Proof of claim: Let γ = ϕ ◦ h + ψ ◦ g, so γ ◦ η = ϕ ◦ h ◦ η + ψ ◦ g ◦ η = ψ ◦ g ◦ η as h ◦ η = 0, and g ◦ η = idN means that ψ ◦ g ◦ η = ψ, so γ ◦ η = ψ. Similarly, γ ◦ f = ϕ ◦ h ◦ f + ψ ◦ g ◦ f = ϕ ◦ h ◦ f = ϕ 0 0 0 And so the diagram commutes. For uniqueness, let γ such that γ ◦η = ψ, γ ◦f = ϕ. But f ◦h+η ◦g = idM , so γ0 = γ0 ◦ (f ◦ h + η ◦ g) = (γ0 ◦ f) ◦ h + (γ0 ◦ η) ◦ g = ϕ ◦ h + ψ ◦ g = γ

iv) M ' L Q N with projections h, g Claim: given a module X ∈ R -Mod, and morphisms of modules ϕ ∈ Hom(X,L), ψ ∈ Hom(X,N), there is a unique γ ∈ Hom(X,M) such that the following diagram commutes:

g L h M N γ ϕ ψ X Proof of claim: let γ = η ◦ ψ + f ◦ ϕ. Then g ◦ γ = g ◦ η ◦ ψ + g ◦ f ◦ ϕ = g ◦ η ◦ ψ = ψ and h ◦ γ = h ◦ η ◦ ψ + h ◦ f ◦ ϕ = h ◦ f ◦ ϕ = ϕ For uniqueness, let γ0 such that h ◦ γ0 = ϕ, g ◦ γ = ψ. But 0 0 0 0 0 γ = idM ◦ γ = (f ◦ h + η ◦ g) ◦ γ = f ◦ (h ◦ γ ) + η ◦ (g ◦ γ ) = f ◦ ϕ + η ◦ ψ = γ 

Now we’re ready to prove the statement for RepK Q :

9 Proof of proposition 2.27. Note that we assume that f is a section, and thus there is a morphism h =

(hi)i∈Q0 ∈ Hom(M,L) such that h ◦ f = idL, and thus in particular, for all i ∈ Q0 we have hi ◦ fi = idLi .

Now we let η = (ηi)i∈Q0 , where ηi is constructed from hi as above. Now, as mentioned earlier, all the desired properties stated in a) and b) will follow as long as η is a well-defined morphism. To show this we

let L = (Li, ρα)i∈Q0,α∈Q1 , M = (Mi, ϕα)i∈Q0,α∈Q1 , and N = (Ni, ψα)i∈Q0,α∈Q1 . We need to show that for all α : i → j ∈ Q1 we have ηj ◦ ψα = ϕα ◦ ηi. But since the properties proven earlier still hold pointwise, i.e. for all i ∈ Q0, Mi ' Li ⊕ Ni, we can represent our morphisms with matrices in following manner: 1 [ ] [ ] fi gi 0 i 0 1 i Li Mi Ni Li Li ⊕ Ni Ni [ ] 0 hi ηi 1 0 i [ 1 ]i ρα ϕα ψα ' ρα ϕα ψα f g 1 j j [ 0 ]j [ 0 1 ]j Lj Mj Ni Lj Lj ⊕ Nj Ni η hj j 0 [ 1 0 ]j [ 1 ]j  a b   a b  1 1 And in particular, ϕα = c d , but since f is a morphism we have ϕα ◦ fi = fj ◦ ρα, so c d [ 0 ]i = [ 0 ]j ρα, a ρα so [ c ] = [ 0 ], so a = ρα, c = 0, similarly for g we obtain gj ◦ ϕα = ψα ◦ gi and thus [ 0 ψα ] = [ c d ]. Therefore we have d = ψα, and finally with help of h we get hj ◦ ϕα = ρα ◦ hi, which means that [ ρα 0 ] = [ a b ], so h i h i b = 0, so that ϕ = ρα 0 Now we have η ◦ ψ =  0  = ρα 0 [ 0 ] = ϕ ◦ η and so our candidate α 0 ψα j α ψα 0 ψα 1 α i morphism is well-defined, and the results follow.  Definition 2.30. Given a representation X, we define the covariant Hom-functor of X, Hom(X, −): RepK Q → VectK by:

• Sending a representation M ∈ RepK Q to the vector space of morphisms Hom(X,M) • Sending a morphism of representations f ∈ Hom(L, M) to the K-linear map Hom(X, f) : Hom(X,L) → Hom(X,M) which, given a morphism of representations ϕ ∈ Hom(X,L), sends it to f ◦ ϕ ∈ Hom(X,M) Proposition 2.31. A sequence f g 0 → L −→ M −→ N in RepK Q is exact if and only if for all representations X ∈ RepK Q the following sequence is exact in VectK : Hom(X,f) Hom(X,g) 0 → Hom(X,L) −−−−−−→ Hom(X,M) −−−−−−→ Hom(X,N) In particular, the covariant Hom-functor Hom(X, −) is left exact. Proof. " =⇒ " First, if ϕ ∈ Hom(X,L) such that Hom(X, f)(g) = f ◦ g = 0 then by universal property of kernels, ϕ = ι ◦ h for a unique h ∈ Hom(X, Ker(f)). But f is mono, so Ker(f) = 0, so h = 0, and thus ϕ = 0 and so Hom(X, f) is mono. Now we need to show Im(Hom(X, f)) = Ker(Hom(X, g)). Clearly, Im(Hom(X, f)) ⊂ Ker(Hom(X, g)) as g ◦f = 0 means that Hom(X, g)(Hom(X, f)(ϕ)) = g ◦f ◦ϕ = 0◦ϕ = 0. We also have Ker(Hom(X, g)) ⊂ Im(Hom(X, f)), since ψ ∈ Ker(Hom(X, g)) ⇐⇒ g ◦ψ = 0 and by universal property of kernels we thus get ∃!h ∈ Hom(X, Ker(g)) such that ψ = ι ◦ h, with ι being the kernel morphism. But Ker(g) = Im(f) and f is epi onto Im(f), so there is a unique isomorphism f¯ : L −→∼ Im(f) and ι ◦ f¯ = f, and finally ψ = f ◦ f¯−1 ◦ h, so ψ is in Im(Hom(X, f)). " ⇐= " First we show that Hom(X, f) is mono implies that f is mono, arguing by contraposition: if f is not mono, then Ker(f) 6= 0, ι 6= 0 (ι being the kernel morphism), but f ◦ ι = 0 = f ◦ 0, so Hom(X, f) is not mono. Next we have Im(f) ⊂ Ker(g) as Hom(X, g)(Hom(X, f)(idL)) = 0 = g ◦ f ◦ idL = g ◦ f, and finally Ker(Hom(X, g)) ⊂ Im(Hom(X, f)) as the kernel morphism of Ker(g), ιg : Ker(g) → M is in Ker(Hom(X, g)), and thus factors through f.  Definition 2.32. Given a representation X, we define the contravariant Hom-functor of X, Hom(−,X): RepK Q → VectK by:

• Sending a representation M ∈ RepK Q to the vector space of morphisms Hom(M,X) • Sending a morphism of representations g ∈ Hom(M,N) to the K-linear map Hom(f, X) : Hom(N,X) → Hom(M,X) which, given a morphism of representations ψ ∈ Hom(N,X), sends it to ψ ◦ g ∈ Hom(M,X) 10 Proposition 2.33. A sequence f g L −→ M −→ N → 0

in RepK Q is exact if and only if for all representations X ∈ RepK Q the following sequence is exact in VectK : Hom(g,X) Hom(f,X) 0 → Hom(N,X) −−−−−−→ Hom(M,X) −−−−−−→ Hom(L, X) In particular, the contravariant Hom-functor Hom(−,X) is left exact. Proof. " =⇒ ": First, g epi means that Hom(g, X) is mono, since by universal property of cokernels, if φ ∈ Hom(N,X) such that Hom(g, X)(φ) = φ ◦ g = 0, φ = h ◦ π for a unique h ∈ Hom(Coker(g),X) (π being the cokernel projection), but Coker(g) = 0) means that h = 0, and hence φ = 0) Secondly, Im(Hom(g, X)) ⊂ Ker(Hom(f, X) as Hom(f, X)(Hom(g, X)(φ)) = φ ◦ g ◦ f = φ ◦ 0 = 0. And finally Ker(Hom(f, X)) ⊂ Im(Hom(g, X)) as φ ∈ Ker(Hom(f, X)) is equivalent to φ ◦ f = 0, which implies that there is a unique morphism h ∈ Hom(Coker(f),X) such that φ = h ◦ π, π being the cok- ernel morphism of Coker(f) (statement following from universal property of cokernels), and Coker(f) ' Coker(Ker(Coker(f))) =: A, since given the kernel morphism ι of Ker(Coker(f)), we have π ◦ι = 0, and thus there is a unique morphism ψ ∈ Hom(A, Coker(f)) such that ψ ◦ ρ = π, ρ denoting the cokernel morphism of A, so φ = h ◦ ψ ◦ ρ, and h ◦ ψ acts as cokernel morphism, and is unique such: κ ∈ Hom(A, X) such that κ ◦ ρ = φ, so κ ◦ ρ = h ◦ ψ ◦ ρ , and thus κ = h ◦ ψ as ρ is epi, and the equality follows by earlier part of this proof. Now Coker(f) = Coker(Ker(Coker(f))) = Coker(Ker(g)) = Ker(Coker(g)) = Ker(0) = N, and thus φ factors through g. " ⇐= ": First Hom(g, X) being mono implies that g is epi. We argue by contraposition: if g is not epi, then Coker(g) 6= 0 =⇒ π 6= 0, but Hom(g, X)(π) = π ◦ g = 0 = 0 ◦ g = Hom(g, X)(0) so Hom(g, X) is not mono. Secondly Im(f) ⊂ Ker(g) as Hom(f, X)(Hom(g, X)(idN )) = idN ◦ g ◦ f = 0 =⇒ g ◦ f = 0 Thirdly Ker(g) ⊂ Im(f): Let π denote the cokernel morphism of Coker(f). By definition we have π ◦ f = 0, so π ∈ Ker(Hom(f, X)). Thus π ∈ Im(Hom(g, X)) and therefore there is a morphism ϕ ∈ Hom(N, Coker(f)) such that π = ϕ ◦ g. Now let ι denote the kernel morphism of Ker(g). We have π ◦ ι = ϕ ◦ g ◦ ι = ϕ ◦ 0 = 0 so Ker(g) ⊂ Ker(Coker(f)) = Im(f).  f g Proposition 2.34. A sequence 0 → L −→ M −→ N → 0 in RepK Q is split exact if and only if for all Hom(X,f) Hom(X,g) X ∈ RepK Q : 0 → Hom(X,L) −−−−−−→ Hom(X,M) −−−−−−→ Hom(X,N) → 0 is exact. Proof. " =⇒ ": Since Hom(X, −) is left exact, to show exactness of the sequence, we only need to show that Hom(X, g) is epi. But the corresponding sequence in RepK Q is split exact, so g is a retraction, and thus there is a morphism η ∈ Hom(N,M) such that g ◦ η = idN . Now let ϕ ∈ Hom(X,N). We have ϕ = g ◦ η ◦ ϕ which means that Hom(X, g) is epi. " ⇐= ": Here we need to both show that g is epi and also that the sequence splits. We let X = N, and since Hom(X, g) is epi, we get that idN = g ◦ η for some η ∈ Hom(N,M). But idN being epi means that g is epi, and also g is a retraction, so the sequence splits.  f g Proposition 2.35. A sequence 0 → L −→ M −→ N → 0 in RepK Q is split exact if and only if for all Hom(g,X) Hom(f,X) X ∈ RepK Q : 0 → Hom(N,X) −−−−−−→ Hom(M,X) −−−−−−→ Hom(L, X) → 0 is exact.

Proof. " =⇒ ": all we need to show is that Hom(f, X) is epi. But the corresponding sequence in RepK Q splits, so f is a section, i.e. there is a morphism η ∈ Hom(M,L) such that η◦f = idL. Given ϕ ∈ Hom(L, X), we have ϕ = ϕ ◦ η ◦ f, thus ϕ ∈ Im(Hom(f, X)) and hence Hom(f, X) is epi. " ⇐= ": In analogy to the proof of the preceding proposition, idL = h ◦ f for some h ∈ Hom(M,L). Thus f is mono and is a section, and the result follows.  3. Projective and Injective Representations

Definition 3.1. Given Q = (Q0,Q1, s, t), and vertices i, j ∈ Q0, a path from i to j is a sequence of arrows n {αk}k=1 such that for all k, t(αk) = s(αk+1), and s(α1) = i, t(αn) = j Definition 3.2. We say that a quiver is acyclic if and only if it contains no cycles, i.e. non-empty paths from i to i, for some i ∈ Q0 11 Definition 3.3. Let Q = (Q0,Q1, s, t). i ∈ Q0 is a source if and only if there is no α ∈ Q1 such that i = t(α). Similarly, i is a sink if and only if there is no α ∈ Q1 such that i = s(α).

