I read this statement in Siegfried Bosch's "Algebraic geometry and commutative algebra", but it is not proved in the book. Basically it says that if $M_*$: $$\cdot\cdot\cdot \rightarrow M_{n+1} \xrightarrow{d_{n+1}} M_n \xrightarrow{d_{n}} M_{n-1} \xrightarrow{d_{n-1}}\cdot\cdot\cdot$$ and $N_*$: $$\cdot\cdot\cdot \rightarrow N_{n+1} \xrightarrow{\delta_{n+1}} N_n \xrightarrow{\delta_{n}} N_{n-1} \xrightarrow{\delta_{n-1}}\cdot\cdot\cdot$$ are chain complexes and $f$ and $g$ are chain complex homomorphisms from $M_*$ to $N_*$ that are homotopic, then for all $n$ the homomorphisms: $$H_n(f):H_n(M_*) \rightarrow H_n(N_*); \ x + im(d_{n+1}) \mapsto f_n(x) + im(\delta_{n+1})$$ and $$H_n(g):H_n(M_*) \rightarrow H_n(N_*); \ x + im(d_{n+1}) \mapsto g_n(x) + im(\delta_{n+1})$$ are the same. Any ideas how that can be proven ?
My approach is this. If $h:M_* \rightarrow N_*$ is the homotopy of $f$ and $g$ then for every $x + im(d_{n+1})$, $$H_n(f)(x + im(d_{n+1})) - H_n(f)(x + im(d_{n+1})) = f_n(x) + im(\delta_{n+1}) - g_n(x) + im(\delta_{n+1}) = f_n(x) - g_n(x) + im(\delta_{n+1}) = h_{n-1}d_n(x) + \delta_{n+1}h_n(x) + im(\delta_{n+1}) = h_{n-1}d_n(x) + im(\delta_{n+1})$$ However I don't know why $h_{n-1}d_n(x) + im(\delta_{n+1})$ has to be equal to $im(\delta_{n+1})$ ?
Remember that homology classes are represented by cycles. The homology group $H_n(M)$ isn't the quotient of $M_n$ by boundaries; it's the quotient of the submodule of cycles by boundaries.
So let $[x] \in H_n(M)$ (if you want you can write $x + \operatorname{im} d_{n+1}$ but this not isn't very practical) be a homology class, represented by the cycle $x \in M_n$ with $dx = 0$. Then $$\require{cancel}g(x) - f(x) = h(\cancel{d(x)}) + d(h(x)) = d(h(x))$$ is a boundary, and so the homology classes $[g(x)]$ and $[f(x)]$ are equal. In other words, $g_*[x] = f_*[x]$ (or if you prefer, $H_n(g)(x + \operatorname{im}(d_{n+1})) = H_n(f)(x + \operatorname{im}(d_{n+1}))$ but again this notation isn't very practical).