Does $\operatorname{Hom}_{\mathbf K (\mathcal{A})}(A, B) = 0$ imply $\operatorname{Hom}_{\mathbf D (\mathcal{A})}(A, B) = 0$?

Question

Let $\mathcal{A}$ be an abelian category, let $\mathbf K(\mathcal{A})$ be the category of cochain complexes modulo homotopy, and let $\mathbf{D}(\mathcal{A})$ be the derived category.

Let $A, B$ be two cochain complexes. Is it true that $\operatorname{Hom}_{\mathbf K(\mathcal A)}(A, B) = 0$ implies $\operatorname{Hom}_{\mathbf D(\mathcal A)}(A, B) = 0$?

Answer 1

I'll do things with chain complexes in homological conventions, but of course it's exactly the same with cochain complexes in cohomological conventions

Let $X,Y\in \mathcal A$ such that $\mathrm{Ext}^1(X,Y) \neq 0$. Then take as complexes $A= X[0]$ ($X$ in degree $0$) and $B = Y[1]$ ($Y$ in degree $1$, homological conventions).

Then $\hom_{K(\mathcal A)}(A,B)$ is a quotient of $\hom_{Ch(\mathcal A)}(A,B) = 0$, so it's itself $0$.

However, one may check that $\hom_{D(\mathcal A)}(A,B) = \mathrm{Ext}^1(X,Y) \neq 0$.