Proposition 3.4. If Q is acyclic then: a) There is a source in Q b) There is a sink in Q

Proof. a) Assume Q has no source. Then for all i ∈ Q0 there is an arrow α ∈ Q1 such that i = t(α). Thus if we start at some fixed vertex, we can go back to another vertex by that arrow and continue so indefinitely, inducing an infinite path in the quiver. But we assume (throughout) that our quiver is finite, so some vertices must have been visited multiple times, thus we can separate a subpath of our path that starts and ends at one of those vertices, and thus is a directed loop, contradicting our hypothesis. b) Similarly, assume Q has no sink. Then, after choosing some i ∈ Q0, we can produce an infinite path by starting with arrow that has its source at i, and continuing with an arrow that has its source at the target of our arrow etc. And thus, just like in a), we obtain directed loops, leading us to contradiction. 

From now on we assume that all quivers are acyclic, unless otherwise noted.

Definition 3.5. Let i ∈ Q0 Define S(i), the simple representation at vertex i as (S(i)j, ϕα)j∈Q0,α∈Q1 , where S(i)j = K if j = i and S(i)j = 0 else, and for all α ∈ Q1 we have ϕα = 0

Proposition 3.6. Given a representation M = (Mi, γα)i∈Q0,α∈Q1 ∈ RepK Q, M is simple (i.e. is not zero and has no non-zero subrepresentations) if and only if it is isomorphic to S(i) for some i ∈ Q0

Proof. " =⇒ ": Choose a sink j in Q. We have two cases: 1) Mj 6= 0 2) Mj = 0 " 1 # 0 For 1) We have a monomorphism ι : S(j) → M, ι = (ιi)i∈Q defined by ιj = . , ιi = 0 for i 6= j, which is 0 . 0 0 S(i)m S(i)j well-defined, as the following square clearly commutes: 0 ιj

γα Mm Mj For 2) We remove j and all the arrows into it (as there are no arrows from it to remove) from Q to obtain Q0 and a representation M 0 of it, and consider the same problem for Q0. If we happen to be in case 1) this time and thus have a subrepresentation S(j0) of M 0, it will also be a subrepresentation of M after we put j and its arrows back to our quiver, since we extend by 0 morphisms for both S(j0) and M 0. This procedure terminates after at most |Q0| steps, and the only way it could avoid case 1) in all steps is by all its vector spaces being 0, and thus M = 0, which is not allowed by the definition of a simple object. " ⇐= ": For N 6= 0 to be a subrepresentation of M ' S(i), i.e. have a monomorphism N → M, the morphism must be pointwise mono, so Nj = 0 for j 6= i and Nj ' K ' S(i)i, and all the arrows being 0 now follows from 0 being a zero object in V ectK 

Definition 3.7. Let i ∈ Q0. Define the projective representation at vertex i, P (i) = (P (i)j, ϕα)j∈Q0,α∈Q1 , n by P (i)j ' K , where n is the number of paths from i to j (from now on referred to as i-j paths), with a canonical basis formally consisting of those paths. Given α : j → k, the map ϕα is defined on that basis by ϕα : p → pα, where p is any of the i-j paths in the basis.

Proposition 3.8. For all i ∈ Q0, P (i) is a projective object of RepK Q , i.e. given M,N ∈ RepK Q , g ∈ Hom(M,N) epi, and h ∈ Hom(P (i),N), there is a (not necessarily unique) morphism f ∈ Hom(P (i),M) 12 such that g ◦ f = h. This is illustrated by the following diagram: P (i) ∃f h

M g N

Proof. Let M = (Mi, γα)i∈Q0,α∈Q1 ,N = (Ni, ψα)i∈Q0,α∈Q1 . Set Qi = {j ∈ Q0 such that there is an i-j path} and set Qi := {j ∈ Q0 such that there is no i-j path}. On Qi the problem is trivial: j ∈ Qi, so P (i)j = 0, so we must have fj = 0 = hj, and thus clearly gj ◦ fj = hj. Our choice is also well-defined: given α : j → m we have fm ◦ ϕα = fm ◦ 0 = 0 = γα ◦ 0 = γα ◦ fj We thus focus on Qi as we define f. First choose fi ∈ Hom(P (i)i,Mi) so that gi ◦ fi = hi. This is possible as vector spaces are themselves −1 projective (more explicitly, let fi(i) ∈ g (h(i))) (note that our canonical basis for P (i)i consists only of dim Mi the empty path i, and thus P (i)i ' K, and Hom(P (i)i,Mi) ' K ' Mi) Later it will be clear that this choice uniquely determines all of f. Let m ∈ Qi and let Rm be the canonical basis for P (i)m. Choose r ∈ Rm and write r = α1α2...αn. Now we

introduce some convenient notation: given a set of K-linear maps {γα}α∈Q1 , let γr denote γαn ◦γαn−1 ◦· · ·◦γα1 , so that γr composes γ along the path r. Now let fm(r) = γr ◦ fi(i). It is now clear that this definition extends to all of Q0, so the distinction between Qi and Qi is superfluous. We now need to show two things: 1) f is well defined, and 2) The diagram above is satisfied pointwise. 1) Let j, m ∈ Q0. We need to show that this commutes:

ϕα P (i)j P (i)m

fj fm

γα Mj Mm

Let Rj be the canonical basis for P (i)j, r ∈ Rj, r = α1α2...αn. Now we have fm ◦ϕα(r) = fm(α1α2...αnα) = γα ◦ (γr ◦ fi(i)) = γα ◦ fj(r) and we’re done. 2) Let m ∈ Q0 We need to show gm ◦ fm = hm. We have (gm ◦ fm)(r) = (gm ◦ γr ◦ fi)(i), which, since g is a morphism, is equal to (ψr ◦ gi ◦ fi)(i) = (ψr ◦ hi)(i) = (hm ◦ ϕr)(i) = hm(r) 

Proposition 3.9. Given a representation M = (Mi, γα)i∈Q0,α∈Q1 ∈ RepK Q, the following vector spaces are isomorphic: Hom(P (i),M) ' Hom(P (i)i,Mi) ' Mi

Proof. Choose a K-linear map ψ ∈ Hom(P (i)i,Mi), and let f ∈ Hom(P (i),M) be a morphism of represen- tations such that fi = ψ, which is then constructed as the morphism f in the preceding proposition, i.e., given a vertex j ∈ Q0 and an i-j path r in the canonical basis Rj for P (i)j, we let fj(r) = γr ◦ fi(i). 0 0 0 Now let f ∈ Hom(P (i),M) be a morphism of representations such that fi = fi. In order for f to be well defined, we in particular need that the following diagram commutes for all vertices j ∈ Q0 and all i-j paths r ∈ Rj: ϕr P (i)i P (i)j

0 0 fi fj

γr Mi Mj 0 0 Hence, under the canonical choice of basis, we need fj ◦ ϕr(i) = γr ◦ fi (i), but we know that ϕr(i) = r 0 0 and that fi = fi and hence we need fj(r) = γr ◦ fi(i). But if that is the case, then for all vertices j 0 0 and all basis elements r ∈ Rj we have fj(r) = fj(r) and so we conclude that f = f. Thus there is a bijection between K-linear maps in Hom(P (i)i,Mi) and morphisms of representations in Hom(P (i),M). That bijection is clearly linear, so we conclude that Hom(P (i)i,Mi) ' Hom(P (i),M). The isomorphism Hom(P (i)i,Mi) ' Mi follows by earlier discussion of dimensions.  0 0 Proposition 3.10. P,P ∈ RepK Q are projective if and only if P ⊕ P is projective. Proof. First, we let ι : P → P ⊕ P 0, π : P ⊕ P 0 → P , ι0 : P 0 → P ⊕ P 0, π0 : P ⊕ P 0 → P 0 be the canonical projections and injections of P and P 0. 13 " ⇐= ": We show that given representations M,N ∈ RepK Q, a morphism h ∈ Hom(P,N) and an epimor- phism g ∈ Hom(M,N), there is a morphism f ∈ Hom(P,M) such that h = g ◦ f. Let γ : P ⊕ P 0 → M be a morphism such that g ◦ γ = h ◦ π, existing due to P ⊕ P 0 being projective. We let f = γ ◦ ι, so that g ◦ f = (g ◦ γ) ◦ ι = (h ◦ π) ◦ ι = h ◦ (π ◦ ι) = h ◦ idP = h Diagrammatically this gives: ∃γ P ⊕ P 0 M g ι π γ◦ι

P f N And P 0 being projective is proven by analogy. 0 0 " =⇒ ": Given M,N ∈ RepK Q, h ∈ Hom(P ⊕ P ,N), g ∈ Hom(M,N) epi, we define γ : P ⊕ P → M so that h = g ◦ γ. Since both P and P 0 are projective, there are morphisms f : P → M, f 0 : P 0 → M such that g ◦ f = h ◦ ι, g ◦ f 0 = h ◦ ι0. And by the coproduct property of P ⊕ P 0, we have a unique γ = (f, f 0) 0 0 0 0 0 0 such that γ ◦ ι = f, γ ◦ ι = f , and since idP ⊕P 0 = ι ◦ π + ι ◦ π , we have g ◦ γ = g ◦ γ ◦ (ι ◦ π + ι ◦ π ) = 0 0 0 0 0 0 g ◦(γ ◦ι)◦π +g ◦(γ ◦ι )◦π = (g ◦f)◦π +(g ◦f )◦π = h◦ι◦π +h◦ι ◦π = h◦idP ⊕P 0 = h Diagrammatically we have: ∃h P ι f

∃γ g P ⊕ P 0 M N ι0

∃h0 P 0



Proposition 3.11. For all i ∈ Q0, P (i) is indecomposable.

Proof. Assume P (i) is not indecomposable, i.e. there are non-zero M = (Mi, γα)i∈Q0,α∈Q1 , N = (Ni, ψα)i∈Q0,α∈Q1 ∈ RepK Q such that P (i) ' M ⊕N. Then in particular we have Mi ⊕Ni ' P (i)i ' K, so either Mi = 0,Ni ' K or Mi ' K,Ni = 0. Without loss of generality assume Mi = 0. But then by proposition 3.9 we know that Hom(P (i),M) ' Mi = 0, so in particular the canonical projection πM : P (i) → M induced by the product property of P (i) ' L ⊕ N must be zero. Hence M = 0, which contradicts our assumption.  Now we introduce some injective representations and present results about them that are dual to those concerning the projective representations discussed above.

Definition 3.12. Let i ∈ Q0. Define the injective representation at vertex i, I(i) = (I(i)j, ϕα)j∈Q0,α∈Q1 , n by I(i)j ' K , where n is the number of j-i paths, with a canonical basis formally consisting of those paths. Let w be a j-i path in the canonical basis for I(i)j. Given α : j → k, we define ϕα(w) = p if w = αp for some k-i path p, and as ϕα(w) = 0 if that is not the case.

Proposition 3.13. For all i ∈ Q0, I(i) is an injective object of RepK Q , i.e. given L, M ∈ RepK Q ,

M = (Mi, ψα)i∈Q0,α∈Q1 , N = (Ni, γα)i∈Q0,α∈Q1 , g ∈ Hom(L, M) mono, and h ∈ Hom(L, I(i)), there is a (not necessarily unique) morphism f ∈ Hom(M,I(i)) such that f ◦ g = h. This is illustrated by the following diagram:

L g M h ∃f I(i)

Proof. First choose fi such that fi ◦ gi = hi (let fi be zero on cokernel of g and equal to hi on its image). Let Wj be the canonical basis for I(i)j consisting of j-i paths. Given m ∈ Mj, define X fj(m) = (fi ◦ ψw)(m)w w∈Wj 14 so " fi◦ψw1 # . fj = .. fi◦ψwn

There are two things to be shown here: 1) f is well-defined, and 2) for all j ∈ Q0 we have fj ◦ gj = hj 1) for α : m → j we need the following diagram to commute:

ψα Mm Mj

fm fj

ϕα I(i)m I(i)j and indeed we have X X (fj ◦ ψα)(x) = (fi ◦ ψw)(ψα(x))w = (fi ◦ ψαw)(x)w w∈Wj w∈Wj and

X 0 X 0 (ϕ ◦ f )(x) = ϕ ( (f ◦ ψ 0 )(x)w ) = (f ◦ ψ 0 )(x)ϕ (w ) α m α 0 i w 0 i w α w ∈Wm w ∈Wm X X = (fi ◦ ψαw)(x)ϕα(αw) = (fi ◦ ψαw)(x)w = (fj ◦ ψα)(x) αw,w∈Wj w∈Wj 2) We have X X (fj ◦ gj)(l) = (fi ◦ ψw ◦ gj)(l)w = (fi ◦ gi ◦ γw)(l)w w∈Wj w∈Wj X X = (hi ◦ γw)(l)w = (fi ◦ gi ◦ γw)(l)w w∈Wj w∈Wj X X = (fi ◦ gi ◦ γw)(l)w = (ϕw ◦ hj)(l)w w∈Wj w∈Wj Now if we denote h (l) as P ρ v we get j v∈Wj v X X X X (ϕw)(hj(l))w = (ρvϕw(v))w = ρww = hj(l) w∈Wj w∈Wj v∈Wj w∈Wj 

