4 Equivalent Categories

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4 Equivalent Categories.

Suppose that R and S are two rings. We want to determine what it would mean for them to have the same representation theories — or for these theories to be equivalent in some meaningful way. What we want should be pretty clear — the two rings have equivalent representation theories when their module (= representation) categories RMod and SMod are “the same”. But that just trades one problem for another — what does it mean for two categories “to be the same”? This is a bit dicey. We’d probably like to say that they are the same if they are isomorphic. But that’s too strong. For example, the category of all nite sets, an enormous sucker, has exactly the same categorical structure as the category of all nite subsets of N, but there is no possible way to build a bijective function between these two categories. As we shall see, natural transformations will come to the rescue, and we will be able to determine an appropriate notion of equivalent categories and hence of equivalent representation theories for rings.

Let C and D be two (additive) categories. A covariant (additive) functor F : C→Dis an equivalence in case there exists a covariant (additive) functor G : D→Csuch that GF and FG are naturally isomorphic, respectively, to the identity functors 1C and 1D. If such an equivalence exists, then we say that the categories are equivalent. This relation is symmetric. Indeed, the two categories are equivalent i there exist (additive) covariant functors

F : C→D and G : D→C and natural isomorphisms

: GF → 1C and : FG → 1D.

This means that assigns to each object C in C an isomorphism C : GF (C) → C in C and assigns to each object D in D an isomorphism D : FG(D) → D in D such that for all C-morphisms f : C → C0, and all D-morphisms g : D → D0, the diagrams GF (-f) 0 FG(-g) 0 GF (C) GF (C ) FG(D) FG(D )

0 0 C C D D ? ? ? ? f - 0 g - 0 C C D D commute. When this happens, we call F and G inverse equivalences for C and D.

When such inverse equivalences exist, it means that all “categorical” information available in one of the categories carries over unchanged to the other. In other words, using only categorical machinery we have no way to distinguish equivalent categories. So now let’s see how to characterize category equivalences. 31 32 Section 4

A covariant functor F : C→Dis faithful in case it is one-to-one on the class of morphisms of C. 0 0 0 The functor F is full in case for each pair C, C in C the map F : MorC(C, C ) → MorD(F (C),F(C )) is surjective. Finally, the functor F is dense in case for each object D in D there is an object C in C with F (C) = D. It turns out that these three properties characterize category equivalences.

4.1. Theorem. Let C and D be (additive) categories and let F : C→Dbe a covariant (additive) functor. Then F is an equivalence i F is full, faithful, and dense.

Proof. (=⇒) Suppose that F is an equivalence and that G, , and are as above. Then the 0 0 map f 7→ GF (f) is a bijection (isomorphism) from MorC(C, C )ontoMorC(GF (C),GF(C )). Thus, 0 0 the map f 7→ F (f) is injective (monic) from MorC(C, C )toMorD(F (C),F(C ), and the map G : 0 0 F (f) 7→ GF (f) from MorD(F (C),F(C )) to MorC(GF (C),GF(C )) is surjective. Using the natural isomorphism gives that f 7→ F (f) is surjective (epic). Thus, F is full and faithful. That it is dense is trivial.

(⇐=) This time suppose that F is full, faithful, and dense. We must build a functor G : D→C and natural isomorphisms ϕ :1D → FG and :1C → GF .

Since F is dense, for each object D in D we can choose an object C in C and an isomorphism

ϕD : D → F (C). Set G(D)=C. Then ϕD is an isomorphism D → FG(D). Now for each morphism g : D → D0 in D, since F is full and faithful, there is a unique morphism f : C = G(D) → C0 = G(D0) for which the diagram g - 0 D D

0 ϕD ϕD ? ? F (f-) 0 FG(D) FG(D ) commutes. Set G(g)=f. By uniqueness it is clear that for each D we have G(1D)=1G(D) and that for every pair of morphisms g and g0 in D,ifg g0 is dened, then G(g g0)=G(g) G(g0). Thus,

G : D→Cis a covariant functor, and ϕ :1D → FG is a natural isomorphism. In the additive case, 0 0 0 0 if g, g ∈ MorD(D, D ), then since F (G(g)+G(g )) = FG(g)+FG(g ), uniqueness again gives that G(g + g0)=G(g)+G(g0), so that G is additive.

Finally, we claim that there is a natural isomorphism :1C → GF . Let C be an object of C. Then there is an isomorphism ϕF (C) : F (C) → FGF(C). But F is full and faithful, so there is a unique morphism C : C → GF (C) with F (C )=ϕF (C). Since ϕF (C) is an isomorphism, there exists a morphism g : FGF(C) → F (C) with

gϕF (C) =1F (C) and ϕF (C)g =1FGF(C).

