I am looking to show that there exists two chain complexes $C_*$ and $C'_*$ and a chain map between them $f_\#\colon C_*\to C'_*$ such that in degree $0$, the map $f_0\colon C_0\to C_0'$ is non-zero yet the induced map on homology is the zero map. My idea was to find a chain complex $C_0'$ such that its homology $H_0'=0$, and hence the induced map has to be zero.
So I took $$C_*:= 0\to 0\to \mathbb Z\\ C_*':=0\to\mathbb Z\stackrel{\operatorname{id}}\to \mathbb Z$$
The homology of the first chain is $H_0=\mathbb Z$, and since the second chain has an isomorphism $\operatorname{id}:\mathbb{Z} \to \mathbb{Z}$, then the homology of the second chain $H_0'=0$. In particular, the identity map induces the zero map. Contradiction? If homology is functorial, the induced map on the identity must be the identity itself. What went wrong?