Give an example of cochain map which is bijective?

Question

I am looking for a injective cochain map $\psi: C^*\rightarrow D^*$ such that the map $\Psi^i$ from $C^i$ to $D^i$ is injective, but the map $\psi^*$ from $H^i(C^*)$ to $H^i(D^*)$ is not injective for any $i\geq0$.

Answer 1

Consider any nonzero module $M$ and the following complexes : $C = \dots \to 0\to 0\to M \to 0\to \dots$ and $D= \dots \to 0\to M\to M \to 0\to \dots$, the only nontrivial map being the identity.

with the (co)chain map that is $$\require{AMScd}\begin{CD}0 @>>> 0 @>>> M @>>>0\\ @VVV @VVV @VVV @VVV \\ 0@>>> M @>>> M @>>> 0\end{CD}$$ the only nontrivial map being again the identity. Then this is of course injective on the (co)chain level, but not on (co)homology, because $H^1(C)= M\neq 0$ and $H^1(D) = 0$

There is only one degree in cohomology on which this map is noninjective, if you want it to be injective on no degree in cohomology (as you asked), you can just shift this example at various degrees and take direct sums.