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Home , Abelian category, Dual (category theory), Homological algebra, Kernel (category theory), Snake lemma

Homological in Abelian Categories

Rylee Lyman March 3, 2019

1 Preliminaries

Recall that a is a collection of objects with arrows between them. We use the word “collection” advisedly, making no claim that we are dealing with sets. We require each object A to have an associated arrow

A 1A A called the identity. We also require composition of arrows to be defined whenever it makes sense, and we require this composition to be associative. Furthermore, given f : A → B, f1A = f = 1Bf We follow the treatment in [Mac71] to which the curious reader is referred. In , a dual statement is one in which all the arrows have been reversed. For example, an arrow m is monic if it is “right-cancellative,” i.e.

mf = mg =⇒ f = g

The dual notion is an epi. An is an arrow that has a two-sided inverse.

Exercise 1. are both monic and epi The converse does not always hold. A standard example is the inclusion Z → Q in the . An object z is initial if, given another object t, there is a unique arrow z A. Dually, an object is terminal if there is a unique arrow to the object from any given object. A zero object is one that is both initial and terminal. Exercise 2. If z and w are both initial, terminal, or zero objects, there exists a unique isomorphism z → w.

We will denote our zero objects as 0, as well as any arrow that factors through the zero object.

1 A for a f : A → B is an object ker f together with an arrow m: ker f → A (sometimes also denoted ker f) such that the following diagram commutes. ker f m A

0 f B We say a diagram commutes when following any directed path in the diagram yields equal arrows, so the content of the statement here is just that fm = 0 as arrows ker f → B. Actually that’s not all that’s required. A kernel must also be universal for this property. That is, for any other choice of object and arrow that makes the diagram commute, there is a unique arrow (indicated as dashed in the figure below) that makes the diagram commute.

K

ker f m A 0 f 0 B The dual notion is a . The following are good practice, and facts we need. Exercise 3. Kernels and , when they exist, are unique up to unique isomorphism. Moreover, kernels are monic and cokernels are epi. Exercise 4. If m: A → B is monic in a category with a zero object, 0 → A is a kernel for m. Dually, if e: A → B is epi in a category with a zero object, B → 0 is a cokernel for e.

Exercise 5. In a category with a zero object, 1B : B → B is a cokernel for 0: A → B. Therefore any cokernel of a zero arrow is an isomorphism. Finally, a (binary) product of two objects A and B is a third object A × B and π1 : A × B → A, π2 : A × B → B, that satisfies the that given maps f : C → A and g : C → B, there exists a unique map f × g : C → A × B that makes the following diagram commute.

C f g f×g A A × B B π1 π2 The dual notion is a , A q B, for which the diagram is

A ι1 A q B ι2 B fqg f g C

2 Exercise 6. Products and , when they exist, are unique up to unique isomorphism.

2 Abelian Categories

Definition 1. An is a category which 1. has a zero object 2. every pair of objects has a product and a coproduct and the unique map [(1A × 0) q (0 × 1B)]: A q B → A × B is an isomorphism. 3. every arrow has a kernel and a cokernel 4. every monic arrow is a kernel and every epi arrow is a cokernel. Because the dual of each statement is itself, proving a statement about abelian categories proves the dual statement as well. Rather than binary co- products and products, we will think instead of an object A ⊕ B that is both a product and a coproduct.

Exercise 7. Given A⊕B in an abelian category, use [(1A×0)q(0×1B)] = 1A⊕B to show that π1ι1 = 1A, π2ι2 = 1B, π1ι2 = 0 and π2ι1 = 0. Conclude that πi and ιi are epi and monic, respectively. Show also that [(1A q 0) × (0 q 1A)] = 1A⊕B.

Exercise 8. The unique map ∆ = (1A × 1A) described below is a monomor- phism. A 1 1 A ∆ A A A ⊕ A A π1 π2 We would like to prove the following Theorem 1. Given arrows f : A → B, g : A → B in an abelian category, there is a third arrow f +g : A → B. This new arrow is uniquely determined by f and g as (f + g) = (f q g)∆, and gives the collection of arrows A → B the structure of an abelian .