Proposition 3.14. Given a representation M = (Mi, ψα)i∈Q0,α∈Q1 ∈ RepK Q, the following vector spaces are isomorphic: Hom(M,I(i)) ' Hom(Mi,I(i)i) ' Mi Proof. The argument is very similar to that given in proof of proposition 3.9: Pick a K-linear map η ∈ Hom(Mi,I(i)i), and let f ∈ Hom(M,I(i) be a morphism of representations such that fi = η, which is then constructed as in the preceding proposition, i.e, given a vertex j ∈ Q0 and a basis element m ∈ Mj, we let f (m) = P (f ◦ ψ )(m)w. Let f 0 ∈ Hom(M,I(i)) be a morphism of representations such that f 0 = f . j w∈Wj i w i i In order to be well-defined, f 0 must satisfy the following commutative diagram:

ψw Mj Mi

0 0 fj fi

ϕw I(i)j I(i)i

0 0 Which in turn implies that fj = fj, and hence f = f . So there is a natural isomorphism Hom(Mi,I(i)i) ' Hom(M,I(i)). Moreover, Hom(Mi,I(i)i) ' Mi by equality of dimensions.  0 0 Proposition 3.15. I,I ∈ RepK Q are injective if and only if I ⊕ I is injective Proof. We denote the projections and injections as we did in the proof of the dual statement (Proposition 0 3.10)" ⇐= ": Given M,N ∈ RepK Q , f ∈ Hom(M,N) mono, g ∈ Hom(M,I). Since I ⊕ I is projective, 15 0 we let γ : N → I ⊕ I be such that γ ◦ f = ι ◦ g, so now we have (π ◦ γ) ◦ f = π ◦ ι ◦ g = idI ◦ g = g so π ◦ γ completes the following diagram, proving the injective property of I: M g f π◦γ N I " =⇒ ": Given M g f N I ⊕ I0 0 0 0 in RepK Q , we use the injectivity of I and I to obtain h : N → I, h N → I such that h ◦ f = π ◦ g, h0 ◦ f = π0 ◦ g, and by product property of I ⊕ I0 we get γ : N → I ⊕ I0 such that π ◦ γ = h, π0 ◦ γ = h0. We claim that γ completes the diagram above, thus proving the injectivity of I ⊕ I0. That is because 0 0 0 0 0 0 γ ◦ f = idI⊕I0 ◦ γ ◦ f = ι ◦ π ◦ γ ◦ f + ι ◦ π ◦ γ ◦ f = ι ◦ h ◦ f + ι ◦ h ◦ f = ι ◦ π ◦ g + ι ◦ π ◦ g = g  Proposition 3.16. I(i) is indecomposable. Proof. In analogy to the proof of indecomposability of P (i), if we assume I(i) ' M ⊕ N, with neither of summands zero, we can assume Mi = 0, and hence have Hom(M,I(i)) ' Mi = 0, hence the canonical injection of M into I(i) is zero, and thus M is zero, contradicting the hypothesis.  We now extend our definition of exactness (and that of homology) to a more general setting:

f g Definition 3.17. In an abelian category, given a sequence A −→ B −→ C such that g ◦ f = 0, we define the homology of the sequence at B as the cokernel of the induced morphism ϕ : Coker(j) → Ker(g), illustrated by the following diagram: ϕ Coker(j) Ker(g)

ρ ψ ι j f g Ker(f) A B C Where ψ is the morphism induced by f due to kernel property of Ker(g), and ϕ is the morphism induced by ψ due to cokernel property of Coker(j). We say that the sequence is exact at B whenever the homology at B is zero.

Definition 3.18. Given M ∈ RepK Q, a projective resolution of M is an exact sequence

· · · → P2 → P1 → P0 → M → 0 where Pl is projective for all l. We define the length of a resolution as greatest n such that Pn 6= 0.

Theorem 3.19. M = (Mi, ψα)i∈Q0,α∈Q1 ∈ RepK Q admits a projective resolution of length 1, γ f · · · → 0 → P1 −→ P0 −→ M → 0

Proof. We see that our resolution is a short exact sequence, so if we manage to find a suitable P0 =

(P0i, ϕα)i∈Q0,α∈Q1 together with an epimorphism f : P0 → M, we will have P1 ' Ker(f), and all that will remain to be done will be to show that Ker(f) is projective. We let dim(M ) M M i k P0 = P (i) i∈Q0 k=1 (superscript meaning "kth copy"; superscripts will be used to index copies throughout this proof). The K-linear maps of P0 will be denoted as ϕ (being - due to the linear maps of its direct summands - a large diagonal matrix of morphisms composing arrows). By definition, given i ∈ Q0, the canonical basis for (P0)i 16 consists of dim(Mj) copies of the set Wji of j-i paths, for all j ∈ Q0. We define f : P0 → M on that basis. l So fix j ∈ Q0, w ∈ Wji, and let (as mentioned earlier) w denote the lth copy of w. Now fix a basis Ej for l Mj, Ej = {e1, e2, ..., edim(Mj )}. We define f by letting fi(w ) = ψw(el) Clearly then, given an arrow κ ∈ Q1, κ : i → s we have l l (ψκ ◦ fi)(w ) = (ψκ ◦ ψw)(el) = ψwα(el) = (fs ◦ ϕα)(w ) Hence the following diagram commutes:

ϕκ (P0)i (P0)s

fi fs

ψκ Mi Mj

and so f is well-defined. Clearly, f is an epimorphism, as the dim(Mj) copies of empty paths j at j are mapped to respective basis elements of Mj by the definition of f. This also gives information about the dimension vector of Ker(f): each non-trivial path in our canonical basis contributes with one extra dimension to the kernel of f. We can already intuitively describe the projective structure of the kernel: we’re left with non-trivial paths in our basis, and by removing the starting arrow out of each path we shift from projective representation (as a direct summand of P0) at some vertex to projective subrepresentations at each vertex that is a target of an arrow starting from our initially chosen vertex. This is done for each arrow, and number of copies of the representation depends of numbers of copies of paths whose starting arrow we forget, thus we have the following hypothesis: M Ker(f) ' dim(Ms(α)) P (t(α))

α∈Q1 And hence we let P = L . dim(M ) P (t(α)) as above. We we denote the K-linear maps of P by ϕ¯. 1 α∈Q1 s(α) 1 By definition, given i ∈ Q0 the canonical basis for P1i is the following set: X {( dim(Ms(α))) copies of Wji | j ∈ Q0} α∈Q1,t(α)=j There is a natural bijection between that set and the set of non-trivial paths ending at i, which, as is l described above, is in bijection with a basis for Ker(f)i. Given j ∈ Q0, α ∈ Q1, α : k → j, let wα be the lth l l copy of the j-i path wα lying in the summand corresponding to α. We send wα to (αwα) (as the ordering comes from numbering the copies of α in P0, the order is naturally preserved). This clearly is a bijection. We denote αwα by w, and use subscripts to denote paths that have ’forgotten’ their starting arrows (represented by the subscript). Now we start defining γ : P1 → P0. We will show three properties of γ: (1) γ is well-defined. (2) Im(γ) ⊂ Ker(f). (3) γ is a monomorphism. Once we have established those three properties we will be able to conclude that γ is an isomorphism between P1 and Ker(f), since the pointwise equality of dimensions combined with γ being pointwise mono makes γ pointwise iso, and as we’ve seen before, a well-defined morphism of representations that is a pointwise isomorphism of vector spaces is an isomorphism of representations.

We define γ by defining it on a general basis element of P1. Hence let i ∈ Q0, α ∈ Q1 such that α : k → j l and let w ∈ Wki be a k-i path such that α is the starting arrow of w. Let wα be the lth copy of w which has ’forgotten’ its starting arrow, as described above. This is our basis element.

l l l It would be natural (and most certainly well-defined) to set γ(wα) = w , but we need γ(wα) to lie in the l kernel of fi, and that is not necessarily the case for w . By definition, if el is the lth element in our chosen l basis Ek for Mk, we have f(w ) = ψw(el) Now we use that w = αwα to obtain X ψw(el) = ψαwα (el) = ψwα (ψα(el)) = ψwα ( λcec). ec∈Ej 17 Note that the linear combination P λ e ) is determined uniquely. ec∈Ej c c

By definition of f we have that ψ (P λ e ) = f (P λ wc ) and finally we conclude that f (wl) = wα ec∈Ej c c i c c α i P c l P c fi( c λcwα), so fi(w − c λcwα) = 0. We thus set l l X c γi(wα) = w − λcwα c

The quest now is to show that γ is well-defined. Thus we let κ ∈ Q1, κ : i → s, and aim to show that the following diagram commutes:

ϕ¯κ P1i P1s

γi γs

ϕκ P0i P0s We have l l X c l X c l X c (ϕκ ◦ γi)(wα) = ϕκ(w − λcwα) = (w )κ − ( λcwα)κ = (wκ) − λc(wκ)α c c c (note that composing with a new arrow added at the end of the path does not change neither the ’forgotten’ starting arrow nor the index of the path, so our subscript and superscript operations commute with ϕ and ϕ¯) and: l l l l X v (γs ◦ ϕ¯κ)(wα) = γs(wακ) = γs((wκ)α) = (wκ) − λv(wκ)α v For the unique set of coefficients {λv} such that l X v fs(wκ) − λv(wκ)α = 0 v But we also have

l X c X fs((wκ) − λc(wκ)α) = ψwκ(el) − ψ(wκ)α ( λcec) = c c X l X c ψκ(ψw(el) − ψwα ( λcec)) = ψκ(fi(w ) − fi( λcwα = ψκ(0) = 0 c c P v P c Hence, by uniqueness mentioned above we must have v λv(wκ)α = c λc(wκ)α and hence γ is well-defined. l Finally, γ is pointwise mono as the first term in γi(wα) is a path longer than the others. Hence if the first terms of γ of two basis elements b1, b2 are non-equal, then γ(b1) 6= γ(b2). And the first term is just the image of the basis element under the "remembering" bijection, hence if b1 6= b2 then their first terms are non-equal, and finally, if b1 6= b2, then γ(b1) 6= γ(b2). 

Definition 3.20. Given M ∈ RepK Q, a projective cover of M is a projective P ∈ RepK Q together with 0 an epimorphism g : P → M such that given another pair of projective P ∈ RepK Q together with an epimorphism g0 : P 0 → M, there is an epimorphism h : P 0 → P such that g ◦ h = g0. Diagrammatically we have P 0

g0 h g P M

Definition 3.21. Dually, an injective envelope of M is an injective I ∈ RepK Q together with a monomor- 0 0 0 phism f : M → I such that given another injective I ∈ RepK Q together with a monomorphism f : M → I , there is a monomorphism h : I → I0. Definition 3.22. A projective resolution

· · · → P2 → P1 → P0 → M → 0 is minimal if P0 → M is a projective cover and so is Pi → Keri−1 for all i > 0. Proposition 3.23. Let (P, g), (P 0, g0) be projective covers of M. Then P ' P 0 18 Proof. Let h : P 0 → P be an epimorphism induced by the projective cover property of P . Then we have the following exact sequence: 0 → Ker(h) −→ι P 0 −→h P → 0 And since P is projective, this sequence splits. We thus have P 0 ' Ker(h) ⊕ P , and by using the projective cover property of P 0 in the same manner we obtain 0 0 0 0 0 P ' P ⊕Ker(h ) ' P ⊕Ker(h)⊕Ker(h ), so Ker(h) = 0 = Ker(h ), and both h and h are isomorphisms.  Proof of the dual statement is dual: Proposition 3.24. Let (I, f), (I0, f 0) be injective envelopes of M. Then I ' I0

Proof. By injective envelope property of I, we have an exact sequence 0 → I −→h I0 −→π Coker(h) → 0, and that sequence splits as I is injective. So I0 ' Coker(h) ⊕ I, and similarly for I0, I ' Coker(h0) ⊕ I0 ' Coker(h0) ⊕ Coker(h) ⊕ I. 0 Thus we have Coker(h) = 0 = Coker(h ), and so the morphisms are iso.  Definition 3.25. Let A = L P (i). F ∈ Rep Q is free if and only if F ' A ⊕ · · · ⊕ A i∈Q0 K

Proposition 3.26. i) M ∈ RepK Q is projective if and only if it is a direct summand of a free representation F . ii) M is projective if and only if it is a direct sum of some P (i), i ∈ Q0 Proof. i) " ⇐= ": follows by repeated application of proposition 3.10. Using said proposition we know that A (being defined as a direct sum of projective indecomposable representations) is projective, hence so is F , and finally, using the proposition for the third time, we get that M, being a direct summand of F , is projective. " =⇒ " follows from the canonical projective resolution of M, 0 → P1 → P0 → M → 0, since P0 clearly is a direct summand of a large enough F (for example of rank equal to the maximal dimension of a space in M), and M is projective, so our sequence splits and M is a direct summand of P0 and thus also a direct summand of that F . ii) Follows directly from i), M is a direct summand of F , and by Krull-Schmidt theorem and indecompos- ability of P (i), direct summands of F are (small enough) direct sums of P (i), i ∈ Q0 

Proposition 3.27. Let M = (Mi, ψα)i∈Q0,α∈Q1 ∈ RepK Q, and let P be a projective representation P ∈ RepK Q. If M is a subrepresentation of P , then M is projective.

