Snake lemma

Given $x\in \ker(f'')\subseteq A''$:

1. pick $a\in A$ with $p(a)=x$;
2. since $f''(x)=0$, commutativity implies $q(f(a))=0$, hence $f(a)\in \operatorname{im}(j)$;
3. choose $b'\in B'$ with $j(b')=f(a)$;
4. define $\delta(x)$ to be the class of $b'$ in $\operatorname{coker}(f')=B'/\operatorname{im}(f')$.

A standard diagram chase shows this is well-defined (independent of choices) and gives exactness.

The map $\delta$ is natural in the diagram, and it is the prototype for the connecting homomorphism appearing in long exact sequences.

1. Induced map on quotients: the “kernel–cokernel” exact sequence. Let $g:M\to M'$ be an $R$-linear map, and let $N\subseteq M$, $N'\subseteq M'$ satisfy $g(N)\subseteq N'$. Consider the commutative diagram with exact rows:
   $$\begin{array}{ccccccccc} 0 &\to& N &\to& M &\to& M/N &\to& 0\\ & & \downarrow g|_N & & \downarrow g & & \downarrow \overline g & & \\ 0 &\to& N' &\to& M' &\to& M'/N' &\to& 0 \end{array}$$
   The snake lemma produces the exact sequence
   $$0\to \ker(g|_N)\to \ker(g)\to \ker(\overline g)\xrightarrow{\delta} \operatorname{coker}(g|_N)\to \operatorname{coker}(g)\to \operatorname{coker}(\overline g)\to 0,$$
   which cleanly measures how kernels/cokernels change when passing to quotients.

1. Boundary map in homology from a short exact sequence of complexes. Given a degreewise short exact sequence of chain complexes
   $$0\to C'_\bullet \to C_\bullet \to C''_\bullet \to 0,$$
   one applies the snake lemma to the diagram of cycles and boundaries in each degree to construct the connecting map
   $$\partial: H_n(C''_\bullet)\longrightarrow H_{n-1}(C'_\bullet),$$
   yielding the usual long exact sequence in homology. (This is the chain-complex analogue of long exact sequences from derived functors.)

1. Computing a boundary map concretely (multiplication on a quotient). Fix integers $m,n$ and consider the diagram
   $$\begin{array}{ccccccccc} 0&\to& \mathbb Z &\xrightarrow{\cdot m}& \mathbb Z &\to& \mathbb Z/m &\to& 0\\ && \downarrow \cdot n && \downarrow \cdot n && \downarrow \cdot n &&\\ 0&\to& \mathbb Z &\xrightarrow{\cdot m}& \mathbb Z &\to& \mathbb Z/m &\to& 0 \end{array}$$
   The connecting map $\delta:\ker(\cdot n:\mathbb Z/m\to \mathbb Z/m)\to \operatorname{coker}(\cdot n:\mathbb Z\to \mathbb Z)\cong \mathbb Z/n$ sends a class $\overline{x}\in\mathbb Z/m$ with $nx\in m\mathbb Z$ to the class of $\frac{nx}{m}$ in $\mathbb Z/n$. This is a concrete instance of how $\delta$ encodes the failure of lifting across the diagram.