About a chain homotopy

Question

Assume that $C=\lbrace C_{q},d_{q}\rbrace$ is a chain complex with each $C_{q}$ a free $R$-module. Let $C^{'}$ be another chain complex. Furthermore, assume that each $H_{q}$ is also free and that we have a chain map $f=\lbrace f_{q}:C_{q}\rightarrow C_{q}^{'}\rbrace$ such that the induced homomorphisms are zero, i.e. $H_{q}(f)=0$ for each $q$.

I would like to prove that $f$ is homotopic to zero. Any ideas/suggestions ?

Thanks :)

Answer 1

Since $H_0(f)=0,$ the chain map $f$ carries $C_0$ into $d'_1(C'_1)$. Hence, there exists $D_0:C_0\to C'_1$ such that $f=d'_1\circ D_0$.

You can go on similarly, and prove the existence of a homotopy by induction.