Then one has injective resolution if the category has enough injectives

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Jan. 12, 2007

4.4. injective. In this section, we are going to define injective objects.

Then one has injective resolution if the category has enough injectives.

Moreover, we will see that injective resolution are convenient for han- dling left exact but not exact functors.

Definition 4.4.1. Let A be an abelian category. An object I ∈ A is injective if for all 0 → A → B and A → I, there exists B → I makes the diagram commute.

Proposition 4.4.2. I is injective if and only if the functor M 7→

HomA(M, I) is exact.

Proof. For every exact sequence 0 → A → B → C → 0, we have exact sequence

Hom(A, I) ← Hom(B, I) ← Hom(C, I) ← 0.

The definition of injective says nothing more than that Hom(B, I) →

Hom(A, I) is surjective. ¤

Exercise 4.4.3. If I is injective, then every sequence 0 → I → B → C → 0 splits.

An abelian category A is said to have enough injectives if for every A ∈ A, there exist an injective object I ∈ A and an injection 0 → A → I.

Suppose now that A has enough injectives. Then for every A ∈ A, one has 0 → A → I^{ı 0} for some injective I⁰. Next look at coker(ı), one has 0 → coker(ı) → I¹ for some injective I¹ and let d⁰ : I⁰ → I¹ be the composition map. Inductively, we obtained a sequence

0 → A → I⁰ → I¹...

It’s easy to see that it’s an exact sequence because it patches short exact sequences 0 → coker(ı_j−1)→ I^{ıj j} → coker(ı_j) → 0.

Before we move on, it worthwhile to think what indeed injective object is and why we expect an abelian category has enough injectives.

Let Ab be the abelian category of abelian groups. A group G is said to be divisible if m : G → G by m : x 7→ mx is surjective for all m 6= 0 ∈ Z. In other words, for x ∈ G, and for all m 6= 0 ∈ Z, there is y ∈ G such that ny = x. We will show that in Ab:

Lemma 4.4.4. G is divisible, then G is injective.

Lemma 4.4.5. Every abelian group can be embedded into a divisible group.

Thus the abelian category Ab has enough injective. It also follows that those natural abelian categories, such as category of R-modules, category of sheaves of abelian groups, has enough injective.

In order to prove the Lemmata, we observe that:

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(1) if G is divisible, so if G/N for any normal subgroup N.

(2) if Gi are divisible for all i, then P

i∈IGi is divisible.

proof of 4.4.4. Suppose that G is divisible and 0 → A⁰ → A is exact with a map f⁰ : A⁰ → G. We need to show that there is f : A → G extending f⁰.

We shall use Zorn’s Lemma. Let Σ = {(B, g)|A⁰ < B < A, g : B → G, g|_A⁰ = f⁰}. There exists a maximal element (M, h) in Σ. One verifies

that M = A. ¤

proof of 4.4.5. G ∼= F/K, F ∼=P

x∈IZx. F ,→^f P

x∈IQx. G ∼= F/K ∼= f (F )/f (K) <P

x∈IQx/f (K) is divisible. ¤

Lemma 4.4.6. Let I^• be an injective resolution of A and J^• an injec- tive resolution of B. If there is ϕ : A → B, then there exists f : I^• → J^• compatible with ϕ.

Moreover any two such f, g : I^• → J^• are homotopic.

Definition 4.4.7. f, g ∈ Hom(K^•, L^•) are homotopic if there are hⁱ : Kⁱ → Lⁱ⁻¹ such that d_Lh + hd_K = f − g.

Injective resolution is very useful in the study of left exact functors which is not exact. More, precise the following Lemma show that injective rsln splits

Lemma 4.4.8. Given 0 → A → B → C → 0, there is an exact sequence of complexes 0 → I^• → J^• → K^• → 0 such that I^• (resp, J^•, K^•) is an injective resolution of A (resp. B, C). Moreover, Jⁱ = Iⁱ⊕ Kⁱ.

Proof. We define I⁰, K⁰ first. Then there is a natural map B → J⁰ :=

I⁰⊕ K⁰. This map is injective.

Then inductively, we get the resolutions. ¤ Warning: J is not I⊕K as complex. For example, the map I⁰⊕K⁰ → I¹⊕ K¹ is of the form (d_I(i⁰) + ∗, d_K(k⁰)) where ∗ is not necessarily zero.

We are now ready to study the left-exact functors. Apply F to 0 −−−→ A −−−→ B −−−→ C −−−→ 0

 y

 y

 y

0 −−−→ I^• −−−→ J^• −−−→ K^• −−−→ 0 We get

0 −−−→ F (A) −−−→ F (B) −−−→ F (C)

 y

 y

 y

0 −−−→ F (I^•) −−−→ F (J^•) −−−→ F (K^•) −−−→ 0

Notice that the bottom row is exact because Jⁱ = Iⁱ ⊕ Kⁱ by our construction, hence F (Jⁱ) = F (Iⁱ) ⊕ F (Kⁱ) for all i.

