6
$\begingroup$

Let $M$ be a model category. I don't assume that $M$ has functorial factorizations or that $M$ is simplicial. Write $M^{c}$ (respectively, $M^{cf}$) for the full subcategory of $M$ on the cofibrant objects (respectively, the cofibrant and fibrant objects).

Does the inclusion $M^{cf} \to M^c$ induce a Dwyer-Kan equivalence on the simplicial localizations at the class of weak equivalences?

Assuming the answer is "yes", does anyone know of a reference? I couldn't find this in the original papers of Dwyer and Kan for general $M$, only when $M$ has functorial factorizations.

$\endgroup$
  • 3
    $\begingroup$ This should follow from 4.4 in "Function complexes in homotopical algebra". That is: the map on simplicial localizations is essentially surjective up to equivalence, so you just need fully faithful, and 4.4 lets you compute the mapping space in the same way for both. $\endgroup$ – Dylan Wilson Feb 21 at 0:09
  • 2
    $\begingroup$ The answer is definitely yes, but I do not know any published reference. What I can say is that any proof that I know of that $M^c$ and $M$ are Dwyer-Kan equivalent may be promoted to a proof that $M^{cf}$ and $M^f$ are Dwyer-Kan equivalent. Inspection of calculus of fractions as provided by 8.1 in “Function complexes in homotopical algebra” is very closed to being a proof. $\endgroup$ – Denis-Charles Cisinski Feb 21 at 9:32
2
$\begingroup$

The only published reference for the full statement I know is Hinich's paper (V. Hinich, Dwyer-Kan localization revisited. Homology Homotopy Appl. 18, No. 1), where it appears as (the dual of) Proposition 1.3.8. Strictly speaking, the proof given there is "this works just like the proof of the case $\mathscr C^c\hookrightarrow\mathscr C$ (of which Hinich gives a new proof in Section 1.3.7), but at least to me it looks like you can indeed directly adapt his argument by just adding the word "fibrant" at appropriate places.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.