Again, since F is full and faithful there is a unique morphism h : GF (C) → C with F (h)=g. Thus,

F (hC )=1F (C) = F (1C ) and F (C h)=1FGF(C) = F (1GF (C). Since F is faithful, this means that Non-Commutative Rings Section 4 33

hC =1C and C h =1GF (C) so that C : C → GF (C) is an isomorphism. We nish by showing that it is natural. Let f : C → C0 be a morphism and consider the diagram

f - 0 C C

C C0 ? ? GF (-f) 0 GF (C) GF (C )

Hit it with F and it commutes. But F is faithful, so the diagram itself commutes.

Although an equivalence of two categories need not be an isomorphism, under which virtually eveything of mathematical interest is preserved, it does preserve the really important categorical stu. For example,

4.2. Corollary. Let F : C→Dbe a categorical equivalence. Then a morphism f : C → C0 in C is a monomorphism, epimorphism, or isomorphism, respectively, i F (f):F (C) → F (C0) is a monomorphism, epimorphism, or isomorphism in D.IfC, D, and F are additive, then for each pair C, C0 of objects in C,

0 0 F : MorC(C, C ) → MorD(F (C),F(C ))

is a group isomorphism, and for each object C in C

F : MorC(C, C) → MorD(F (C),F(C))

is a ring isomorphism.

Proof. Suppose that f : C → C0 is a monomorphism. Let g, g0 : D → F (C) be morphisms in D with F (f)g = F (f)g0. Since F is dense, there is some C00 and some isomorphism g00 : F (C00) → D. So gg0 and g0g00 are morphisms F (C00) → F (C). Since F is full, there exist morphisms h, h0 in C with F (h)=gg00 and F (h0)=g0g00.SoF (fh)=F (f)F (h)=F (f)gg00 = F (f)g0g00 = F (f)F (h0)=F (fh0). But F is faithful, so fh = fh0 and since f is monic, h = h0.Sogg00 = g0g00. But g00 is an isomorphism, so g = g0. Thus, F (f) is monic. We’ll leave the rest of this to the reader.

We want to apply this to the representation theory of rings. Two rings R and S are (Morita) equivalent in case the additive categories RMod and SMod are equivalent. Not at all surprisingly, if F : RMod → SMod is an additive equivalence, then F is an exact functor.

4.3. Theorem. If R and S are equivalent rings with F : RMod → SMod an additive equivalence, then F is exact.

f g Proof. Let 0 → M 0 → M → M 00 → 0 be exact over R. Then by Corollary 4.2, F (f) is monic 34 Section 4 and F (g) is epic. Also, F (g) F (f)=F (g f)=F (0) = 0, and thus, Im F (f) Ker F (g). So, it will suce to show that Ker F (g) Im F (f). Let K = Ker F (g). Since F is dense, there is an H in RMod with F (H) = K. Since F is also full and faithful, there exist R-homomorphisms for which the diagram

? - 0 f- g- 00 - 0 M M M 0

h = = ? ? ? - - g- 00 - 0 HMM 0 commutes and has exact rows and columns. Thus, h is an isomorphism, so that F (h):F (M 0) → K is an isomorphism, so that Ker F (g)=ImF (f).

A property of modules is said to be a Morita invariant if it is preserved by additive equivalences. Our next result is the key to identifying a whole lot of important Morita invariants.

4.4. Theorem. If R and S are equivalent rings with F : RMod → SMod an additive equivalence,

then for each module RM the lattice of submodules of RM and the lattice of submodules of SF (M) are isomorphic.

Proof. Let RM be a left R-module. For each N M, let N : N,→ M be the inclusion monomorphism, and let (N)=ImF (N ) F (M). We claim that the map : N 7→ (N) from the submodule lattice of M to that of F (M) is a lattice isomorphism.

To begin with is injective. For if (N)=(N 0), then there is an S-isomorphism h : F (N) → 0 F (N ) with h F (N 0 )=F (N ). But since F is full, there must be an R-homomorphism, necessarily 0 an isomorphism by Corollary 4.2, g : N → N with F (g)=h. Then since F is faithful, g N 0 = N , and so N = N 0.

pppppppppppppg - 0 ppppppppppppppppph = F (g) - 0 N N F (N) F (N ) @ Q Q @R © 0 Qs + N N F (N ) F (N 0 ) M F (M)

We claim next that is surjective. Indeed, let K be a submodule of F (M) with, say, K : K,→ F (M) the inclusion mono. Since F is dense, there is some RH and some S-isomorphism f : F (H) → K.

Since F is full, there is an R-monomorphism g : H → M with F (g)=K f : F (H) → F (M). Thus, (Im g)=K.

