The long exact sequence in homology

Step 1 (constructing $\partial$). Let $[a''] \in H_n(A'')$ with representative cycle $a'' \in A''_n$, $d(a'')=0$. Since $A_n \to A''_n$ is surjective, lift $a''$ to some $a \in A_n$. Then $d(a) \in A_n$ maps to $d(a'')=0$ in $A''_{n-1}$, so by exactness $d(a) = \iota(a')$ for a unique $a' \in A'_{n-1}$ (using injectivity of $\iota: A' \to A$). Define $\partial[a''] := [a']$.

Step 2 ($a'$ is a cycle, and $\partial$ is well defined). Since $\iota$ is injective and chain maps commute with $d$, $\iota(d a') = d(\iota a') = d(d a) = 0$, so $d a' = 0$ and $[a']$ is a genuine homology class. If a different lift $\tilde a$ of $a''$ is chosen, $\tilde a - a$ maps to $0$ in $A''_n$, so $\tilde a - a = \iota(c)$ for some $c \in A'_n$; then the two resulting classes in $H_{n-1}(A')$ differ by $[d(c)] = 0$. A similar check shows $\partial$ is independent of the cycle representative of $[a'']$ (replacing $a''$ by $a'' + d(b'')$ changes $a$ by a boundary, hence $a'$ by a boundary too).

Step 3 (exactness at $H_n(A)$ and $H_n(A'')$). Exactness at $H_n(A)$ (image of $H_n(A')\to H_n(A)$ equals kernel of $H_n(A)\to H_n(A'')$) and at $H_n(A'')$ (image of $H_n(A)\to H_n(A'')$ equals kernel of $\partial$) both follow by direct diagram chasing on representatives, using only that $\iota$ is injective, the quotient map is surjective, and $\operatorname{im}\iota = \ker(\text{quotient})$ at the chain level — the same style of argument as Steps 1–2, applied one degree at a time.

Step 4 (exactness at $H_{n-1}(A')$). Finally one checks $\operatorname{im}\partial = \ker(H_{n-1}(A')\to H_{n-1}(A))$: if $\partial[a'']=[a']$ then $\iota(a')=d(a)$ is a boundary in $A$, so $[a']$ maps to $0$ in $H_{n-1}(A)$, giving $\operatorname{im}\partial \subseteq \ker$; conversely if $\iota(a')=d(a)$ for some $a\in A_n$, running Step 1 backwards shows $a''=$ image of $a$ satisfies $\partial[a'']=[a']$, giving the reverse inclusion. This completes exactness at every spot, and the whole construction is exactly the classical snake lemma.