Before we can, we need some preliminaries. Exercise 9. In an abelian category, a monic arrow is the kernel of its cokernel. Dually, an epi is the cokernel of its kernel. Proposition 1. In an abelian category, an arrow that is both monic and epi is an isomorphism. (Proposition 1.5.1 in [Bor94].) Proof. In view of the previous exercises, note that if f : A → B is both monic and epi, then 0 → A is a kernel of f, and since f is epi, f = coker(ker f) = coker 0, which is an isomorphism.

3 Exercise 10. Show that ι1π1 + ι2π2 = 1A⊕B, and that of arrows distributes over composition, i.e. that (f + g)h = (fh) + (gh), similarly k(f + g) = (kf) + (kg). Use the above to show that f q g = fπ1 + gπ2, f × g = ι1f + ι2g and thus in particular ∆ = ι1 + ι2. Use symmetry of the diagram for ∆ to show addition of arrows is commutative. Exercise 11. Show that (A ⊕ B) ⊕ C and A ⊕ (B ⊕ C) are isomorphic, and thus that addition of arrows is associative. Proposition 2. 0: A → B is the identity for addition of arrows.

Proof. We note that the following diagram commutes because π1ι2 = 0 and π1ι1 = 1 A ι1 A ⊕ A ι2 A

fπ1 f 0 B and conclude that fπ1 = f q 0. Now observe that fπ1∆ = fπ1(1A × 1A) = f1A = f. So far we have constructed a commutative structure on arrows A → B, and to do it we have only needed and zero objects. It is the existence of inverses for which we will need the extra hypotheses.

Proposition 3. In an abelian category with a A ⊕ B, ι1 = ker π2, π2 = coker ι1, and similarly for pairs with indices swapped.

Proof. Suppose we have f : C → A⊕B with π2f = 0. Now, f = (ι1π1+ι2π2)f = ι1π1f +ι20 = ι1π1f. since ι1 is monic, the factorization (π1f) is unique, proving that ι1 is a kernel for π2. Now if g : A⊕B → C is such that fι1 = 0, f = f(ι1π1+ι2π2) = 0π1+fι2π2 = fι2π2, and since π2 is epic, this factorization is unique. Proposition 4. In an abelian category, every pair of arrows A → B has an equalizer and a . Moreover, the following are equivalent: 1. m is monic 2. ker m = 0 3. mf = 0 =⇒ f = 0. Indeed, dually if fe = 0 =⇒ f = 0, then e is epi. (Propositions 1.5.3 and 1.5.4 in [Bor94].) Proof. by an equalizer for f, g : A → B we mean an arrow m: M → A such that fm = gm, such that for any other arrow c: C → A with fc = gc, there is a unique arrow (dashed), which makes the diagram commute.

f M m A B g c C

4 As usual, a coequalizer is the dual notion, and when equalizers or exist they are unique up to unique isomorphism. So given f, g : A → B, consider the map f × g : A → B ⊕ B together with ∆: B → B ⊕ B, and write F = coker ∆(f × g). Since ∆ is monic, ∆ = ker(coker ∆), so we get a map e: ker F → B such that the diagram commutes.

ker F e B

ker F ∆ f×g A B ⊕ B

F coker ∆ coker ∆

Indeed this is a universal filling of the square (a pullback), for if we have another object and pair of arrows A c C d B such (f × g)c = ∆d, then F c = coker ∆(f × g)c = (coker ∆)∆d = 0, so c factors through ker f and one checks that the remaining upper triangle commutes.

