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My first question in this forum. Suppose we have a Quillen equivalence $L: M\leftrightarrow N: R$ between two model categories (say left proper cofibrantly generated cellular or locally presentable). Let $S$ a set of morphisms in $M$ and $L(S)=T$ the image of S in $N$.

I was wondering if $L: \mathbf{L}_{S}M\leftrightarrow \mathbf{L}_{T}N: R$ is a Quillen equivalence? $\mathbf{L}_{S}M$ is the left Bousfield localization of $M$ at $S$ and $\mathbf{L}_{T}N$ is the left Bousfield localization of $N$ at $T=L(S)$

Remark: $L(S)$ is the ordinary image not the derived one (as in David White answer), there is no cofibrant replacement.
Edit: I was wondering if there is a clear answer to my question ? The only reference I have found is here, page 5017 just after theorem 5.7 But I'm still not satisfied since there is no proof and the cited version uses Hirschhorn book where the author (Hirschhorn) deals with the derived image. I will be happy for any clear answer, thank you.

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    $\begingroup$ This is not true if $L$ does not preserve weak equivalences. Take $S=\{f\}$ where $f$ is a weak equivalence and $L(f)$ isn't. Then $L_SM=M$ but $N\rightleftarrows L_TN$ is not a Quillen equivalence. $\endgroup$
    – Marc Hoyois
    Commented Aug 28, 2016 at 17:59
  • $\begingroup$ Thanks for the useful counterexample. I'm surprised that in the paper I have mentioned there is such imprecision! $\endgroup$
    – Nguyen lan Lee
    Commented Aug 29, 2016 at 5:35

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Yes. This is Theorem 3.3.20 in Hirschhorn's book, if by $T$ you mean the left derived maps of $L(S)$, e.g. cofibrant replacements of the maps $L(S)$.

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  • $\begingroup$ In my initial question $T=L(S)$ not the derived image as you assumed. $\endgroup$
    – Nguyen lan Lee
    Commented Aug 15, 2016 at 7:05
  • $\begingroup$ You can take the cofibrant replacement of maps in $L(S)$ to get a new set of maps $T'$. It is easy to see that $L_{T'}N$ is the same as $L_{L(S)}N$. I used that trick in my thesis, and Hovey has used it in several of his papers. $\endgroup$
    – David White
    Commented Aug 15, 2016 at 13:15
  • $\begingroup$ thanks @davidwhite do you a precise statement in your references ? $\endgroup$
    – Nguyen lan Lee
    Commented Aug 15, 2016 at 14:05
  • $\begingroup$ It's just the 2 out of 3 property. Given a map $f:A\to B$, let $Qf:QA\to QB$ be its cofibrant replacement. Picture the square featuring $Qf$ and $f$. The legs are cofibrant replacement maps for A and B, so they are weak equivalences. If you invert Qf, then it's a new weak equivalence. By the 2 out of 3 property, f is also a new weak equivalence. $\endgroup$
    – David White
    Commented Aug 16, 2016 at 1:13
  • $\begingroup$ But still I don't understand why you can conclude that $L_{L(S)}N$ is equivalent to $L_{T^{'}} N$ (I'm using your definition of $T^{'}$). $\endgroup$
    – Nguyen lan Lee
    Commented Aug 16, 2016 at 12:37

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