Let $k$ be a commutative ring, and let $V$ be a complex of $k$-modules (more in general, we can take an $\mathcal A$-dg-module, where $\mathcal A$ is a dg-category. We can construct the bar resolution $B(V)$ of $V$, which comes with a quasi-isomorphism $B(V) \to V$ (here is a related post). My question is the following:

Does $B(-)$ define a dg-functor from the dg-category of complexes to itself? Do the maps $B(V) \to V$ define a natural transformation of dg-functors?

I think that $B$ should at least be a (ordinary) functor from the (model) category of complexes to itself. It would be very nice if it were actually dg-functorial. Or, perhaps, is it $A_\infty$-functorial?

  • $\begingroup$ It's a fibrant replacement, so it's functorial. $\endgroup$
    – user40276
    Jun 3, 2016 at 21:08
  • 3
    $\begingroup$ It should be a cofibrant replacement, but this does not answer the question. I don't ask for functoriality, but for dg-functoriality, namely enriched functoriality. $\endgroup$ Jun 3, 2016 at 21:38
  • $\begingroup$ Yes, sorry, I have misread the title. Anyway, I'm almost sure it's $A_{\infty}$ functorial because I have seen this implicitly used a lot of times in integration of representations. $\endgroup$
    – user40276
    Jun 4, 2016 at 13:20


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