d C

ker F e B c ker F ∆ f×g A B ⊕ B

F coker ∆ coker ∆

We claim that the map ker F is the desired equalizer. Indeed, f ker F = π1(f ×g) ker F = π1∆e = e, and similarly g ker F = π2(f ×g) ker F = π2∆e = e. If c is another map with fc = gc, then (fc) × (gc) = (f × g)c factors as ∆fc, so F c = coker ∆(f × g)c = (coker ∆)∆fc = 0, so c factors through ker F , as desired. By , we have also constructed coequalizers. Note that our proof actually shows that equalizers are kernels, and thus monic, so coequalizers will be cokernels, and thus epi. Certainly if m is monic, then ker m = 0, and mf = 0 =⇒ f = 0. Moreover, if mf = 0 =⇒ f = 0, m ker m = 0 tells us that ker m = 0. So supposing ker m = 0, we would like to show m is monic. Take parallel arrows f and g with mf = mg, and let e be a coequalizer for f and g, so that we have the following .

f · · e · g m ·

5 Notice that e, being epi, is the cokernel of ker e, but since m ker e factors through e, m ker e = 0, so ker e = 0, since it then factors through ker m = 0. Therefore e, as a cokernel of a zero arrow, must be an isomorphism, and we have f = e−1ef = e−1eg = g, showing m is monic. Now we begin constructing an inverse.

Proposition 5. In an abelian category, coker ∆ι1 : A → A is an isomorphism. (Lemma 1.6.1 in [Bor94].) Proof. Naturally, we prove that it is monic and epi. To show that it is monic, suppose coker ∆ι1f = 0. Since ∆ = ker(coker ∆), we conclude ι1f = ∆g for some g. Notice that g = π2∆g = π2ι1f = 0f = 0, so ∆0 = 0 = ι1f. But ι1 is monic, so we conclude f = 0, and thus that coker ∆ι1 is monic. Suppose now that f coker ∆ι1 = 0. Since π2 is a cokernel for ι1, we conclude that f coker ∆ = gπ2. Notice that g = gπ2∆ = f(coker ∆)∆ = f0 = 0, so f coker ∆ = 0π2 = 0. But since coker ∆ is epi, we conclude f = 0, and thus that coker ∆ι1 is epi.

Now, empowered with the knowledge that coker ∆ι1 is an isomorphism, let r be an inverse for it, so that given f : A → B, f = fr coker ∆ι1, and define −f = fr coker ∆ι2.

Proposition 6. In an abelian category, with the notation −f = fr coker ∆ι2 as above, f + (−f) = 0 as maps A → B. Proof. All we need to do is check the algebra: f + (−f) = (f q (−f))∆

= (fπ1 + (−f)π2)∆

= (fπ1 + fr coker ∆ι2π2)∆

= (fr coker ∆ι1π1 + fr coker ∆ι2π2)∆

= fr coker ∆(ι1π1 + ι2π2)∆ = fr(coker ∆)∆ = fr0 = 0. This concludes the proof of Theorem 1.

3 Diagram Chasing

Many first courses on present the proof of the Snake Lem- mma constructing the map and checking for exactness by using what it does to elements. As it turns out, this method can be extended to work in arbitrary abelian categories. Firstly, however, we need some more results about pullbacks. Proposition 7. In an abelian category, pullbacks and pushouts exist. Further- more, if one of the given arrows in a pullback square is epi, its opposite arrow is epi and the square is a pushout as well as a pullback. Likewise, pullbacks of monics are monic. (Proposition 1.7.6 of [Bor94])

6 f g Proof. We start with a diagram A C B. Write p = fπ1 − gπ2 : A ⊕ B → C, and consider ker p. Since 0 = p ker p = (fπ1 − gπ2) ker p = f(π1 ker p) − g(π2 ker p), the following diagram commutes (where we’ve written h = π1 ker p and k = π2 ker p)

ker p k B

h g A C f

0 0 So suppose now that we have A h D k B with fh0 = gk0, and con- 0 0 0 0 sider h × k = ι1h + ι2k . Notice that

