Proof of the Long Exact Sequence via the Snake Lemma

We prove the Long Exact Sequence in Cohomology by reducing it to the Snake Lemma.

Setup

Let $0 \to A^\bullet \xrightarrow{f} B^\bullet \xrightarrow{g} C^\bullet \to 0$ be a short exact sequence of complexes. This means $0 \to A^n \xrightarrow{f^n} B^n \xrightarrow{g^n} C^n \to 0$ is exact for each $n$ .

Step 1: Identify Cocycles and Coboundaries

ProofStep 1

For each $n$ , define $Z^n(A) = \ker(d_A^n : A^n \to A^{n+1})$ (cocycles) and $B^n(A) = \mathrm{im}(d_A^{n-1} : A^{n-1} \to A^n)$ (coboundaries). We have $B^n \subseteq Z^n$ since $d^n \circ d^{n-1} = 0$ , and $H^n(A) = Z^n(A) / B^n(A)$ .

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Step 2: The Key Diagram

ProofStep 2

Consider the commutative diagram for each $n$ :

\begin{array}{ccccccccc} & & A^n / B^n(A) & \xrightarrow{\bar{f}^n} & B^n / B^n(B) & \xrightarrow{\bar{g}^n} & C^n / B^n(C) & \to & 0 \\[6pt] & & \downarrow\scriptstyle{\bar{d}_A^n} & & \downarrow\scriptstyle{\bar{d}_B^n} & & \downarrow\scriptstyle{\bar{d}_C^n} & & \\[6pt] 0 & \to & Z^{n+1}(A) & \xrightarrow{f^{n+1}} & Z^{n+1}(B) & \xrightarrow{g^{n+1}} & Z^{n+1}(C) & & \end{array}

where $\bar{d}^n$ denotes the map induced by $d^n$ on the quotient $A^n / B^n \to A^{n+1}$ , landing in $Z^{n+1}$ since $d^{n+1} \circ d^n = 0$ .

Top row is exact on the right: Since $g^n$ is surjective, $\bar{g}^n$ is surjective.

Bottom row is exact on the left: Since $f^{n+1}$ is injective, its restriction to $Z^{n+1}(A) \to Z^{n+1}(B)$ is injective.

Kernels: $\ker(\bar{d}_A^n) = Z^n(A) / B^n(A) = H^n(A)$ (since $\bar{d}^n[a] = 0 \iff d^n(a) = 0 \iff a \in Z^n$ ). Similarly for $B$ and $C$ .

Cokernels: $\mathrm{coker}(\bar{d}_A^n) = Z^{n+1}(A) / \mathrm{im}(\bar{d}_A^n) = Z^{n+1}(A) / B^{n+1}(A) = H^{n+1}(A)$ . Similarly for $B$ and $C$ .

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Step 3: Apply the Snake Lemma

ProofStep 3

Applying the Snake Lemma to the diagram in Step 2, we obtain the exact sequence:

$H^n(A) \to H^n(B) \to H^n(C) \xrightarrow{\delta^n} H^{n+1}(A) \to H^{n+1}(B) \to H^{n+1}(C)$

Since this holds for every $n$ , splicing these sequences together gives the long exact sequence:

$\cdots \to H^n(A) \xrightarrow{H^n(f)} H^n(B) \xrightarrow{H^n(g)} H^n(C) \xrightarrow{\delta^n} H^{n+1}(A) \to \cdots$

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Step 4: The Connecting Homomorphism Explicitly

ProofStep 4

In $R\text{-}\mathbf{Mod}$ , $\delta^n$ acts as follows. Given $[c] \in H^n(C)$ (so $c \in C^n$ with $d_C^n(c) = 0$ ):

Lift: Choose $b \in B^n$ with $g^n(b) = c$ .
Differentiate: Compute $d_B^n(b) \in B^{n+1}$ . We have $g^{n+1}(d_B^n(b)) = d_C^n(g^n(b)) = d_C^n(c) = 0$ .
Lift to $A$ : By exactness, $d_B^n(b) = f^{n+1}(a)$ for a unique $a \in A^{n+1}$ .
Verify cocycle: $f^{n+2}(d_A^{n+1}(a)) = d_B^{n+1}(f^{n+1}(a)) = d_B^{n+1}(d_B^n(b)) = 0$ . Since $f$ is injective, $d_A^{n+1}(a) = 0$ , so $a \in Z^{n+1}(A)$ .
Define: $\delta^n([c]) = [a] \in H^{n+1}(A)$ .

Independence of the lift $b$ follows from the Snake Lemma proof. $\square$

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Naturality

Theorem4.8Naturality of the connecting homomorphism

Given a morphism of short exact sequences of complexes:

\begin{array}{ccccccccc} 0 & \to & A^\bullet & \to & B^\bullet & \to & C^\bullet & \to & 0 \\ & & \downarrow\scriptstyle{\alpha} & & \downarrow\scriptstyle{\beta} & & \downarrow\scriptstyle{\gamma} & & \\ 0 & \to & A'^\bullet & \to & B'^\bullet & \to & C'^\bullet & \to & 0 \end{array}

the connecting homomorphisms satisfy $\delta' \circ H^n(\gamma) = H^{n+1}(\alpha) \circ \delta$ . This gives a morphism of long exact sequences.

ExampleApplication to Ext

For a SES $0 \to M' \to M \to M'' \to 0$ and any module $N$ , the connecting homomorphism $\delta : \mathrm{Hom}(M', N) \to \mathrm{Ext}^1(M'', N)$ sends a map $\varphi : M' \to N$ to the extension class obtained by pushing out the original SES along $\varphi$ .

ExampleApplication to sheaf cohomology

For $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ on $X$ , the connecting homomorphism $\delta : H^0(X, \mathcal{F}'') \to H^1(X, \mathcal{F}')$ sends a global section $s \in \Gamma(\mathcal{F}'')$ to the obstruction class for lifting $s$ to $\Gamma(\mathcal{F})$ .