Clarify about the connecting homomorphisms and the snake lemma

Question

In class we said (without proof) that given a an exact sequence of chain complexes $$0\to \mathcal A\xrightarrow{f} \mathcal B\xrightarrow g \mathcal C\to 0$$ one can construct the long exact sequence $$\dots \xrightarrow{\delta_{q+1}} H(A_q)\xrightarrow{f_q} H(B_q) \xrightarrow{g_q} H(C_q)\xrightarrow{\delta_q} H(A_{q-1})\xrightarrow{f_{q-1}}\dots \ ,$$ where the $\delta _q$ are the connecting homomorphisms. Now, I read on wikipedia how these homomorphisms are obtained, but I don't understand well why the snake lemma works in this context. In fact the vertical maps of the lemma here are the boundary homomorphisms of the corresponding complex (let's call them $\alpha_q$, $\beta_q$ and $\gamma_q$ for respectively $A_q$, $B_q$ and $C_q$); and the lemma builds a homomorphism from $\text{ker}(C_q)$ to $\text {coker }(\alpha_q)$. However, shouldn't we need an homomorphism that goes in $H(A_{q-1})\subseteq\text {coker }(\alpha_q)$? My idea is that when the vertical maps are the boundary homomorphisms one can prove that $\text {Im} (\delta_q)\subseteq \text{ker} (\alpha_{q-1})$, but I first want to know if I'm approaching things from the right point of view. Thanks for any clarify

Answer 1

Apply the snake lemma not to $$\begin{array}{ccccccccc} & & A_q & \rightarrow & B_q & \rightarrow & C_q & \rightarrow & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & A_{q-1} & \rightarrow & B_{q-1} & \rightarrow & C_{q-1}, & & \end{array}$$ but to $$\DeclareMathOperator{\coker}{coker} \begin{array}{ccccccccc} & & \coker(\alpha_{q+1}) & \rightarrow & \coker(\beta_{q+1}) & \rightarrow & \coker(\gamma_{q+1}) & \rightarrow & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & \ker(\alpha_{q-1}) & \rightarrow & \ker(\beta_{q-1}) & \rightarrow & \ker(\gamma_{q-1}). & & \end{array}$$

Make sure you see where the maps come from, that the diagram commutes, and that the rows are exact.