Proof. Consider the following short exact sequence in RepK Q : 0 → M −→ι P −→π P/M → 0 where ι is the canonical injection and π is the canonical projection. By proposition 3.19 we know that there 0 0 are projective representations P0,P1,P0,P1 such that the following commutative diagram exists: 0 0

0 M ι P π P/M 0

0 P0 P0

0 P1 P1

0 0 Now by Horseshoe lemma (proof of which can be found in [6], p.349) we can complete the "missing" second and third rows of the diagram so that the rows split, and that the resulting middle column is a projective 19 resolution of P . Using that P itself is projective, we conclude that the middle column also is split. Hence the morphisms f,¯ g¯ in the diagram below are induced by f and g being sections. 0 0 0

0 M ι P π P/M 0

0 b 0 P0 Q0 P0 0

a g g¯ f 0 0 P1 Q1 P1 0 f¯

0 0 0 In particular we have

f¯◦ g¯ ◦ b ◦ a = f¯◦ g¯ ◦ g ◦ f = f¯◦ id ◦ f = f¯◦ f = id 0 . Q1 P1 Thus a is a section, and by proposition 2.27 we conclude that the first column is a split exact sequence, 0 0 hence P0 = M ⊕ P1 and by proposition 3.10 we know that M is projective. 

We thus have established that RepK Q is what is called a hereditary category. This has immediate consequences for some of the homological algebra we have managed to defined for quiver representations. Namely, recall that given a projective resolution of M,

· · · → P2 → P1 → P0 → M → 0 and cochain complex given by Hom(−,X) of the sequence:

0 → Hom(M,X) → Hom(P0,X) → · · · we can, independently of chosen resolution, obtain the Exti(M,X), i ≥ 0 by taking the cohomology of the so called deleted complex

0 → Hom(P0,X) → Hom(P1,X) → · · · but since we know that every quiver representation has a projective resolution of length 1, we obtain the following result: Proposition 3.28. Ext2 vanishes for quiver representations. Note that this can be established for any abelian category C that satisfies the following conditions: a) It is hereditary b) It has enough projectives, i.e. given M ∈ C, there is a projective object P with an epimorphism P → M f This is easily seen by exactness of 0 → Ker(f) → P −→ M → 0, yielding a projective resolution of length 1, putting us in the same situation as we had for quiver representations. Lemma 3.29. Additive functors preserve finite biproducts. (i.e. the coinciding notions of products and coproducts, see [5], corollary 8.2.4) Proof. Let C, D be additive (i.e. Ab-Enriched, with finite coproducts) categories, and let F : C → D be an additive functor. (i.e. for morphisms f, g in C, we have F (f + g) = F (f) + F (g) whenever f + g is well- defined.) Let M,M 0 ∈ C, and let M ⊕ M 0 be their biproduct. We show that F (M ⊕ M 0) = F (M) ⊕ F (M 0) by showing that F (M ⊕ M 0) satisfies the universal property of product of F (M) and F (M 0), and thus is the biproduct of those objects. 0 0 Let N ∈ D, f1 ∈ Hom(F (M),N), f2 ∈ Hom(F (M ),N). Let f := f1 ◦ F (π) + f2 ◦ F (π ). First, we have: 0 0 f ◦ F (ι) = (f1 ◦ F (π) + f2 ◦ F (π )) ◦ F (ι) = f1 ◦ F (π ◦ ι) + f2 ◦ F (π ◦ ι) = f1 ◦ F (idM ) + f2 ◦ F (0) = f1 20 0 f ◦ F (ι ) = f2 is obtained in the same fashion. So f satisfies the necessary conditions for being the product morphism, and now we only need to show it’s unique such. Thus let f 0 also satisfy those conditions. We get 0 0 0 0 0 0 0 0 f = f ◦ idF (M⊕M 0) = f ◦ F (ι ◦ π + ι ◦ π ) = (f ◦ F (ι)) ◦ F (π) + (f ◦ F (ι )) ◦ F (π ) = f1 ◦ π + f2 ◦ π = f  Definition 3.30. Let Qop denote the opposite quiver of Q i.e a quiver whose set of vertices is the same as that of Q, and its set of arrows consists of arrows of Q with their direction reversed. Let D be the contravariant op ∗ functor: Rep Q → Rep Q sending representations M = (M , ϕ ) to (M , ◦ϕ ) op op K K i α i∈Q0,α∈Q1 i α i∈Q0 ,α∈Q1 where M ∗ denotes the dual space of M and ◦ϕ denotes the pullback of ϕ, and sending the morphisms γ = (γ ) : M → N to Dγ : DN → DM, where Dγ is defined as (Dγ ) op := (◦γ ) op . i i∈Q0 j j∈Q0 j j∈Q0

Proposition 3.31. Let Proj Q denote the subcategory of RepK Q of projective representations of Q, and Inj Q the subcategory of RepK Q of injective representations of Q. D is a contravariant equivalence of categories (a so called duality). In particular we have D(PQ(i)) = IQop (i). Thus, by additivity of D and lemma 3.29, its restriction to Proj Q is a duality between Proj Q and Inj Qop.

op op op Proof. We show that i) DQ ◦ DQ = idRepK Q which is equivalent to DQ ◦ DQ = idRepK Q , and thus the categories are equivalent. ∗ 2 Note that given a linear map f, its pullback is exactly its dual f . Now denote DQop ◦ DQ as D . Given 2 ∗∗ ∗∗ M = (Mi, ϕα)i∈Q0,α∈Q1 ∈ RepK Q, we have D M = (Mi , ϕα )i∈Q0,α∈Q1 ' M (by isomorphisms, both point- and arrow-wise), and similarly for f ∈ Hom(M,N), D2f = f ∗∗ ' f, so we’re done with that part. Now we show D(PQ(i)) ' IQop (i). First note that for m ∈ Q0, there is a canonical choice of basis for DPQ(i)m, namely the basis dual to the ∗ canonical basis for PQ(i)m. Such a basis consists of a set of 1-forms {w | w ∈ Wim (set of i-m paths) }, ∗ ∗ ∗ where w is defined on the canonical basis by letting w (w) = 1 and letting w (η) = 0 for all η ∈ Wim : η 6= w. op op o o Let DPQ(i) = (DPQ(i)j, ◦ϕα)j∈Q0,α∈Q1 , IQ = (IQ (i)j, ψα)j∈Q p0,α∈Q p1 . Given α : j → m in Q1 and an ∗ ∗ ∗ i-j path t in Q, we have ((◦ϕα) ◦ ω )(t) = ω ◦ ϕα(t) = ω (tα) = δω(tα), giving the zero map if ω does not end with α and otherwise checking whether t is equal to ω if we forget the α at the end of it. Let this form ∗ −1 −1 be denoted by ω α. We define f : DPQ(i) → IQop (i) by ω → ω , where ω denotes the path "opposite to ω". Clearly it is pointwise mono and epi (being a natural bijection between i-m paths in Q and m-i paths in Qop), and in order to prove that it is an isomorphism we only need to show that it is well-defined, that is −1 op fj ◦ (◦ϕα) = ψα−1 ◦ fm (note that α ∈ Q1 corresponds to α ∈ Q1 ). Choose an i-m path ω in canonical ∗ −1 −1 −1 −1 basis for DPQ(i)m. We have (ψα ◦ fm)(ω ) = ψα(ω ) = ω − α (which is 0 if ω does not start at α−1 ⇐⇒ ω ends on α, else, by definition, one just cuts off the starting arrow α−1). Clearly we also have −1 −1 (fj ◦ (◦α))(ω) = fj(ω α) = ω − α , so our isomorphism is well defined.  We make one more observation about D, which will allow us to justify our earlier "omission" of injective resolutions. Proposition 3.32. An equivalence of Abelian Categories is exact. Proof. Let C, D be abelian categories and let F : C → D be an equivalence between them. By our definition of exactness, it is not hard to see that in order to establish the exactness of F , we only need to show that it preserves kernels and cokernels (as then the diagram defining the homology as in definition 3.17 is preserved under F ). The claim is:

f F (ι) F (f) Ker(f) ι L M F (Ker(f)) F (L) F (M) F ∀h −→ ∀h0=F (h) ∃!ϕ ∃!F (ϕ) ∀X ∀X0,X0 ' F (X) Given X0 ∈ D, we have X0 ' F (X) for some X ∈ C as F is essentially surjective. Similarly, an arbitrary h0 ∈ Hom(X0,F (L)) in D can be (under that isomorphism) expressed as F (h) for some h ∈ Hom(X,L) in C as F is full. Finally, F (ι) ◦ F (ϕ) = F (ι ◦ ϕ) = F (h), so F (ι) clearly satisfies the relation required to be a morphism induced by kernel property, and it is unique such since ϕ was unique such in C. The proof for cokernels is analogous.  21 Thus F is exact, and we can justify the following claim:

Lemma 3.33. M ∈ RepK Q has an injective resolution of length 1. op Proof. Take the standard projective resolution of DM in RepK Q ,

0 → P1 → P0 → DM → 0 and now apply D to that sequence to obtain

0 → M → DP0 → DP1 → 0

By results above we know that DP0 and DP1 are injective, and that the sequence is exact.  Now we introduce an auxiliary, "special hom-functor", which instead of sending quiver representations to op vector spaces (the hom-sets), sends RepK Q to RepK Q . Recall that we called a representation free if and only if it was isomorphic to A ⊕ · · · ⊕ A. We define our functor using Hom(−,A). op Definition 3.34. Hom(−,A) : RepK Q → RepK Q is defined as follows: Given X ∈ RepK Q , let op Hom(−,A)(X) = Hom(X,A) = (M , ϕ −1 ) op −1 op , where M = Hom(X,P (i)) for all i ∈ Q , and i α i∈Q0 ,α ∈Q1 i 0 −1 op now, using the fact that Hom(P (i),P (j)) has a basis consisting of all j-i paths, given α : j → i ∈ Q1 0 we let ϕαop : Hom(X,P (j)) → Hom(X,P (i)) take f to α ◦ f. Now for morphisms, given g ∈ Hom(X,X ) op 0 we let Hom(−,A)(g) = ◦g. Clearly for α as earlier we have ((◦gi) ◦ ϕαop )(f) = (α ◦ f) ◦ g = α ◦ (f ◦ g) = (ϕαop ◦(◦gj))(f), so it is well defined, and just like for the usual hom-functor we have (◦(f ◦g)) = (◦g)◦(◦f), so we have a contravariant functor. Now we have the necessary tools to introduce the Nakayama functor, but there are still two remarks to be made. Proposition 3.35. Hom(−,A) is zero on non-projective indecomposable summands. L Proof. Let M ∈ RepK Q , and let M = Mk be a decomposition of M into indecomposable sum- k L mands. Since Hom(−,A) is additive (being a hom-functor), we have Hom(−,A)M = k Hom(Mk,A), and Hom(Mk,A) 6= 0 if and only if Mk is projective, since for a non-zero morphism in Hom(Mk,P (j)), the sequence 0 → Ker(f) → Mk → Im(f) → 0 splits (as Im(f) = Ker(Coker(f)) splits, since Im(f) is a subrepresentation of Pj and thus is projective), and thus Mk = Ker(f) ⊕ Im(f), so either Ker(f) = 0 or Im(f) = 0, but f 6= 0 implies that Im(f) 6= 0, so Ker(f) = 0, thus Mk ' Im(f), and f is injective, and so Mk is a subrepresentation of P (j), and thus Mk is projective.  Remark 3.36. Hom(−,A) restricted to Proj Q is a duality Proj Q → Proj Qop

Proof. On objects we have Hom(−,A)(PQ(i)) = PQop (i): a canonical basis for (Hom(−,A)(PQ(i)))k = Hom(PQ(i),PQ(k)) consists of the set of all k-i paths. Map those to their "opposites" in the representation of the opposite quiver PQop (i)k. Call that map γ. γ is clearly a bijection of bases and thus a pointwise isomorphism. To show that it’s well-defined we have, given a k-i path ω in Q and an arrow α : j → −1 −1 −1 k ∈ Q1, (γj ◦ (◦α))(ω) = γj(αω) = (αω) = ω α = (ϕα−1 ◦ γk)(ω), where ϕα−1 denotes the linear −1 −1 −1 op map of PQop along α , attaching α to ω . Thus we have (HomQop (−,A ) ◦ HomQ(−,A))(PQ(i))) = op opop (HomQop (−,A )(PQop (i)) = PQ(i), using the fact that Q = Q. On morphisms we only need to study what happens to morphisms between the indecomposable projectives, since Hom is additive. Take two indecomposable projectives in Proj Q, P (j),P (m), and let ω be the morphism corresponding to the m-j path ω. ω is sent to ◦ω under Hom(−,A), and then to ωop under the isomorphism described above, and thus again, if we first apply HomQ(−,A) and then its (as is being proven) quasi-inverse op −1 −1 HomQop (−,A ) on ω, we obtain (ω ) = ω. 