Finally, for submodules N,N0 of M we have N N 0 i there is some monomorphism j : N → N 0 0 with N = N 0 j i there is some monomorphism g : F (N) → F (N ) with F (N )=F (N 0 ) g i Non-Commutative Rings Section 4 35

(N) (N 0).

4.5. Corollary. Each of the following properties of a module is a Morita invariant. That is, if F is an equivalence from the category RMod of left R-modules to the category

SMod of left S-modules, then one of the following applies to RM i it applies to SF (M):

(a) simple; (d) artinian; (g) nitely generated;

(b) semisimple; (e) noetherian; (h) projective;

Proof. By Corollary 4.2, epimorphisms and monomorphisms are preserved, and splitting is func- torial, so projective and injective are Morita invariants. All the rest of these are characterized in the lattice of submodules.

So what isn’t a Morita invariant? By and large properties that deal with the elements or with the arithmetic of the ring are not invariants. Once we have found another characterization of equivalence, some of these will become quite clear. But some properties that are not invariants are commutativity, division ring, existence of nilpotent elements — all dealing with the arithmetic of the ring. A possibly surprising non-invariant is that of being a free module. Projectivity is, but freeness is not.

Exercises 4.

4.1. Finish the proof of Corollary 4.2 by proving that if F : C→Dis an equivalence and if f : C → C0 is a surjective, then so is F (f):F (C) → F (C0).

4.2. Prove that the category sets of all nite sets is equivalent to its full subcategory C with object class the collection of all nite initial subsets of N.

4.3. Let C be a category. Choose an object from each isomorphism class of objects of C. (Yes, we are using a monster version of the Axiom of Choice here.) For each object A denote the representative of its class by Aˆ.So Aˆ = Bˆ ⇐⇒ A = B.

Let Cˆ be the full subcategory of C with object class the representatives Aˆ of the objects of C.We call this the skeleton of C. Using the word “the” is premature, but in Exercise 4.4 we will prove that there is essentially one skeleton.

(a) Find a skeleton for sets whose objects are all subsets of N; 36 Section 4

(b) Find a skeleton for ab the category of nitely generated abelian groups.

4.4. Two categories C and D are isomorphic in case there exist covariant functors, S : C→Dand

T : D→Csuch that TS =1C and ST =1D.

(a) Show that there exists a category with two objects that is not isomorphic to any of its skeletons.

(b) Show that if Cˆ and C are two skeletons of the category C, then they are isomorphic.

4.5. Prove that two categories C and D are equivalent i they have isomorphic skeletons.

4.6. Let C and D be two (additive) categories. A contravariant (additive) functor F : C→Dis a duality in case there exists a contravariant (additive) functor G : D→Csuch that GF and FG

are naturally isomorphic, respectively, to the identity functors 1C and 1D. Prove that for a pair C and D of categories the following statements are equivalent:

(a) There is a duality F : C→D;

(b) There is an equivalence H : Cop → D ;

4.7. Let F : Top → Sets be the forgetful functor from the category of topological spaces to the category of sets. On the other hand, let T : Sets → Top be the functor that assigns to each set S the discrete space T (S)onS, and to each f : S → S0 in Sets, the continuous function T (f)=f : T (S) → T (S0)inTop.

(a) Show that the functor F : Top → Sets is faithful and dense but not full.

(b) Show that the functor T : Sets → Top is faithful and full but not dense.

4.8. Let F : C→Dand G : D→Cbe two functors. These two functors then determine two bi-functors

op op MorD(F (), ):C D→Sets and MorC(,G()) : C D→Sets.

We say that (F, G)isanadjoint pair in case there is a natural isomorphism

: MorC(F (), ) → MorD(,G()).

Alternatively, we say that F is a left adjoint of G and that G is a right adjoint of F . Of course, Non-Commutative Rings Section 4 37

this means that for each f : C0 → C in C and each g : D → D0 in D, the diagram

Mor(F (f),g) - 0 0 MorD(F (C),D) MorD(F (C ),D )

(C,D) (C0,D0)

? ? Mor(f,G(g)) - 0 0 MorC(C, G(D)) MorC(C ,G(D ))

commutes. That is, for each ϕ ∈ MorD(F (C),D),

G(g) (C,D)(ϕ) f = (C0,D0)(g ϕ F (f)).

(a) Let F : C→Dand G : D→Cbe inverse equivalences. (See Exercise 0.16.) Prove that both (F, G) and (G, F ) are adjoint pairs.

(b) Let K be a eld and let V be a K-vector space. Consider the additive functors

F = HomK (V,):KMod → KMod and G =() K V : KMod → KMod .

Show that (F, G) is an adjoint pair.