Let A C B be abelain groups, then we have exact sequence: 0 → A → B → B/A → 0

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Jan. 5, 2007 4.3. complexes, exact sequences. .

Definition 4.3.1. By a short exact sequence, we mean an exact se- quence 0 → A → B → C → 0.

Example 4.3.2.

1. Let A, B be abelian groups, then we have exact sequence:

0 → A→ A ⊕ B^ıA → B → 0.^pB

2. Let A C B be abelain groups, then we have exact sequence:

0 → A → B → B/A → 0.

3. Let ϕ : B → C be a surjective homomorphism, then we have exact sequence:

0 → ker(ϕ) → B → C → 0.

¤ Given a long exact sequence K^• = (Kⁱ, di), it can be decomposed into short exact sequences

0 → ker(dⁱ) = im(dⁱ⁻¹) → Kⁱ → im(dⁱ) = ker(dⁱ⁺¹) → 0.

Therefore, short exact sequences play the most important role in our studies.

Given a morphism φ ∈ Hom(K^•, L^•) of complexes, one can define its kernel, image, cokernel, in a natural way. Thus we can formulate a new category Kom(A), whose objects are complexes over A and morphisms are morphism of complexes.

Exercise 4.3.3. Kom(A) is an abelian category in which A is a sub- category.

Let K^• be a complex. We let Zⁱ := ker(dⁱ), called the i-th cocycle and Bⁱ := im(dⁱ⁻¹), called the i-th coboundary. Then Hⁱ(K^•) :=

Zⁱ/Bⁱ is called the i-th cohomology of K^•. Cohomology can be viewed as a tool detecting the non-exactness of complexes.

Given two complexes K^•, L^•, a morphism of complexes φ ∈ Hom_A(K^•, L^•) consists of morphisms φⁱ : Kⁱ → Lⁱ such that φⁱ⁺¹◦ dⁱ_K = dⁱ_L◦ φⁱ for all i. Another way to put it is the following diagram commute:

−−−→ Kⁱ −−−→ K^{diK i+1} −−−→

φⁱ



y ^φi+1

 y

−−−→ Lⁱ −−−→ L^{diL i+1} −−−→

One can easily checked that there is an induced map Hⁱ(φ) : Hⁱ(K^•) → Hⁱ(L^•) for all i. Moreover, if φ, ψ are morphism of complexes, then Hⁱ(ψ) ◦ Hⁱ(φ) = Hⁱ(ψ ◦ φ) for all i whenever it make sense.

Before we move on, we discuss the following useful lemmas:

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Lemma 4.3.4 (Snake Lemma). Given a diagram A⁰ −−−→ A −−−→ A^{f 00} −−−→ 0

d⁰



y ^d



y ^d00

 y 0 −−−→ B⁰ −−−→ B −−−→ B^{g 00}

with each rows are exact. Then there is a well-defined map δ : ker(d⁰⁰) → coker(d⁰) such that we have an exact sequence

ker(d⁰)→ ker(d) → ker(d^{f 00})→ coker(d^{δ 0}) → coker(d) → coker(d^{g¯ 00}).

If moreover that f : A⁰ → A is injective, then f : ker(d⁰) → ker(d) is injective. And if g : B → B⁰⁰ is surjective, then ¯g : coker(d)→coker(d⁰⁰) is surjective.

Proof. The proof consists of various diagram chasing. We leave it to

the reader. ¤

Corollary 4.3.5. Keep the notation as above. If both d⁰, d⁰⁰ are injec- tive (resp. surjective) then so is d.

Assume that f is injective and g is surjective. If any two of d⁰, d, d⁰⁰ are isomorphism. So is the third one.

Lemma 4.3.6 (Five Lemma). Given a diagram

A₁ −−−→ A₂ −−−→ A₃ −−−→ A₄ −−−→ A₅

d1



y ^d2



y ^d3



y ^d4



y ^d5

 y B₁ −−−→ B₂ −−−→ B₃ −−−→ B₄ −−−→ b₅ with each rows are exact.

If d₁ is surjective (resp. injective) and d₂, d₄ are injective (resp.

surjective), then d₃ is injective (resp. surjective).

In particular, if d₁, d₂, d₄, d₅ are isomorphic, then so is d₃.

