I was asked to prove that if $A^\bullet$ is a bounded above cochain complex of abelian groups, then $A^\bullet$ is quasi-isomorphic to the sum of its cohomology groups.
As $A^\bullet$ is bounded above there exists a cochain complex $P^\bullet$ of projectives and a quasi-isomorphism $P^\bullet \to A^\bullet$. Furthermore, $\mathbb Z$ is a PID so $0 \to Z^n(P^\bullet) \to P^n \to B^{n+1}(P^\bullet) \to 0$ splits and $P^n \cong Z^n(P^\bullet) \oplus B^{n+1}(P^\bullet)$. The projection onto the first summand gives a quasi-isomorphism $P^\bullet \to H^\bullet(P^\bullet) \cong H^\bullet(A^\bullet)$ where $H^\bullet(A^\bullet)$ is the complex $H^\bullet(A^\bullet) = (\cdots \to H^n(A^\bullet) \to H^{n+1}(A^\bullet) \to \cdots)$ with zero differentials.
However, I am unable to construct a quasi-isomorphism $H^\bullet(A^\bullet) \to A^\bullet$ or $A^\bullet \to H^\bullet(A^\bullet)$. I somehow doubt that if such a quasi-isomorphism exists, one can construct it similar as above:
Consider the complex $A^\bullet = (0 \to 0 \to \mathbb Z / 2 \mathbb Z \to 0)$. The complex consisting of its cohomology groups also is $0 \to 0 \to \mathbb Z / 2 \to 0$ which clearly is quasi-isomorphic to $A^\bullet$. But if we take the complex $P^\bullet = (0 \to \mathbb Z \xrightarrow{2} \mathbb Z \to 0)$, which also is quasi-isomorphic to $A^\bullet$, a quasi-isomorphism between $A^\bullet$ and $H^\bullet (A^\bullet)$ can't factor through $P^\bullet$.
I couldn't find a counterexample or proof for the original claim, however.