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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?

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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.

Though what you could do is find some kind of projective resolution of the chain complexes and then the induced map you will get should have a quasi inverse. You could look for model categories(the projective model structure on chain complexes) cofibrant replacement and the Whitehead theorem.

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  • $\begingroup$ 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. $\endgroup$ – Bobby Dec 11 '16 at 10:40
  • $\begingroup$ @Bobby Maybe you can tell me what you mean by homotopy morphisms? $\endgroup$ – user60589 Dec 11 '16 at 11:03
  • $\begingroup$ 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. $\endgroup$ – Bobby Dec 11 '16 at 11:32
  • $\begingroup$ @Bobby It is obvious since there are no nonzero homomorphisms $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}$. $\endgroup$ – user144221 Dec 11 '16 at 15:47
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    $\begingroup$ @Bobby You should take look at model categories. For cofibrant fibrant objects it holds. $\endgroup$ – user60589 Dec 11 '16 at 16:29

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