Proof. Decompose the sequence into short exact sequences. ¤ An immediate application is the following:

Proposition 4.3.7. Given an exact sequence 0 → A → B^f → C → 0,^g the following are equivalent:

(1) there is h : C → B such that gh = 1C. (2) there is l : B → A such that lf = 1_A.

(3) the sequence is isomorphic to 0 → A→ A ⊕ C^ıA → C → 0.^pC Such sequence is called split.

If the sequence split, then in particular, B ∼= A ⊕ C.

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Proof. Given h : C → B, we can construct the following commutative diagram:

0 −−−→ A −−−→ A ⊕ C^ıA −−−→ C −−−→ 0^pC



y ^1A



y ^{f pA+hpC}



y ^1C

 y

 y 0 −−−→ A −−−→^f B −−−→ C −−−→ 0^g

By Five Lemma, f p_A+ hp_C is an isomorphism. Hence those two se- quences are isomorphic.

On the other hand, if the two sequence are isomorphic. That is we have the following commutative diagram, which is invertible:

0 −−−→ A −−−→ A ⊕ C^ıA −−−→ C −−−→ 0^pC

1A



y ^φ



y ^1C

 y

0 −−−→ A −−−→^f B −−−→ C −−−→ 0^g Let h = φ ◦ ı_C : C → B, then gh = gφı_C = 1_Cp_Cı_C = 1_C.

The proof for other equivalence is similar. ¤ Theorem 4.3.8. Given a short exact of complexes, then it induces a long exact sequences of cohomology.

Proof. This can be proved directly, or by Snake Lemma.

We briefly sketch the proof by using Snake Lemma here.

First look at the diagram

0 −−−→ Aⁱ⁻¹ −−−→ Bⁱ⁻¹ −−−→ Cⁱ⁻¹ −−−→ 0

 y

 y

 y

0 −−−→ Aⁱ −−−→ Bⁱ −−−→ Cⁱ −−−→ 0

Then we have exact sequence Aⁱ/Bⁱ(A^•) → Bⁱ/Bⁱ(B^•) → Cⁱ/Bⁱ(C^•) → 0 by looking at cokernel of the maps.

Next we look at the diagram

0 −−−→ Aⁱ⁺¹ −−−→ Bⁱ⁺¹ −−−→ Cⁱ⁺¹ −−−→ 0

 y

 y

 y

0 −−−→ Aⁱ⁺² −−−→ Bⁱ⁺² −−−→ Cⁱ⁺² −−−→ 0

Then we have exact sequence 0 → Zⁱ⁺¹(A^•) → Zⁱ⁺¹(B^•) → Zⁱ⁺¹(C^•) by looking at kernels.

These two exact sequences fit into a commutative diagram

Aⁱ/Bⁱ(A^•) −−−→ Bⁱ/Bⁱ(B^•) −−−→ Cⁱ/Bⁱ(C^•) −−−→ 0

d¯ⁱ_A

 y

 y

 y 0 −−−→ Zⁱ⁺¹(A^•) −−−→ Zⁱ⁺¹(B^•) −−−→ Zⁱ⁺¹(C^•)

One can check that ker( ¯dⁱ_A) = Hⁱ(A^•) and coker( ¯dⁱ_A) = Hⁱ⁺¹(A^•).

And similarly for B^• and C^•. Hence by Snake Lemma, we are done. ¤

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Definition 4.3.9. Let F : A → B be a functor between two abelian categories. We say that F is exact if for an exact sequence K^• over over A, F (K^•) is exact over B.

Exercise 4.3.10. Show that F is exact if and only if for any short exact sequence 0 → A → B → C → 0 in A, the induced sequence 0 → F (A) → F (B) → F (C) → 0 is exact in B.

Definition 4.3.11. Keep the notation as above. We say that F is left- exact (resp. right-exact) if for any short exact sequence 0 → A → B → C → 0 in A, the induced sequence 0 → F (A) → F (B) → F (C) (resp.

F (A) → F (B) → F (C) → 0) is exact in B.

Unfortunately, most natural functors are left-exact (or right-exact) but not exact. We list some of them:

Example 4.3.12.

Let X be a topological space. Let Sh_X be the category of sheaves on X, which is an abelian category. The global section functor Γ(X, ·) :

ShX → Ab is left exact but not exact. ¤

Example 4.3.13.

Let Ab be the category of abelian groups. Fixed M ∈ Ab, we consider Hom(M.·) : Ab → Ab by A 7→ Hom(M, A). This is left-exact but nor

right exact. ¤