Definition 3.37. The Nakayama functor is defined as ν : RepK Q → RepK Q , ν = D ◦ Hom(−,A) Some properties of ν follow from the properties of D and Hom(−,A) discussed earlier. First of all, since op Hom(−,A) is a duality Proj Q → Proj Q , with Hom(−,A)(PQ(i)) = PQop (i), and D is a duality mapping projectives to injectives, we have: Proposition 3.38. The restriction of ν to projective representations is an equivalence Proj Q → Inj Q with ν(P (i)) = I(i) 22 And since Hom is left exact and D is exact and contravariant, we have: Proposition 3.39. The Nakayama functor ν is right exact.

Definition 3.40. Given M ∈ RepK Q let

p1 p0 0 P1 P0 M 0

i0 i1 0 M I0 I1 0 be minimal projective respectively minimal injective resolutions of M. (those exist by I.5.10 in [1]) We define:

(1) The Auslander-Reiten translate of M, τM, as Ker(νp1) in the sequence

ι p1 p0 π 0 → Ker(νp1) −→ νP1 −→ νP0 −→ νM −→ Coker(νp0) obtained by applying nu to the projective resolution above

(2) Let ORepK Q denote the set of objects of RepK Q (the set of all representations of Q). We define The

Auslander-Reiten translation as the map τ : ORepK Q → ORepK Q by M → τM. −1 −1 (3) The inverse Auslander-Reiten translate of M, τ M, as Coker(ν i1) in the sequence

−1 ι −1 i0 −1 i1 −1 π −1 0 → Ker(ν i0) −→ ν I0 −→ ν I1 −→ ν M −→ Coker(ν i1) obtained by applying ν−1 to the injective resolution above. −1 −1 (4) The inverse Auslander-Reiten translation τ : ORepK Q → ORepK Q by M → τ M Note that this is well-defined by propositions 3.23 and 3.24, since as a consequence of those, minimal resolutions are unique up to isomorphism.

4. Irreducible morphisms and almost split sequences Towards the end of the last section we have payed significant attention to a particular quiver representa- tion, A = L P (i). We will now reveal why that representation plays a special role. i∈Q0 Definition 4.1. Given a quiver Q, we let KQ denote the algebra defined as follows: as a K-vector space, we let KQ be the K-linearization of the set Q¯ of paths in Q, and the product is defined (on the basis consisting of paths) by composition of paths; if the composition is impossible, the product is P P P defined to be 0. More formally we have ( ω∈Q¯ λωω) · ( κ∈Q¯ ακ) = ω,κ δt(ω),s(κ)λωακ ωκ Lemma 4.2. As composing paths is associative, the path algebra is an associative algebra. Moreover, under our assumptions about the quiver (being finite, acyclic), this algebra is unital (its unit being P  and i∈Q0 i finite-dimensional (its dimension being equal to the number of paths in the quiver). Before the significance of A is revealed, we need one more result: Proposition 4.3. The category mod-KQ of right finite dimensional KQ-modules is equivalent to the cate- gory RepK Q of finite-dimensional representations of Q. Proof. Let M ∈ mod-KQ. Note that M = L M as the identity of the path algebra is given by i∈Q0 i P  , and i 6= j implies that M ∩ M = 0 as i∈Q0 i i j 2 m = m1i = m2j, so m1i = m1i = m = m2(ji) = m20 = 0 and that for α : i → j ∈ Q1 the map · α : M → M is 0 on Mj for j 6= i and im(· α) ⊂ Mj. And by the definition of M we know that · α is K-linear. Thus · α : Mi → Mj is a linear map. Now we will define the equivalence F : mod-KQ → RepK Q . Let F send M to quiver representation (Mi, · α)i∈Q0,α∈Q1 and for a module morphism ϕ ∈ Hom(M,N) (being a K-linear map commuting with the KQ-action) we have ϕ(Mi) ⊂ ϕ(M)i, so ϕ can be viewed as a set of maps ϕi : Mi → Ni, and all of them commute with mod-KQ action, so if we instead view that set as a morphism of quiver representations F (M),F (N), our map commutes with arrows, and thus is well-defined. Clearly now F is a functor. Now we construct its quasi-inverse: 23 Let M = (M , ψ ) ∈ Rep Q. Now let F −1(M) = L M and endow it with following action: i α i∈Q0,α∈Q1 K i∈Q0 i for m ∈ M, let ( m, if m ∈ Mi mi = 0, otherwise

, and for m ∈ Mi, α : i → j ∈ Q1 let mα = ψα(m) (for other m let mα = 0). This clearly defines a KQ-module, and since F only "disconnects" a module, which is later "joined" back together in the same −1 fashion by F , these are clearly quasi-inverses.  Proposition 4.4. A = L P (i) = F (KQ ), where KQ denotes the path algebra as a module over i∈Q0 KQ KQ itself.

Proof. By definition of F we have a canonical basis for F (KQKQ)i consisting of paths ending at i, which is exactly the basis for Ai (since basis for P (j)i consists of paths from j to i, but here we take the direct sum over all j). Thus F (KQKQ)i = Ai. And for an arrow α : i → j ∈ Q1, the linear map corresponding to it acts by composing with α, by definition, thus we also obtain arrow-wise equality, and thus obtain our result. 

Proposition 4.5 (II.1.12 in [1]). Let |Q0| = n. Then

 1(KQ)1 0 ... 0  2(KQ)1 2(KQ)2 ... 0 KQ '  . . .   . . .  n(KQ)1 n(KQ)2 ... n(KQ)n

Proof. Since the quiver is acyclic, we can enumerate Q0 so that α : i → j ∈ Q0 implies j ≥ i. Now let W denote the set of all paths in Q, and, just as before, let Wij denote the set of all i-j paths in Q. Clearly then W = ` W . Now for x = P λ ω we can rewrite it as x = P P λ ω, and that element we j≥i ij ω∈W ω j≥i ω∈Wij ω uniquely express as the lower triangular matrix X = (P λ ω) . Now all we need to do is to show ω∈Wij ω j≥i that our bijection χ : x → X is a morphism of algebras. For x, y ∈ KQ we have X X X x = λωω, y = κηη, so xy = λωκηωη

j≥i,Wij j≥i,Wij j≥k≥i,ω:j→k,η:k→i and fixing j, i in that expression yields χ(xy)ij as a sum over k. On the other hand n X X (χ(x)χ(y))ij = λωκηωη k=1 ω:j→k,η:k→j and we get entry-wise matrix equality by noting that in order for an element in the sum to be non-zero we need j ≥ k ≥ i. Other criteria follow entry-wise.  The remaining results of this chapter are proven within the framework of modules over some finite- dimensional algebra A, but by the equivalence shown above, all the results follow for the category of finite-dimensional representations of an acyclic quiver Q. We could thus even restrict the generality of our considerations and think of A as the path algebra of Q. Definition 4.6 (IV.1.1 [1]). Let L, M, N ∈ mod-A.

(1) f ∈ Hom(L, M) is left minimal if for every h ∈ End(M) we have: if h ◦ f = f, then h is an automorphism. (2) g ∈ Hom(M,N) is right minimal if for every h ∈ End(M) we have: if g ◦ h = g then h is an automorphism. (3) f ∈ Hom(L, M) is left almost split if f is not a section and for all h ∈ Hom(L, X): if h is not a section, then there is a morphism h0 ∈ Hom(M,X) such that h0 ◦ f = h. (4) g ∈ Hom(M,N) is right almost split if g is not a retraction and for all h ∈ Hom(X,N): if h is not a retraction, then there is a morphism h0 ∈ Hom(X,M) such that g ◦ h0 = h. (5) f ∈ Hom(L, M) is left minimal almost split if it is left minimal and left almost split. (6) g ∈ Hom(M,N) is right minimal almost split if it is right minimal and right almost split. 24 Proposition 4.7 (IV.1.2 in [1]). (1) If A-module homomorphisms f : L → M, f 0 : L → M 0 are left minimal almost split, then there is an isomorphism h : M → M 0 such that f 0 = h ◦ f (2) If A-module homomorphisms g : M → N, g0 : M 0 → N are right minimal almost split, then there is an isomorphism h : M → M 0 such that g0 = g ◦ h Proof. (1) By "almost split" property of f we know that there is ϕ : M → M 0 such that ϕ ◦ f = f 0. By the same property of f 0 we know that there is ψ : M 0 → M such that ψ ◦ f 0 = f, thus ψ ◦ ϕ ◦ f = ψ ◦ f 0 = f, and ϕ ◦ ψ ◦ f 0 = ϕ ◦ f = f 0, and by "minimal" property of f and f 0 we know that ψ ◦ ϕ respectively ϕ ◦ ψ are isomorphisms. Now ϕ ◦ ψ is epi, hence so is ϕ, and ψ ◦ ϕ being mono means that so is ϕ, thus ϕ is mono and epi and thus an isomorphism, and by its definition we have f 0 = ϕ ◦ f. (2) Similarly to 1), we first use the "almost split" property to obtain ϕ : M → M 0, ψ : M 0 → M such that g = g0 ◦ ϕ, g0 = g ◦ ψ, which yields g ◦ ψ ◦ ϕ = g, g0 ◦ ϕ ◦ ψ = g0, and thus, by the "minimal" property we have ϕ ◦ ψ iso and ψ ◦ ϕ iso, and thus ψ satisfies said conditions.  Proposition 4.8 (IV.1.3 in [1]). f (1) If L −→ M is left almost split, then L is indecomposable. g (2) If M −→ N is right almost split, then N is indecomposable. Proof.

(1) Assume L is not indecomposable, L ' L1 ⊕ L2,L1,L2 6= 0. Recall that then idL = ι1 ◦ π1 + ι2 ◦ π2, ι, π denoting the canonical inclusions and projections. By "almost split" property of f we know that there is ϕ1 such that ϕ1 ◦ f = π1, and ϕ2 such that ϕ2 ◦ f = π2, since clearly none of the γ projections is a section (as long as the other summand is non-zero). But then define M −→ L as γ = ι1 ◦ ϕ1 + ι2 ◦ ϕ2. We have

γ ◦ f = (ι1 ◦ ϕ1 + ι2 ◦ ϕ2) ◦ f = ι1 ◦ (ϕ1 ◦ f) + ι2 ◦ (ϕ2 ◦ f) = ι1 ◦ π1 + ι2 ◦ π2 and thus f is a section, contradicting the definition of f. (2) This is analogous to 1): assume N is not indecomposable, N ' N1 ⊕ N2. Then since g is almost split we have ψ1, ψ2 such that g ◦ ψ1 = ι1, g ◦ ψ2 = ι2 (as inclusions are not retractions). Then by η letting N −→ M, and letting η = ψ1 ◦ π1 + ψ2 ◦ π2, we get g ◦ η = idN , and thus g is a retraction, contradicting the definition of g.  f g Definition 4.9. A short exact sequence 0 → L −→ M −→ N → 0 in mod-A is an almost split sequence if f is left minimal almost split and g is right minimal almost split.

f g f 0 g0 Remark 4.10. Let 0 → L −→ M −→ N → 0 and 0 → L0 −→ M 0 −→ N 0 → 0 be two almost split exact sequences. Then the following statements are equivalent: (1) The two sequences are isomorphic (2) L ' L0 (3) N ' N 0 Proof. (1) implies (2) and (3) by definition. We show that (2) implies (1). (3) implies (1) in a very similar manner. f 0◦u Let L −→∼ L0. Then clearly L −−−→ M 0 is left minimal almost split, and thus, by proposition 4.7, there u is an isomorphism t : M −→∼ M 0 such that t ◦ f = f 0 ◦ u. Note that by exactness of sequences we have p N ' Coker(f), N 0 ' Coker(f 0), and thus g0 ◦ t induces a morphism N −→ N 0 such that p ◦ g = g0 ◦ t, by the p0 cokernel property of N. Similarly, g ◦ t−1 induces a morphism N 0 −→ N such that p0 ◦ g0 = g ◦ t−1, by the cokernel property of N 0. Using those two expressions we conclude that p ◦ g ◦ t−1 = (p ◦ p0) ◦ g0 = g0 ◦ (t ◦ t−1) = g0 25 0 0 0 0 and thus p ◦ p = idN 0 as g is epi. Similarly we obtain p ◦ p = idN , so p and p are mutually inverse isomorphisms that also commute with g, g0, t and t−1, and thus we have:

f g 0 L M N 0

u t p f 0 g0 0 L0 M 0 N 0 where the vertical arrows are isomorphisms.  We shift our focus from the almost split sequences and morphisms for a moment and instead introduce a closely related notion of irreducible morphisms: Definition 4.11. A morphism f : M → N is irreducible if: (1) f is neither a section nor a retraction (2) if f = f1 ◦ f2 for some f2 ∈ Hom(M,X), f1 ∈ Hom(X,N), then either f1 is a retraction or f2 is a section. Proposition 4.12. Given a morphism f : X → Y , if f is irreducible, then it is either a monomorphism or an epimorphism. Proof. Assume that f is not an epimorphism. Consider the canonical factorization of f through its image, f 0 X −→ Im(f) −→ι Y Since f is not an epimorphism, neither is ι, and thus ι can’t be a retraction. Thus, since f is irreducible, f 0 is a section, and thus f 0 is mono and so is f. Now assume that f is not a monomorphism. We use the first isomorphism theorem to reverse the order of the universal properties we used in the first argument and obtain the following diagram: f Ker(f) ι X Y ∃!g π

Coker(ι) Now since f is not mono, neither is π, so π is not a section and thus g is a retraction, so g, and hence also f, is an epimorphism.  In order to prove the first basic fact about irreducible morphisms, we first show some statements about pullbacks and pushouts. Lemma 4.13. (1) Given a pullback square A t B u v C w D in mod-A, if v is an monomorphism, then so is u. Similarly, if v is a epimorphism, then so is u. (2) Dually, the same statements hold for pushout squares.

