$B$ a chain complex of free abelian groups and $C$ a chain complex such that $H_n(C)=0$. Then any chain map from $B$ to $C$ homotopic to the zero map

Solution 1:

Assuming your chain complexes are bounded, it is standard to construct all necessary maps by induction. Start with $s_0: B_0 \dashrightarrow C_1.$ For this we need $d \circ s_0 = f_0.$ Moreover, $C_1 \to C_0$ is surjective. Since $B_i$ are free, we can always solve equations like $d \circ s_0 = f_0$ by "cancelling" $d$: $s_0(b):= d^{-1}(f_0(b))$, where $d^{-1}$ takes any preimage, which exists due to surjectivity.

We can continue: for $s_1: B_1 \dashrightarrow C_2$ we'd need $d \circ s_1 = f_1 - s_0 \circ d.$ So we can repeat the same process as long as for every $b \in B_1$ there is $d^{-1}(f_1 - s_0 \circ d)(b).$ Now use exactness of $C$: this preimage exists as long as $d(f_1 - s_0 \circ d)(b) = 0.$ Show this, using that $s_0$ already satisfies the chain homotopy equation.

And so on.