0 0 0 0 0 0 p(h × k ) = (fπ1 − gπ2)(ι1h + ι2k ) = fh − gk = 0

0 0 0 0 0 So we conclude that h × k factors as (ker p)e, and we check h = π1(h × k ) = 0 π1(ker p)e = he, and similarly k = ke. Therefore the diagram above is indeed a pullback. Now suppose that f is epi, we’ll first show that p is too. Indeed, suppose we have x with xp = 0. Then xf = xpι1 = 0, so since f is epi, we conclude x = 0, and conclude p is epi (so in particular p = coker(ker p)). Now we would like to show that k = π2(ker p) is epi, so choose y with 0 = yk = yπ2 ker p, so we get a factorization through the cokernel of ker p, yπ2 = zp. Note that zf = zpι1 = yπ2ι1 = 0, so we conclude that z = 0, since f is epi. On the other hang yπ2 = zp = 0, and π2 is also epi, so y = 0, proving that k is epi. Indeed, we are prepared to show that the square is a pushout. Suppose we 0 0 f g 0 0 0 0 have B E C with f h = f π1 ker p = g π2 ker p = g k. Therefore 0 0 0 0 we conclude (f π1 − g π2) ker p = 0, and thus f π1 − g π2 factors as ep, and we see that 0 0 0 ef = epι1 = (f π1 − gπ2)ι1 = f π1ι1 = f and similarly

0 0 0 0 eg = −epι2 = −(f π1 − g π2)ι2 = −(−g )π2ι2 = g which shows the square is also a pushout. Before we go further, let us also prove VIII.4.2 in [Mac71]: Proposition 8. The kernel of the bottom and top arrows in a pullback diagram are isomorphic as objects. Proof. First, we notice that p ker p = g0 = 0, so by universality of the pullback,

7 we get a map k that makes the diagram commute.

ker p 0 k q · · ker p f g p · ·

Notice that fk = ker p implies that k is monic, for if kx = ky, then fkx = (ker p)x = (ker p)y = fky, which implies x = y. Suppose we have x with qx = 0. Then gqx = pfx = 0, so we have y with (ker p)y = fx. Notice that fx = (ker p)y = fky and qky = 0 = qx by assumption, so by universality of the pullback, we conclude x = ky, proving that k is a kernel of q, and thus that ker q and ker p are isomorphic as objects. Armed with this knowledge, we can now define an equivalence relation on the set of arrows with common . MacLane [Mac71] in VIII.4 calls these “members,” while Borceux [Bor94] in 1.9 calls them “pseudo-elements.” We will just write “elements,” but follow MacLane’s notation. Two arrows x and y with common codomain A will be called equivalent or equal as elements, written as x ≡ y if there exists a commutative diagram as below with p and q epi. q · ·

p y

· x A

We write x ∈e A for “x is an arrow with codomain A.” This relation is manifestly reflexive and symmetric, and using that the pullback of an epi is an epi, one shows Exercise 12. If x ≡ y and y ≡ h, then x ≡ h. Moreover if we have f : A → B, then fx ≡ fy.

Exercise 13. All zero arrows 0 ∈e A are equivalent. If a ≡ 0, then a = 0. What’s more, we have Theorem 2. ([Mac71] VIII.4.3, [Bor94] 1.9.4)

1. f : A → B is monic if an only if for all x ∈e A, fx ≡ 0 implies x ≡ 0. 2. f : A → B is monic if and only if for all x, y ∈e A, fx ≡ fy implies x ≡ y. 3. f : A → B is epi if and only if for all y ∈e B, there exists x ∈e A with fx ≡ y. 4. f : A → B is a zero arrow if and only if for all x ∈e A, fx ≡ 0. f g 5. A B C is exact if and only if gf = 0 and for every y ∈e B with gy ≡ 0, there exists x ∈e A with fx ≡ y.

8 6. Given f : A → B and x, y ∈e A with fx ≡ fy, there is z ∈e A with fz ≡ 0 and the property that for any g : A → C with gx ≡ 0 has gy ≡ gz and for any h: A → D with hy ≡ 0 has hx ≡ −hz. Proof. We leave the forward directions of each of the if and only if statements as exercises to the reader. We prove the reverse implications.