Proof. First we show the statement for monomorphisms: Let X ∈ mod-A, g, g0 ∈ Hom(X,A) such that u ◦ g = u ◦ g0. We will show that then g = g0, and thus u is mono. We have u ◦ g = u ◦ g0, and thus w ◦ u ◦ g = w ◦ u ◦ g0, and by commutativity of the square also v ◦ t ◦ g = v ◦ t ◦ g0, and thus, since v is a monomorphism, t ◦ g = t ◦ g0. But then both g and g0 satisfy the properties of the pullback morphism for t ◦ g ∈ Hom(X,B), u ◦ g ∈ Hom(X,C), but such morphism is unique, and thus g = g0. Surprisingly, the analogous statement for epimorphisms is not as natural, and in the opposite to that for monomorphisms, it does not hold for general abelian categories. We prove the statement as follows: Note that under our equivalence of categories established in proposition 4.3, mod-A has pullbacks, and these 26 pullbacks look just like those for K-modules (since all we do is "gathering" the pointwise K-module pullbacks of the quiver into one KQ-module), that is, given B C

v w D in mod-A, its pullback is defined as {(x, y) ∈ B ⊕ C | v(x) = w(y)}. Now we have ∀c ∈ C : ∃b ∈ B : w(c) = v(b) by choosing b ∈ v−1(w(c)), which is non-empty since v is epi. Thus u(P ) = {c ∈ C | ∃b ∈ B : w(c) = v(b)} = C and hence u is epi. The proof for pushouts squares is dual, and dually to pullbacks, it is epimorphisms that are stable un- der pushouts in more general settings, whereas monomorphisms require the underlying category to have additional properties.  Proposition 4.14. Given the diagram

0 0 L r V 0 v V 0

0 1L g v f g 0 L M N 0 g V 0 is isomorphic to the pullback of V −→v N ←− M. Dually, given the diagram f g 0 L M N 0

0 u f 1N 0 0 U u U 0 t N 0 f U 0 is isomorphic to the pushout of U ←−u L −→ M. In order to prove that result, we first state and show a lemma: Lemma 4.15 (Theorem II.6.2 in [4]). Let P α A

β ϕ ψ B X be a pullback square in an abelian category C. Let J be the kernel of β and let µ be its canonical inclusion morphism into P . Then Ker(ϕ) = J and the canonical inclusion of Ker(ϕ) is equal to α ◦ µ α α α◦µ Proof. First, α ◦ µ is mono, since the inclusion [ β ]: P → A ⊕ B is mono and so is µ, hence [ β ] ◦ µ = [ 0 ] is mono and thus α ◦ µ is mono. Moreover, ϕ ◦ α ◦ µ = ψ ◦ β ◦ µ = 0. Now we only need to show that (J, α ◦ µ) satisfies the universal property of kernels. So let Z ∈ C, and τ ∈ Hom(Z,A) be a morphism such that ϕ ◦ τ = 0. Then we have ϕ ◦ τ = ψ ◦ 0 = 0 and that commutative square induces a unique morphism σ : Z → P by the pullback property of P such that α ◦ σ = τ and β ◦ σ = 0. By the kernel property of µ, the second equality induces a unique morphism γ such that σ = µ ◦ γ. Thus we have τ = αµγ, and γ is the unique morphism satisfying that equality.  Now we prove the preceding proposition. Proof of proposition 4.14. Let P  V ϕ v g M N 27 g be the pullback square of V −→v N ←− M. By lemma 4.15 we know that ϕ induces an isomorphism Ker() ' Ker(g) ' L. We thus have the following short exact sequence: 0 → L −→ι P −→ V → 0, where ι is the kernel morphism of L as the kernel of . By pullback property of P there is a morphism ω ∈ Hom(V 0,P ) such that ϕ ◦ ω = g0 and v0 =  ◦ ω. Moreover, since r is the kernel morphism of v0, we have  ◦ ω ◦ r = v0 ◦ r = 0, so Im(ω ◦ r) ⊂ Ker(). We also have ϕ ◦ ω ◦ r = g0 ◦ r = f = ϕ ◦ ι. Since ϕ induces an isomorphism Ker() → Ker(g), we can treat ϕ like a monomorphism and thus conclude that ω ◦ r = ι. We then have the following commutative diagram with exact rows:

0 0 L r V 0 v V 0

idL ω idV 0 L ι P  V 0

and since both idL and idV are isomorphisms, then, by five lemma, so is ω. Proof of the dual statement for pushouts is dual.  f g Proposition 4.16 (IV.1.7, [1]). Let 0 → L −→ M −→ N → 0 be a nonsplit short exact sequence in mod-A. Then (1) f is irreducible if and only if, for every module V , and every morphism v ∈ Hom(V,N), there exists v1 ∈ Hom(V,M) such that v = g ◦ v1 or there exists v2 ∈ Hom(M,V ) such that g = v ◦ v2. (2) g is irreducible if and only if, for every module U, and every morphism u ∈ Hom(L, U), there exists u1 ∈ Hom(M,U) such that u = g ◦ u1 or there exists u2 ∈ Hom(U, M) such that g = u ◦ u2. Proof. (1)" =⇒ ": Diagrammatically we can represent our premises as

V v f g 0 L M N 0 Let P be the pullback of V v g M N and let g0 ∈ Hom(P,V ), u ∈ Hom(P,M) denote the morphisms completing the pullback square. Since g is epi, by lemma 4.13 we find that g0 is epi. There is also a morphism f 0 ∈ Hom(L, P ) induced by pullback property of P applied on square L 0 V f v g M N f 0 is mono as we have u ◦ f 0 = f and f is mono. Thus the following diagram commutes:

f 0 g0 0 L P V 0

idl u v f g 0 L M N 0

Moreover, its rows are exact. By definition of f 0 we know g0 ◦ f 0 = 0, so Im(f 0) ⊂ Ker(g0). Now let x ∈ Ker(g0). Then (g ◦ u)(x) = 0, and by exactness of the lower row we have u(x) ∈ Im(f), and as f is mono, there is a unique l ∈ L : f(l) = u(x), and so (u ◦ f 0)(l) = f(l) = u(x), hence u(f 0(l)) = u(x) and as f 0 is mono, we know that f 0(l) = x. So Ker(g0) ⊂ Im(f 0) and thus the lower row is exact. Now since f is irreducible, either f 0 is a section or u is a retraction. In the first case we know by 28 proposition 2.27 that g0 is a retraction and our desired morphism is u ◦ g¯0 (g¯0 denoting the retraction morphism of g0), and in the second case we have g0 ◦ u¯ (u¯ denoting the retraction morphism of u.) " ⇐= ": Since the given sequence is not split, f is not a retraction nor a section (thus satisfying one of the two defining properties of an irreducible morphism). Now suppose that there is a module X and morphisms f2 ∈ Hom(L, X), f1 ∈ Hom(X,M) such that f = f1 ◦ f2. Because f is mono, so is f2 and we obtain the following commutative diagram with exact rows (note: rightmost vertical morphism is obtained by the universal property of cokernels):

f2 π 0 L X Coker(f2) 0

idL f1 v f g 0 L M N 0 Now we can show that f satisfies the second property defining an irreducible morphism, that is, either f2 is a section, or f1 is a retraction. For if there is v1 ∈ Hom(Coker(f2),M) such that v = g ◦ v1,

then v ◦ idCoker(f2 = g ◦ v1 is a commutative square containing M → N ← Coker(f2), and thus

by pullback property of X we have γ ∈ Hom(Coker(f2),X) such that π ◦ γ = idCoker(f2). In other words π is a retraction, and thus f2 is a section. Else if there is v2 ∈ Hom(M, Coker(f2)) such that g = v ◦ v2, then v ◦ v2 = g ◦ idM is the commutative square that by pullback property of X induces a morphism ω ∈ Hom(M,X) such that idM = f1 ◦ ω and we obtain our result immediately as then f1 is a retraction. (2) Analogous to 1)  This leads us to first connection between the irreducible morphisms and the almost split sequences: Proposition 4.17. (1) If f : L → M is an irreducible monomorphism, then Coker(f) is indecomposable. (2) If g : M → N is irreducible epimorphism, then Ker(g) is indecomposable. Proof. We aim to use the preceding lemma. Assume that Coker(f) is not indecomposable, Coker(f) = N1 ⊕N2 ,N1,N2 6= 0. We then have the following nonsplit (for if it were, f would be a section, contradicting its irreducibility) short exact sequence: f ρ 0 → L −→ M −→ Coker(f) → 0 and if we add the decomposition of Coker(f) to the picture, we obtain the following diagram:

N1

ι1 f ρ 0 L M Coker(f) 0

ι2

N2 Now we can view that diagram as union of two diagrams as the one in proposition 4.16, and thus for each i ∈ {1, 2} we either have a morphism hi ∈ Hom(M,Ni) such that ιi ◦hi = ρ or a morphism gi ∈ Hom(Ni,M) such that ιi = ρ ◦ gi. But in both cases the first alternative yields a contradiction, as ιi ◦ hi = ρ implies that ιi is epi. But ιi is also mono, and thus is an isomorphism, which contradicts the assumption that Ni 6= 0. Thus we have g1, g2 as described above. But then by coproduct property of Coker(f) we know that there is a morphism [g1, g2] : Coker(f) → M and then ρ ◦ [g1, g2] = idCoker(f), so ρ is a retraction and thus f is a section, contradicting its irreducibility. Thus Coker(f) is indecomposable. 2) can be proven by dualizing the proof above.  The following proposition establishes a further connection between the two central notions of this chapter. Proposition 4.18. 29 (1) Let f : L → M be left minimal almost split. Then f is irreducible. (2) Let g : M → N be right minimal almost split. Then g is irreducible. Proof. By definition f is not a section. By proposition 4.8, L is indecomposable. We also know that f is not an isomorphism, as then f would be both a section and a retraction. Assume f is a retraction. f Then the following short exact sequence splits: 0 → Ker(f) −→ι L −→ M → 0, and thus L = M ⊕ Ker(f), with M, Ker(f) 6= 0 (if Ker(f) was 0, f would be both epi and mono and thus iso). That contradicts L being indecomposable. So f is not a retraction. Now we move on to the second property: assume f = f1 ◦ f2 for some X ∈ mod-A, f1 ∈ Hom(X,M), f2 ∈ Hom(L, X). Assume f2 is not a section (if it is, the proof ends here). Then as f is left almost split, there is g ∈ Hom(M,X) such that g ◦ f = f2, so f = f1 ◦ f2 = (f1 ◦ g) ◦ f, and now by minimality of f we know that f1 ◦ g must be an automorphism. Call −1 −1 −1 it h. Then f1 ◦ (g ◦ h ) = (f1 ◦ g) ◦ h = h ◦ h = idM , so f1 is a retraction. And thus either f2 is a section or f1 is a retraction, and so f is irreducible. Proof of (2) is similar.  Moreover, the following weak converse of proposition 4.18 holds: Proposition 4.19. (1) Let f ∈ Hom(L, M) be left minimal almost split. Then, given f 0 : L → M 0, f 0 is irreducible if and only if M 0 6= 0, and there is a module M 00 such that M ' M 0 ⊕ M 00 together with morphism 00 00 h f 0 i f : L → M such that f 00 is left minimal almost split. (2) Let g ∈ Hom(M,N) be right minimal almost split. Then, given g0 : M 0 → N, g0 is irreducible if and only if M 0 6= 0, and there is a module M 00 such that M ' M 0 ⊕ M 00 together with morphism g00 : M 00 → N such that [ g0 g00 ] is right minimal almost split.

Proof. See IV.1.10 in [1].  Proposition 4.20. Let f g 0 L M N 0 u v w f g 0 L M N 0 be a commutative diagram with exact non-split rows.

