This is a point of stupid confusion for me.
Let $s:\mathcal{A}\to\mathcal{B}$ be an equivalence of abelian categories. Does this functor induce a triangulated equivalence $\overline{s}: \mathbb{D}^b(\mathcal{A})\to \mathbb{D}^b(\mathcal{B})$ of bounded derived categories?
My intuition says that this is not true in general, since it is possible that for some $i$, $\mathrm{Ext}^i_{\mathcal{A}}(A,B)$ and $\mathrm{Ext}^i_{\mathcal{B}}(s(A),s(B))$ are not isomorphic. As I understand, it then follows that $\mathrm{Hom}_{\mathbb{D}^b(\mathcal{A})}(A,B[i])$ and $\mathrm{Hom}_{\mathbb{D}^b(\mathcal{B})}(s(A),s(B)[i])$ are not the same. This would mean the derived categories are not equivalent right?
On the other hand it doesn't feels right that the derived category not respect equivalence. Is the remedy that I should be thinking about exact equivalences of abelian categories?