Let $\mathcal{A}$ be an abelian category, the "complexes category", "homotopy category" and "derived category" denoted by $\mathcal{C(A)}$ , $\mathcal{K(A)}$ and $\mathcal{D(A)}$ respectively.
It is well known that
$$ f = g\ in\ \mathcal{C(A)} \Longrightarrow f = g\ in\ \mathcal{K(A)} \Longrightarrow f = g\ in\ \mathcal{D(A)} \Longrightarrow f\ and\ g\ induce\ the\ same\ morphisms\ of\ cohomology.$$
All three of these directions of implication are irreversible. It is easy to find counterexamples for the first two irreversible directions, but I have not found a counterexample for the irreversibility of the last direction.
So the question arises: Are there two morphisms $f$ and $g$ between complexes which induce the same morphisms of cohomology but are not equivalent in the "derived category" , i.e. , there is not an quasi-isomorphism $t$ such that $ft = gt$ or $tf = tg$ in "homotopy category" $\mathcal{K(A)}$?
This problem is related with the existence of non-split extensions:
If $0 \to N \to E \to M \to 0$ is a non-split extension in $\mathcal{A}$, we obtain a non-zero map $f\colon M \to N[1]$ in $D(\mathcal{A})$. But clearly, $f$ is the zero map on cohomology.