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Proposition 4.4.9. Let RⁱF (A) := Hⁱ(F (I^•)). Then we have:

(1) R⁰F (A) = A.

(2) there is a long exact sequence

0 → F (A) → F (B) → F (C) → R¹F (A) → R¹F (B) → ...

Proof. It’s easy to see that ker(F (I⁰) → F (I¹)) ∼= F (A) by the left exactness. And the second statement follows from the long exact se- quence of cohomology of short exact sequence of complexes. ¤ Exercise 4.4.10. Show that RⁱF (A) is well-defined. That is, indepen- dent of choice of injective resolution.

4.5. derived category. In this section, we are going to describe de- rived category briefly. It’s a category over which cohomology theory can be defined and convenient to operate.

Exercise 4.5.1. If h is homotopic to 0, denoted h ∼ 0, then f h ∼ 0, hg ∼ 0 for all f, g whenever it makes sense.

So we can think of the class of homotopic equivalence as an ideal.

Let K(A) be the category whose objects are complex in A and mor- phisms are morphism in A quotient homotopic equivalence. More precisely, Hom_K(A)(K^•, L^•) consists of homotopic equivalent class of Hom_Kom(A)(K^•, L^•).

Then in K(A), injective resolution is unique (up to isomorphism).

Definition 4.5.2. Given a complex K^• = (Kⁱ, dⁱ_K), we define K[n]^• such that K[n]ⁱ = Kⁿ⁺¹, dⁱ_K[n] = (−1)ⁿdⁿ⁺ⁱ_K .

And given a morphism f : K^• → L^•, we define a complex C(f )^•, called the mapping cone of f , by C(f )ⁱ = Kⁱ⁺¹⊕Lⁱ and dⁱ_C(kⁱ⁺¹, lⁱ) = (−dⁱ⁺¹_K (kⁱ⁺¹), f (kⁱ⁺¹) + dⁱ_L(lⁱ)).

Example 4.5.3.

If K^• = K, L^• = L, then C(f ) = 0 → K → L → 0. ¤ Example 4.5.4.

If f = 0, then C(f ) = K^•⊕ L^•.

Definition 4.5.5. Given a morphsim f : K^• → L^•, we define a com- plex Cyl(f )^• such that Cyl(f )ⁱ = Kⁱ⊕Kⁱ⁺¹⊕Lⁱ. And dⁱ_Cyl(kⁱ, kⁱ⁺¹, lⁱ) = (dKkⁱ − kⁱ⁺¹, −dKkⁱ⁺¹, f (kⁱ⁺¹) + dLlⁱ).

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Theorem 4.5.6. We have the following diagram

0 −−−→ L^• −−−→ C(f )^• −−−→ K[1]^{δ •} −−−→ 0

α

 y

 y

0 −−−→ K^• −−−→ Cyl(f )^{f¯ •} −−−→ C(f )^{π •} −−−→ 0

=



y ^β

 y K^• −−−→^f L^•

Such that each row is exact. α, β are quasi-isomorphisms. Moreover, βα = 1_L and αβ ∼ 1_{Cyl(f )}.

Proof. All the above maps are the natural ones. One has to check that all the maps indeed gives morphism of complexes and the diagram commutes. We leave the detail to the readers.

The homotopy is defined by hⁱ(kⁱ, kⁱ⁺¹, lⁱ) = (0, kⁱ, 0). ¤ Theorem 4.5.7. Given an exact sequence 0 → K^• → L^• → M^• → 0, we have the following commutative diagram with each vertical map being quasi-isomorphic.

0 −−−→ K^• −−−→^f L^• −−−→ M^{g •} −−−→ 0 x

 ^β

x

 ^γ

x



0 −−−→ K^• −−−→ Cyl(f )^{f¯ •} −−−→ C(f )^{π •} −−−→ 0 ,

where γ(kⁱ⁺¹, lⁱ) = g(lⁱ).

The second row is called distinguished triangle.

Derived category D(A) is the category localizing K(A) with respect to quasi-isomorphisms. That is, a morphism Hom_D(A)(X, Y ) in D(A) is a roof (t, f ) where t : Z^• → X^•is a quasi-isomorphism and f : Z^• → Y^• is a morphism in K(A). Then in this setting, a quasi-isomorphism s : X^• → Y^• has inverse (s, 1_X) ∈ Hom_D(A)(Y^•, X^•).

Derived category has the universal property that any functor F : Kom(A) → D sending quasi-isomorphism into isomorphism can be uniquely factored through D(A).

Note that a cohomology (homology) theory on A is nothing but a functor F : Kom(A) → Kom(Ab) and thus factors through derived category.