Theorem: The short exact sequence $0 \to A_* \xrightarrow{i} B_* \xrightarrow{j} C_* \to 0$ induces a long exact sequence $\cdots \to H_n(A) \xrightarrow{i_*} H_n(B) \xrightarrow{j_*} H_n(C) \xrightarrow{\partial_*} H_{n-1}(A) \to \cdots$

Construction of $\partial_*$:

Let $[c] \in H_n(C)$ be represented by a cycle $c \in C_n$ with $\partial^C c = 0$. Since $j_n$ is surjective, choose $b \in B_n$ with $j_n(b) = c$. Then $j_{n-1}(\partial^B b) = \partial^C (j_n(b)) = \partial^C c = 0$ so $\partial^B b \in \ker j_{n-1} = \operatorname{im}\, i_{n-1}$. Since $i_{n-1}$ is injective, there exists a unique $a \in A_{n-1}$ with $i_{n-1}(a) = \partial^B b$.

We claim $a$ is a cycle: $i_{n-2}(\partial^A a) = \partial^B(i_{n-1}(a)) = \partial^B(\partial^B b) = 0$, and injectivity of $i_{n-2}$ gives $\partial^A a = 0$.

Define $\partial_*[c] = [a] \in H_{n-1}(A)$.

Well-definedness of $\partial_*$:

Suppose $b' \in B_n$ also satisfies $j_n(b') = c$. Then $j_n(b - b') = 0$, so $b - b' = i_n(a_0)$ for some $a_0 \in A_n$. Then the corresponding elements differ by $\partial^B(b - b') = i_{n-1}(\partial^A a_0)$, so the two choices of $a$ differ by the boundary $\partial^A a_0$, giving the same homology class.

If $c = \partial^C c'$ for some $c' \in C_{n+1}$, choose $b' \in B_{n+1}$ with $j_{n+1}(b') = c'$. Then $j_n(\partial^B b') = c$, so we may take $b = \partial^B b'$, giving $\partial^B b = 0$ and hence $a = 0$. Thus $\partial_*$ vanishes on boundaries.

Exactness at $H_n(B)$ (i.e., $\ker j_* = \operatorname{im}\, i_*$):

If $[b] = i_*[a]$, then $b = i(a) + \partial^B b'$, so $j_n(b) = j_n(\partial^B b') = \partial^C(j(b'))$, giving $j_*[b] = 0$. Conversely, if $j_*[b] = 0$, then $j(b) = \partial^C c'$ for some $c'$. Lift $c'$ to $b' \in B_{n+1}$ and note $j(b - \partial^B b') = 0$, so $b - \partial^B b' = i(a)$ for some cycle $a$, giving $[b] = i_*[a]$.

Exactness at $H_n(C)$ (i.e., $\ker \partial_* = \operatorname{im}\, j_*$):

If $[c] = j_*[b]$ with $\partial^B b = 0$, then $\partial_*[c] = 0$ since we can choose the lift $b$ with $\partial^B b = 0$, giving $a = 0$. Conversely, if $\partial_*[c] = 0$, then $a = \partial^A a'$, so $\partial^B b = i(\partial^A a') = \partial^B(i(a'))$. Then $b - i(a')$ is a cycle in $B_n$ mapping to $c$, giving $[c] = j_*[b - i(a')]$.

Exactness at $H_{n-1}(A)$ (i.e., $\ker i_* = \operatorname{im}\, \partial_*$):

Given $\partial_*[c] = [a]$, we have $i(a) = \partial^B b$, so $i_*[a] = 0$. Conversely, if $i_*[a] = 0$, then $i(a) = \partial^B b$ for some $b \in B_n$. Set $c = j(b)$; then $\partial^C c = j(\partial^B b) = j(i(a)) = 0$ and $\partial_*[c] = [a]$.

This completes the proof of exactness at every node. $\square$