Example

Let $u : M \to N$ and $v : N \to P$ be homomorphisms of $A$-modules such that $v \circ u = 0$; let $C$ be the complex such that $C_2 = M, C_1 = N, C_0 = P, C_i = 0$ for $i \neq 0, 1, 2$, $d_2 = u, d_1 = v, d_i = 0$ for $i \neq 1, 2$. Then $C$ has zero homology if and only if the sequence $0 \to M \xrightarrow{u} N \xrightarrow{v} P \to 0$ is exact. It is homotopic to zero if and only if this sequence is split. Indeed, saying that $C$ is homotopic to zero means that there exist $A$-homomorphisms $s : P \to N$ and $t : N \to M$ such that $v \circ s = 1_P, s \circ v + u \circ t = 1_N, t \circ u = 1_M$; this implies that the sequence is split; conversely, if $s$ is an $A$-linear section of $v$, one defines $t$ by $u \circ t = 1_N - s \circ v$, which is possible since $v \circ (1_N - s \circ v) = v - v \circ s \circ v = 0$.