1. Take ker f ∈e A. Then f(ker f) ≡ 0 tells us ker f = 0, so f is monic. 2. Similarly, take x = ker f and y = 0. 3. Consider 1B ∈e B. By assumption there is some x ∈e A with fx ≡ 1B, i.e. there are p and q epi with fxp = q. If zf = 0, then certainly zfxp = zq = 0, so z = 0 and we conclude f is epi. 4. Consider 1A ∈e A. f1A = f ≡ 0 tells us exactly that f is a zero arrow. 5. Here we must remind the reader that by the of f : A → B, we mean the arrow ker(coker f) with codomain B. We have also the coker(ker f), and it turns out there is an isomorphism between the corre- sponding objects. Indeed, in abelian categories, every arrow f factors as f = (im f)(coim f), and in general any factorization as epi-followed-by- monic is naturally isomorphis to this factorization. A is exact when (as you expect), the image of one arrow is the kernel of the next. Because gf = 0, im f factors as im f = (ker g)m. Dually to the argument in 3. for this reason we conclude m is monic.

Now, take ker g ∈e B. By assumption, there is x ∈e A with fx ≡ ker g. That is, there are epis p and q with

(ker g)q = fxp = (im f)(coim f)xp = (ker g)m(coim f)xp

And since ker g is monic, we have m at the head of a composition equal to an epi arrow, so we conclude m is epi, and thus ker g and im f are isomorphic, proving exactness. 6. The negative sign suggests that we want to use subtraction of arrows. Indeed, fx ≡ fy tells us we have epis p and q with fxp = fyq and we take z = yq − xp. Clearly fz = fyq − fxp = 0. Now, if we have gx ≡ 0, then gxp = 0, so gz = gyq ≡ gy. Similarly if hy ≡ 0, then hyq = 0, so hz = −hxp ≡ −hx. Borceux notes that this does not give us an structure on the collection of elements of A, since minus the identity is an isomorphism, we always have x ≡ −x = x(−1), but in general we do not have x + x ≡ 0. To give the reader an idea of the flavor, we paraphrase MacLane’s proof in [Mac71] VIII.4.5 of the .

j 0 A i B C 0 α β γ 0 D k E ` F 0

9 Theorem 3 (Snake Lemma). ([Mac71] VIII.4.5) Given a commutative diagram of short exact as above, there is another commutative diagram with exact rows (including the snake) and exact columns as below.

j 0 ker α i∗ ker β ∗ ker γ

ker α ker β ker γ j 0 A i B C 0 α β δ γ

0 D k E ` F 0

coker α coker β coker γ coker α k∗ coker β `∗ coker γ 0

Proof. Commutativity of the squares gives βi ker α = kα ker α = 0, so we can factor ker α = (ker β)i∗. As in the proof of the above proposition, we see immediately that i∗ is monic. Dually, `∗ exists and is epi. Existence of j∗ and k∗ follows similarly. We note that (ker γ)j∗i∗ = ji ker α = 0, so since ker γ is monic, j∗i∗ is 0. A similar argument shows `∗k∗ = 0. Now, consider x ∈e ker β with j∗x ≡ 0. Notice that 0 = (ker γ)j∗x = j(ker β)x, so by exactness, there is y ∈e A with iy ≡ (ker β)x. But kαy = βiy ≡ β(ker β)x ≡ 0, so since k is monic, we conclude αy ≡ 0, and by exactness conclude there is z ∈e ker α with y ≡ (ker α)z, so

(ker β)x ≡ iy ≡ i(ker α)z = (ker β)i∗z and because ker β is monic, we conclude x ≡ i∗z, showing exactness at ker β. The proof at coker β is dual. Following MacLane, we show that δ arises “naturally,” (that is, by using universals) so we need not argue for its existence based on its action on elements. j ker γ First, we form the pullback of B C ker γ Which will give us a monic and an epi, so in particular the arrow with codomain ker γ will not be j∗. We also form the pushout of coker α coker α D k E. Since i = im i = ker j, we can use the proposition earlier to claim that A is the kernel of this pushout as well, and dually that F is its cokernel.

