Is quasi-isomorphism an equivalence relation? Let $E^\bullet$ and $F^\bullet$ be complexes on an abelian category; what does it mean to say that $E^\bullet$ and $F^\bullet$ are quasi-isomorphic?
Does it only mean that there is a map of complexes $f:E^\bullet \to F^\bullet$ that induces isomoprhisms between the cohomology objects?
Or does it also guarantee the existence of a map of complexes $g:F^\bullet \to E^\bullet$ inducing the inverses of $H^pf:H^p(E^\bullet)\to H^p(F^\bullet)$?
Put in another way: is quasi-isomorphism an equivalence relation?

$\def\ZZ{\mathbb Z}$The relation $E \sim F$ defined by «there exists a morphism $E\to F$ inducing an isomorphism in homology» is not an equivalence relation because it is not symmetric (it is relfexive and transitive)
For example, there is a morphism from
$$\cdots 0\to \ZZ\xrightarrow2\ZZ\to0\to\cdots$$
to the complex
$$\cdots 0\to 0\to\ZZ/2\ZZ\to0\to\cdots$$
inducing an isomorphism in homology, but there is no non-zero morphism in the other direction.
The useful relation is the symmetric closure of this relation.