(1) If L is indecomposable and w is an automorphism, then u and hence v are automorphisms. (2) If N is indecomposable and u is an automorphism, then w and hence v are automorphisms.

f g Theorem 4.21 (IV.1.13, [1]). For a short exact sequence 0 → L −→ M −→ N → 0 in mod-A, the following statements are equivalent: (1) The given sequence is almost split. (2) L is indecomposable, and g is right almost split. (3) N is indecomposable, and f is left almost split. (4) The homomorphism f is left minimal almost split. (5) The homomorphism g is right minimal almost split. (6) L and N are indecomposable, and f and g are irreducible. Proof. To outline the proof’s structure, we start by listing the implications that will be proved in order to establish the equivalence (in the order in which they appear in the proof): (1) implies (4); (1) implies (5); (1) implies (2); (1) implies (3); (1) implies (6); (5) implies (2); (4) implies (3); (2) implies (3); (3) implies (2); (2) and (3) imply (1); and finally (6) implies (2). First two implications hold trivially by the definition of almost split sequences. Second two implications are an immediate consequence of proposition 4.8, and (1) implies (6) as an immediate consequence of propositions 4.8 and 4.18. (5) implies (2): If g is right minimal almost split, then g is irreducible by proposition 4.18. Now since the sequence is exact, 30 we have L ' Ker(g) and we also know that g is an epimorphism and thus by proposition 4.17 we know that L is indecomposable, and the implication follows. (4) implies (3): Dual to the preceding implication. (2) implies (3): We need to show that N is indecomposable, f is not a section, and given module U and u ∈ Hom(L, U) such that u is not a section, there is some u0 ∈ Hom(M,U) such that u = u0 ◦ f. First two parts of the statement are obtained immediately: N is indecomposable by proposition 4.8, and f is not a section as g is right minimal almost split and thus is not a retraction. We prove the third part by contraposition. Thus, given u ∈ Hom(L, U) such that there is no u0 ∈ Hom(M,U) such that u = u0 ◦ f, we will show that u must be a section. Just as we did when proving proposition 4.17, we use the stability of epi- and monomorphisms under pullbacks and pushouts (lemma 4.13). Let V be the pushout of the triangle of U, L, and M. We then have the following commutative diagram with exact rows:

f g 0 L M N 0

u v idN 0 U h V k N 0

Note that both the rows are non-split: first row is not split as g is right almost split so g is not a retraction, and second row is not split, for if h was a section with its section morphism h¯, then h¯ ◦ v contradicts our assumption about u. So in particular, k is not a retraction. Now since g is right almost split, there is a morphism v¯ ∈ Hom(V,M) such that k = g ◦ v¯. We thus have g ◦ v¯ ◦ h = k ◦ h = 0 and thus by the kernel property of L, there is a unique morphism u¯ such that f ◦ u¯ =v ¯ ◦ h. Therefore the following commutative diagram with exact rows exists:

f g 0 L M N 0

u¯◦u v¯◦v idN f g 0 L M N 0

and since idN is an automorphism, by proposition 4.20 we know that so is u¯ ◦ u, hence u is a section. Dually, (3) implies (2). (2) and (3) imply (1): All that needs to be shown is minimality of both f and g. Take h ∈ End(M) such that h ◦ f = f. We want to show that h is an automorphism. We arrive at that conclusion immediately by applying proposition 4.20 on the following diagram:

f g 0 L M N 0

idL h idN f g 0 L M N 0

From the same diagram, by analogous reasoning, we obtain the same result for g. Thus both f and g are minimal. (6) implies (2): L is indecomposable by assumption, and thus we only need to prove the second part of (2), namely that g is right almost split. Since g is irreducible, it is not a retraction. All that remains to be proved is that given v : V → N, where v is not a retraction, there exists v0 : V → M such that g ◦ v0 = v. We thus let v be as above, and assume that V is indecomposable. We may do that, for if it is not such, we perform the procedure below on each of the direct summands of V and "glue together" the obtained morphisms by the product property of V to obtain the desired morphism. Since f is irreducible, by proposition 4.16, either there is a morphism v0 : V → M such that v = g ◦v0 and we’re done, or there is a morphism h : M → V such that g = v ◦ h. But then since g is irreducible, either v is a retraction or h is a section, and by assumption v is not a retraction. Thus h is a section, and as such it is also a monomorphism. But then we have the following split short exact sequence: 0 → M −→h V −→π Coker(h) → 0 and so V ' Coker(h) ⊕ M. But V is indecomposable, and thus Coker(h) = 0 and so h is epi. So h is both mono and epi, and thus is an −1 −1 isomorphism. Then we have g ◦ h = v and thus h is the desired morphism, and we’re done.  31 5. Auslander-Reiten Translation Revisited In this chapter we will continue establishing interesting properties of the Auslander-Reiten translation defined in the end of chapter 2. We will also continue using the language of finite dimensional algebras rather than that of quiver representations, but by now it should be clear that all results here have a direct correspondent in the realm of quiver representations. Throughout this chapter we will assume that A is a path algebra of a finite acyclic quiver. The reason is that, as we have shown earlier, all the representations of such quiver, and thus all modules over such an algebra, have a projective resolution and an injective resolution both of length one. The corresponding theory for algebras and quivers not having that property is a bit more advanced, although similar. First thing we will do is reformulating our definition of Auslander-Reiten translation by changing the order of the steps we took. Recall that our definition of τM for M ∈ RepK Q included four steps: (1) Take a minimal projective resolution of M:

p1 p0 0 → P1 −→ P0 −→ M → 0 (2) Apply the left exact contravariant functor Hom(−,A) (to simplify the notation we will from now on refer to that functor as (−)t, so it sends a representation M to M t and a morphism f to f t) on said resolution to obtain: t t t p0 t p1 t π t 0 → M −→ P0 −→ P1 −→ Coker(p1) → 0 (3) Apply the contravariant exact functor D on the sequence resulting from the preceding step and obtain ι νp1 νp0 0 → Ker(νp1) −→ νP1 −−→ νP0 −−→ νM → 0

(4) Obtain the Auslander-Reiten translate as τM := Ker(νp1) But it should be clear that this process commutes with that obtained by switching steps (3) and (4), that t is, we first "choose" Coker(p1). We call it the transpose of M, denoting it by Tr M. Now apply D to obtain τM = D(Tr M). In accordance with our earlier definition of the Nakayama functor ν = D Hom(−,A), for −1 op −1 −1 which we’ve defined an inverse ν = HomAop (−,A )D = Hom(DA, −), we define τ as τ = Tr D The benefit of that change of perspective is that now it will be easier to describe properties of τ using properties of Tr and the fact that D is an exact functor (moreover, an equivalence).

Proposition 5.1 (IV.2.1, [1]). Let M be an indecomposable A-module. (1) The left A-module Tr M has no nonzero projective direct summands. t t t p0 t p1 t (2) If M is not projective, then the sequence 0 → M −→ P0 −→ P1 → Tr M → 0 obtained by applying Hom(−,A) on the minimal projective resolution of M is a minimal projective resolution of Tr M. (3) M is projective if and only if Tr M is zero. If M is not projective, then Tr M is indecomposable and Tr2 M ' M (4) If M and N are indecomposable and not projective, then M ' N if and only if Tr M ' Tr N

Proof. First we show that Tr M = 0 if and only if M is projective. If M is projective, then clearly P1 = 0 and t t t t P1 = 0, and thus p1 is epi, and so Tr M = Coker(p1) = 0. And if Tr M = 0, then p1 is epi, and the sequence t splits (as P1 is projective), so p1 is a retraction. Now since Hom(−,A) is a contravariant equivalence when restricted to projective modules, we know that p1 is a section, and thus the sequence splits and so M is isomorphic to a direct summand of P0 and thus is projective. Now we prove (2): since M is projective indecomposable, we know that Hom(M,A) is zero (by remark 3.35). Thus the first two objects of our sequence are 0, hence we can ignore the first one and obtain

t t p1 t 0 → P0 −→ P1 → Tr M → 0 This clearly is a projective resolution of Tr M. Now we show that it is minimal. Assume it is not. Then (as a t 0 00 t 0 00 00 ∼ 00 corollary of theorem I.5.8 in [1]) we have P1 ' E1 ⊕E1 and P0 ' E0 ⊕E0 with an isomorphism v : E0 −→ E1 t u 0 t such that p1 = [ 0 v ]. Note that as a consequence of (1) and proposition 3.35 we have (Tr M) = 0. Now 32 we apply Hom(−,A) to the sequence above, and use the fact that it is an equivalence when restricted to projective modules to obtain

tt t 0 t 00t p1 0 t 00t 0 → (Tr M) → E1 ⊕ E1 −−→ E0 ⊕ E0 → Coker(p1) which is clearly equal (ignoring the superfluous zero term in the beginning) to

0 t 00t p1 0 t 00t 0 → E1 ⊕ E1 −→ E0 ⊕ E0 → Coker(p1) and that in turn is isomorphic to p1 p0 0 → P1 −→ P0 −→ M 00t 00t But then it is clear that this resolution is not minimal, as we can eliminate E1 and E0 from their respective sums, since they together with the isomorphism vt between them have no contribution to the cokernel of the morphism, since

h t i h t i Coker ([ u 0 ])t) = Coker u 0 = E0 t ⊕ E00t/ Im u 0 = (E0 t/ Im(ut)) ⊕ (E00t/ Im(vt) ' E0 t/ Im(ut) 0 v 0 vt 1 1 0 vt 1 1 1 as vt is an isomorphism. Thus we conclude that our initial minimal resolution is not minimal, which is obviously a contradiction. Now to prove (3) we use (2) by observing that there is a commutative diagram with exact rows and isomor- phisms as vertical columns:

p1 p0 0 P1 P0 M 0

ϕ ψ tt tt tt p1 tt p0 2 0 P1 P0 Tr M 0 And we can now define an isomorphism γ using a standard diagram chasing argument: given m ∈ M, let 0 −1 tt p, p ∈ p0 (m). We want to define γ(m) as p0 ◦ ψ(p) but in order to have that well-defined, we need to show tt tt 0 0 0 p0 ◦ ψ(p) = p0 ◦ ψ(p ). Note that p − p ∈ Ker(p0) and thus p − p ∈ Im(p1), and since p1 is mono, there is −1 0 a unique q ∈ p1 (p − p ). By exactness and commutativity of the diagram we have tt tt tt tt 0 p0 ◦ p1 ◦ ϕ(q) = 0 = p0 ◦ ψ ◦ p1(q) = p0 ◦ ψ(p − p ) and the result follows. Now since the diagram is symmetric, we can perform the same procedure to obtain the inverse of γ, and thus γ is an isomorphism. Now we show that Tr M is indecomposable. We achieve that by showing that if M is decomposable, then so is Tr M. Thus let M = M 0 ⊕ M 00. By horseshoe lemma we obtain the minimal projective resolution of M by taking the ’direct sum’ (letting the modules in resolution being the direct sums of those in corresponding resolutions and letting the morphisms be diagonal) of the minimal projective resolutions of M 0 and M 00. Applying Hom(−,A) on that resolution preserves the "direct sum" structure of the resolution described t 0 00 t 0 00 t above, since Hom commutes with direct sums. But now since p1 :(P0 ⊕ P0 ) → (P1 ⊕ P1 ) is diagonal, we know that its cokernel can be decomposed into a direct sum of the cokernels of its "diagonal entries". Thus Tr M is indecomposable. Now we apply the statement we’ve proven to our situation. We know that M is indecomposable. But M ' Tr2 M, and so Tr2 M = Tr(Tr M) is indecomposable. And thus Tr M too is indecomposable. Finally we move on to (4). Clearly M ' N implies Tr M ' Tr N. But then also Tr M ' Tr N implies Tr(Tr M) ' Tr(Tr N) and thus M ' N  As mentioned earlier, the statements about Tr readily yield properties of τ and τ −1: Corollary 5.2. Let M,N be indecomposable A-modules. (1) The module τM is zero if and only if M is projective. (2) If M is a nonprojective module, then τM is indecomposable noninjective and τ −1τM ' M. (3) If M and N are nonprojective, then M ' N if and only if there is an isomorphism τM ' τN. Dual versions of each of those three statements hold for τ −1 (those are obtained by swapping "injective" with "projective" and τ with τ −1), and are also proven by dualizing the proofs below. Proof. 33 (1) Since D is an equivalence, D(M) = 0 implies M = 0. And since τM = D Tr M, we know that τM is zero if and only if Tr M is zero, which it is if and only if M is projective by the preceding proposition. (2) D is an equivalence, so τM is indecomposable if and only if Tr M is such. But we know that if M is indecomposable, then so is Tr M. Moreover, τM = D Tr M is noninjective as by proposition 3.31 we know that D Tr M is injective if and only if Tr M is projective, which it can not be, as then Tr2 M would be zero, and by proposition 5.1 we know that Tr2 M ' M. Finally we have τ −1τM = (Tr D)(D Tr)M = σD2 Tr M = Tr2 M ' M, as D is a duality. (3) Clearly M ' N implies τM ' τN. Conversely, τM ' τN is by definition equivalent to D Tr M ' D Tr N, and since D is an equivalence, this implies that Tr M ' Tr N and thus M ' N by proposition 5.1.  Proposition 5.3. Let M be a module over A. Then ν−1A = Hom(DA, τM) = 0 and dually νA = Hom(τ −1M,A) = 0. Proof. We only prove the first part of the statement. Note that just as ν is right exact, so ν−1 is left exact. So given an exact sequence f 0 → Ker(f) −→ι L −→ M in mod-A we have that −1 ν−1f 0 → ν−1 Ker(f) −−−→ν ι ν−1L −−−→ ν−1M is exact, and thus ν−1 Ker(f) ' Ker(ν−1f). We thus say that left exact functors commute with kernels. In our case we take the minimal projective resolution

p1 p0 0 → P1 −→ P0 −→ M → 0 −1 of M, apply ν to it and by definition have τM = Ker(νp1) so −1 −1 −1 −1 ν (τM) = ν (τM) = ν Ker(νp1) = Ker(ν νp1) = Ker(p1) −1 (as ν ν = id on the full subcategory of projective modules), which is equal to zero.  We have now seen that τ can yield new indecomposable modules from old ones, and we have introduced the important notion of irreducible morphisms. We finish this chapter with a presentation of some results that will shed some light on the role of τ in the structure of the category mod-A, also involving the irreducible morphisms. Regrettably, some of the results, most notably theorems 5.4 and 5.6 (we hope that their fundamental impor- tance for the theory presented in the remaining part of the text still is conveyed in the text), are presented without proof. These are provided fully (and also for a more general case than that treated here) in [1]. Theorem 5.4 (Auslander-Reiten Formulas, IV.2.14 in [1]). Let M and N be modules over A. There exists K-linear isomorphisms:

1 ExtA(M,N) ' D HomA(N, τM) 1 −1 ExtA(M,N) ' D HomA(τ N,M) And both the isomorphisms are functorial in both variables. Corollary 5.5. There also exist following K-linear isomorphisms: (1) For all modules M,N we have: −1 HomA(N, τM) ' HomA(τ N,M) (2) If N is indecomposable nonprojective, we have:

HomA(τN, τM) ' HomA(N,M) (3) If M is indecomposable noninjective, we have: −1 −1 HomA(τ N, τ M) ' HomA(N,M) 34 Proof. (1) is a direct consequence of the preceding theorem, since D is an equivalence. To prove (2), recall that if N is indecomposable nonprojective, we have τ −1τN ' N. Thus Hom(N,M) ' Hom(τ −1(τN),M), and this in turn, by (1) is isomorphic to Hom(τN, τM). Similarly, if M is indecomposable noninjective, then ττ −1M ' M and Hom(N,M) ' Hom(N, τ(τ −1M)) and again by (1) that is isomorphic −1 −1 to Hom(τ N, τ M)  Theorem 5.6 (IV.3.1 in [1]). (1) For any indecomposable non-projective module M, there exists an almost split sequence 0 → τM → E → M → 0 (2) For any indecomposable non-injective module N, there exists an almost split sequence 0 → N → F → τ −1N → 0

To present some of the results below, we need to introduce two more concepts: Definition 5.7. Given a module M, we define its radical Rad M as the intersection of all the maximal submodules of M. Definition 5.8. Given a module M, we define its socle Soc M as the submodule of M generated by all simple submodules of M. A notion related to the radical of a module described above is that of radical of algebra, and we introduce it for the particular case of the algebra Hom(M,N) (where M and N are indecomposable A-modules, and the product operation is given by composition of morphisms.) Definition 5.9. Given two indecomposable A-modules M and N, we define Rad(M,N) as the set of morphisms from M to N that are not isomorphisms. Proposition 5.10. If M and N are indecomposable, the set Rad(M,N) forms a vector space, which clearly then is a subspace of Hom(M,N). It is equal to Hom(M,N) if M is not isomorphic to N, and if M is isomorphic to N, then Rad(M,N) is the Jacobson radical of the endomorphism algebra End(M) (since we can, without loss of generality, assume equality M = N). This is the case as the endomorphism algebra of an indecomposable module is local and the Jacobson radical of a local algebra consists exactly of non-unit elements in the algebra. A more thorough exposition of the topic can be found in Chapter I of [1]. There is an important characterization of irreducible morphisms. To formulate it, we need to give another definition: Definition 5.11. Given two indecomposable modules M and N, we define Rad2(M,N) as the subspace of Rad(M,N) formed by the set of morphisms Xn { gi ◦ fi | fi ∈ Rad(M,Xi), gi ∈ Rad(Xi,N) for some indecomposable modules Xi and some n ∈ }. i=1 N Now we state (unfortunately without proof) the aforementioned characterization of irreducible morphisms: Proposition 5.12 (IV.1.6 in [1]). Let M, N be indecomposable A-modules. A morphism f : M → N is irreducible if and only if it lies in Rad(M,N) \ Rad2(M,N) This means that further study of irreducible morphisms will, in a sense, give us information about all non-isomorphisms (which will turn out to be of particular interest) that is complete up to the information about Rad2(M,N). This motivates the following definition: Definition 5.13. Let M and N be indecomposable A-modules. By Irr(M,N) we denote the quotient vector space Rad(M,N)/ Rad2(M,N). We call it (in a bit misleading fashion) the space of irreducible morphisms from M to N. We also state (also without proof) an important result about such spaces: 35 Proposition 5.14. Let M be an indecomposable module, and let 0 → τM → E → M → 0 be the almost-split sequence ending with M (which exists by 5.6, and moreover exists uniquely, due to 4.10). Then the dimensions of spaces Irr(τM, E) and Irr(E,M) are equal. Dually, the same holds for the almost-split sequence beginning with M.

Proof. A direct consequence of proposition IV.4.1 in [1]  Definition 5.15. The Auslander-Reiten quiver Γ(mod-A) of the algebra A is defined as follows: (1) The points of Γ(mod-A) are the isomorphism classes of indecomposable A-modules. (2) The arrows M → N represent basis vectors of space Irr(M,N) of irreducible morphisms from M to N. (which in particular can be identified with some set of irreducible morphisms such that the equivalence classes of those morphisms are a basis) Now we present some basic properties of the Auslander-Reiten quiver of A. Proposition 5.16. The set of immediate predecessors of an indecomposable module M (or rather its iso- class) in the Auslander-Reiten quiver Γ(mod-A) consists of (1) The indecomposable direct summands of Rad M if M is projective (2) The indecomposable direct summands of the middle term of the almost split exact sequence ending with M, if M is non-projective.

Proposition 5.17. The set of immediate successors of an indecomposable module M in the Auslander-Reiten quiver Γ(mod-A) consists of (1) The indecomposable direct summands of M/ Soc M if M is injective (2) The indecomposable direct summands of the middle term of the almost split exact sequence starting with M, if M is non-injective.

The preceding two propositions are a consequence of combining multiple statements proven in chapters IV.2, IV.3 and IV.4 in [1]. Since M, Rad M and Soc M all are finite-dimensional, a consequence of those propositions is that Γ(mod-A) is a locally finite quiver, that is, for any point in Γ(mod-A), both the set of its direct predecessors and the set of its direct successors are finite, or, equivalently, the set of all its neighbors (the union of the two aforementioned sets) is finite. Remark 5.18. Γ(mod-A) of A has no loops. Proof. A loop X → X in Γ(mod-A) would correspond to an irreducible morphism ϕ : X → X. By proposition 4.12 we know that ϕ must either be mono or be epi, but since ϕ is in particular an endomorphism of a finite dimensional vector space (the underlying space of the indecomposable module X), it is mono if and only if it is epi, hence it must be both, and hence also an isomorphism. But then ϕ is both a section and a retraction, both contradicting ϕ being irreducible.  Proposition 5.19. If A is representation-finite (that is, there are only finitely many isoclasses of indecom- posable A-modules), then Γ(mod-A) has no multiple arrows. Theorem 5.20. Let Γ0 denote the full subquiver of Γ(mod-A) corresponding to the full subcategory of pro- jective modules, and let Γ00 denote the full subquiver of Γ(mod-A) corresponding to the full subcategory of injective modules. The Auslander-Reiten translation τ is a translation of the Auslander-Reiten quiver Γ(mod-A), i.e. the map 0 00 0 τ : Γ(mod-A)0 \ Γ0 → Γ(mod-A)0 \ Γ0 is a bijection such that for any vertex x ∈ Γ(mod-A)0 \ Γ0 there is also a bijection between the set of arrows having x as their target and the set of arrows having τx as their source. Proof. By theorem 5.6, for any non-projective indecomposable module M there is an almost split sequence 0 → τM → E → M → 0, and by remark 4.10 we know that that sequence is unique up to isomorphism. By corollary 5.2 we know that τM is indecomposable and non-injective. By existence of τ −1 together with 36 −1 00 0 τ τM ' M we know that τ is surjective onto Γ(mod-A)0 \ Γ0 and since τM ' τM holds if and only if M ' M, we know that it is also injective. Thus τ is a bijection of said sets. The bijection of sets of arrows follows from proposition 5.14.  Now we give an example of an Auslander-Reiten quiver: Example 5.21. Let Q = 1 → 2 → 3 and let A be its path algebra. From earlier we know all the indecomposable projective, indecomposable injective and simple modules. We use the notation we’ve used earlier, so for example by P (2) we mean the indecomposable projective representation at vertex 2, and we let it be represented by the following canonical representative: 0 → K −−→idK K It is fairly clear that Rad P (1) = P (2), Rad P (2) = P (3) and that Soc I(3) = P (3), Soc I(2) = S(2), and hence also I(3)/ Soc I(3) ' I(2),I(2)/ Soc I(2) ' I(1). Now, using propositions 5.4, 5.16, 5.17, we obtain a full subquiver of Γ(mod-A): P (1)

P (2) I(2)

P (3) I(1)

And moreover, since the radicals and quotients described above are indecomposable, this quiver contains all predecessors of P (3), P (2), P (1), and I(1). Similarly, it also contains all successors of P (3), P (1) (which is equal to I(3)) and I(2). As a consequence of proposition IV.3.11 in [1], the following sequence is almost split: 0 → P (2) → P (1) ⊕ S(2) → I(2) Hence, by 5.17 P (1) and S(2) are the only successors of P (2), and dually, by 5.16, these modules are the only predecessors of I(2). Now we will show the following: (1) P (2) is the only predecessor of S(2) (2) I(2) is the only successor of S(2) To prove (1), we note that the following sequence: 0 → P (3) −→i P (2) → S(2) → 0 (where i denotes the inclusion morphism) is the standard projective resolution of S(2) given in 3.19, and moreover, since all the modules in the sequence are indecomposable, it is even the minimal projective resolution of S(2), hence also our starting point in finding τS(2). We start by applying Hom(−,A) and, using 3.35 and 3.36, we obtain Hom(i,A) 0 → PQop (2) −−−−−−→ PQop (3) → Coker(Hom(i, A))

where Hom(i, A) is mono, and hence Coker(Hom(i, A)) ' IQop (3). We now apply D on that cokernel to obtain (using 3.31) τS(2) ' P (3). Now, by 5.6, there is an almost split sequence starting with P (3) and ending with S(2), which is also the unique almost split sequence ending with S(2), by 4.10. But the left morphism in that sequence must be irreducible, and so by our earlier result about the successors of P (3), we know that the middle term of the sequence is P (2), hence being the only predecessor of S(2). (2) is proven in exact analogy by using the fact that 0 → S(2) → I(2) → I(1) is the minimal injective resolution of S(2). We need to make one more remark: since τ gives us a way to go "left" in our quiver, and the only way to terminate that procedure is to have τM = 0 for some module, we know that each connected component of 37 Γ(mod-A) must contain some projective indecomposable modules, since these are the only ones yielding 0 after τ was applied. And since all of the indecomposable projective representations of the quiver appeared in the full subquiver we gave above, we know that Γ(mod-A) is connected. In fact, Γ(mod-A) is the following quiver: P (1)

P (2) I(2)

P (3) S(2) I(1)

References [1] Ibrahim Assem, Daniel Simson, and Andrzej Skowroński. Elements of the representation theory of associative algebras. Vol. 1, volume 65 of London Mathematical Society Student Texts. Cambridge University Press, Cambridge, 2006. Techniques of representation theory. [2] Maurice Auslander, Idun Reiten, and Sverre O. Smalø. Representation theory of Artin algebras, volume 36 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1997. Corrected reprint of the 1995 original. [3] Peter Gabriel. Unzerlegbare Darstellungen. I. Manuscripta Math., 6:71–103; correction, ibid. 6 (1972), 309, 1972. [4] P. J. Hilton and U. Stammbach. A course in homological algebra, volume 4 of Graduate Texts in Mathematics. Springer- Verlag, New York, second edition, 1997. [5] Masaki Kashiwara and Pierre Schapira. Categories and sheaves, volume 332 of Grundlehren der Mathematischen Wis- senschaften [Fundamental Principles of Mathematical Sciences]. Springer-Verlag, Berlin, 2006. [6] Joseph J. Rotman. An introduction to homological algebra. Universitext. Springer, New York, second edition, 2009. [7] Ralf Schiffler. Quiver representations. CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC. Springer, Cham, 2014.

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