P e ker γ 0 ker e m ker γ j 0 A i B C 0 α β γ 0 D k E ` F 0

coker α e0 coker m0 0 0 coker α m P 0

10 Note that the column is not exact, and it is here that we will look for our snake. Indeed, by commutativity of the diagram, e0βm ker e = m0(coker α)α = 0, so e0βm factors through e = coker(ker e) as e0βm = ηe. On the other hand, (coker m0)ηe = (coker m0)e0βm = γ(ker γ)e = 0, so we also have a factorization ηe = m0δ, since m0 = ker(coker m0). Notice that we have indeed found a map δ : ker γ → coker α! Let us see what δ does to an element x ∈e ker γ. Since j is epi, there is an element y ∈e B with jy ≡ (ker γ)x Notice that since `βy = γjy ≡ γ(ker γ)x ≡ 0, exactness gives us z ∈e D with kz ≡ βy. We claim that (coker α)z ∈e coker α satisfies δx ≡ (coker α)z. We record this schematically below.

x ∈e ker γ

y (ker γ)x

z βy 0

(coker α)z

So, to prove that this is actually what δ does, consider m0δx. Indeed, since P is a pullback and by our above discussion jy ≡ (ker γ)x—meaning up to precomposing with epi arrows we have a commutative square—we get an element x0 ∈e P with ex0 ≡ x and mx0 ≡ y. So we conclude

0 0 0 0 0 0 m δx ≡ m δex0 ≡ e βmx0 ≡ e βy ≡ e kz = m (coker α)z and since m0 is monic, we conclude that δx ≡ coker αz as desired. Indeed, this actually shows that this description of δ is well-defined up to equality as elements of coker α, independent of all the choices we made in the description! We use this description of δ to prove exactness. Indeed, if we have t ∈e ker β, j∗(ker γ)t = j(ker β)t, in our diagram chase, we can choose y = (ker β)t but βy = 0 tells us we must choose z = 0, so δj∗t = 0 for any t ∈e ker β. On the other hand, if δx ≡ 0, in our description, our corresponding choice of z must have (coker α)z ≡ 0, so by exactness we can choose a ∈e A with αa ≡ z. Notice that βia = βαa ≡ kz ≡ βy, so take their difference (as in the final part of Theorem 2) b ∈e B. By definition, βb ≡ 0, so by exactness we can write b ≡ (ker β)t for some t ∈e ker β. This is exactly what we wanted! For (ker γ)j∗t ≡ jb, and since jia ≡ 0, jb = jy = (ker γ)x by definition! So we conclude (because ker γ is monic) that j∗t ≡ x, as required, showing exactness at ker γ. Now all that remains is to check exactness at coker α. Take x ∈e ker γ. k∗δx ≡ k∗(coker α)z = (coker β)kz ≡ (coker β)βy ≡ 0, so k∗δ = 0.

11 0 Now take t ∈e coker α. Notice that since k is monic, the pushout P is 0 0 0 also a pullback. Since e is epi, there is v ∈e E with e v ≡ m t, which, up to composing on the right with epis, gives us a filling of the pullback, so we get 0 z ∈e D with kw ≡ v and coker αz ≡ t. But since P is a pushout, there is 0 0 0 f : P → coker β with coker β = fe and k∗ = fm , so if we take t with k∗t ≡ 0, 0 0 (coker β)v = fe v ≡ fm t = k∗t ≡ 0, so by exactness, we can write v ≡ βy. Notice, γjy = `βy ≡ `v ≡ `kz ≡ 0, so there is x ∈ ker γ with ker γx ≡ jy. So we have found a candidate x ∈e ker γ, and since we know that even though our particular choices for y and z are quite convenient, we do have δx coker αz ≡ t, we are finished.

References

[Bor94] Francis Borceux. Handbook of categorical algebra. 2, volume 51 of En- cyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 1994. Categories and structures.

[Mac71] Saunders MacLane. Categories for the working mathematician. Springer-Verlag, New York-Berlin, 1971. Graduate Texts in Mathe- matics, Vol. 5.

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