Let $C_*$ and $D_*$ be chain complexes and let $f:C_*\to D_*$ be a chain map. Since $f$ is a chain map, its image $f(C_*)$ is a subcomplex of $D_*$.
My question is now the following:
Assume that its induced map between the homology groups $f_*:H_*(C_*)\to H_*(D_*)$ is the zero map, or as a slightly stronger assumption that $f$ is chain homotopic to zero. Is it true that $H_*(f(C_*))=0$?
Removing the assumptions on $f$ one could also ask more generally if it holds that $f_*(H_*(C_*))\stackrel{?}{=}H_*(f(C_*))$.
Thanks a lot in advance!
In general the answer is no and a counterexample is the following: consider the two complexes \begin{align*} C_{\bullet}& :\,\, \dots \to 0\to 0 \to \mathbb{Z} \to 0 \to \dots \\ D_{\bullet}& :\,\, \dots \to 0\to \mathbb{Z} \overset{id}{\to} \mathbb{Z} \to 0 \to \dots \\ \end{align*} and the morphism given by the inclusion $f\colon C_{\bullet} \hookrightarrow D_{\bullet}$. Then $f$ is homotopic to $0$ ( as a homotopy we can take the map that is $0$ everywhere and the identity at $\mathbb{Z}\to \mathbb{Z}$). However the complex $f(C_{\bullet})$ cannot have zero cohomology, as it is isomorphic to $C_{\bullet}$, that is not exact at $\mathbb{Z}$.