# Invert a quasi isomorphism of chain complexes

Is there an explicit way to invert a quasi-isomorphism of two chain complexes?

In case of homotopy algebras ($A_\infty$, $L_\infty$ ect.) there is an explicit way to invert any quasi isomorphism, if we are willing to work with infinity-morphisms. The inversion is then basically done by the homotopy transfer theorem.

Is something like this available for plain chain complexes?

Just think of $\Bbb Z/2 \Bbb Z$ and of $\Bbb Z \xrightarrow{\cdot 2} \Bbb Z$ as chain complexes, so that both only have non-vanishing homology in degree zero.
The the quotient map $\Bbb Z \to \Bbb Z/2 \Bbb Z$ induces a quasi isomorphism, but obviously there is no nontrivial map in the other direction.
• Why is it obvious, that there is no map in the other direction? Homotopy morphisms can look very different from the morphisms of the original category. See infiniy morphisms of $A_\infty$-algebras for example. – Bobby Dec 11 '16 at 10:40
• For example for $A_\infty$-algebras $A \to B$, these are sequences of linear maps $f_n: A^{\otimes n} \to B$, homogeneous of degree $n-1$, such that they satisfy some interchange-relations with the $A_\infty$-structure. The geneal definition is, that these are the morphisms, which have inverses to quasi-isomorphisms. Therfore we change the ambient categorical setting. In the case of dg-associative algebras, to invert quasi isomorphisms we have to go to the category of $A_\infty$-algebras with their infinity morphisms. – Bobby Dec 11 '16 at 11:32
• @Bobby It is obvious since there are no nonzero homomorphisms $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}$. – user144221 Dec 11 '16 at 15:47