Homology of a chain complex

Remarks

The complex is exact at $C_n$ if and only if $H_n(C) = 0$. A chain map $f \colon C_\bullet \to D_\bullet$ sends cycles to cycles and boundaries to boundaries, so it induces morphisms $H_n(f) \colon H_n(C) \to H_n(D)$, and $H_n$ is a functor. For a cochain complex $(C^\bullet, d)$ the same construction gives the cohomology $H^n(C) = \ker d^n / \im d^{n-1}$.

Example: let $C_1 = C_0 = \Z$ with $d_1$ multiplication by $2$, and $C_n = 0$ otherwise. Then $Z_1 = \ker d_1 = 0$, so $H_1(C) = 0$; and $Z_0 = \Z$, $B_0 = 2\Z$, so $H_0(C) = \Z/2\Z$.

Non-example: for the sequence $\Z \xrightarrow{1} \Z \xrightarrow{1} \Z$, the image $\Z$ of the first map is not contained in the kernel $0$ of the second. So $\ker / \im$ is not defined; the condition $d^2 = 0$ is what makes homology make sense.