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I believe the following statement is true under certain technical conditions.

Let $$L\colon C \rightleftharpoons D\colon R$$ be a Quillen equivalence. Suppose we are given objects $c\in C$, $d\in D$, and an equivalence $c\to R(d)$. (Perhaps we need to assume that $d$ is fibrant, but $c$ doesn't have to be cofibrant.) Then one has a Quillen equivalence of under-categories: $\hat L\colon c/C\rightleftharpoons d/D\colon \hat R$, where $\hat R$ is an obvious functor sending $d\to d_1$ to the composition $c\to R(d)\to R(d_1)$; and $\hat L$ sends $c\to c_1$ to $$ d\to \mathrm{colim}\bigl(d\leftarrow L(c)\rightarrow L(c_1)\bigr). $$

I wonder what are the conditions under which this is true? Any reference? In fact I am interested in the case C and D are categories of bimodules over operads.

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2 Answers 2

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If $c$ is cofibrant and $d$ is fibrant, then $c \to R(d)$ is a weak equivalence if and only if $L(c) \to d$ is. Now, we can prove that if $L(c) \to d$ is a weak equivalence, $c$ is cofibrant, and either $d$ is cofibrant or $D$ is left proper, then $\hat{L} \dashv \hat{R}$ is a Quillen equivalence. First, note that $\hat{L} \dashv \hat{R}$ is always a Quillen adjunction. Let us prove that it is a Quillen equivalence under these assumptions. Since $R$ reflects weak equivalences between fibrant objects, so does $\hat{R}$. Thus, we just need to prove that the map $c_1 \to RL(c_1) \to RSL(c_1) \to RS(d_1)$ is a weak equivalence for every cofibration $i : c \to c_1$, where $S$ is a fibrant replacement functor and $L(c_1) \to d_1$ is the pushout of $f : L(c) \to d$ along $L(i) : L(c) \to L(c_1)$. The composition of the first two maps is a weak equivalence since $L \dashv R$ is a Quillen equivalence and $c_1$ is cofibrant. To prove that the last map is a weak equivalence, note that $S$ preserves weak equivalences and $R$ preserves weak equivalences between fibrant objects. Thus, we just need to prove that $L(c_1) \to d_1$ is a weak equivalence. Since it is a pushout of a weak equivlence along a cofibration, this follows from our assumptions.

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  • $\begingroup$ In the case I am interested in, $L(c)\to d$ is not a weak equivalence, only $c\to R(d)$ is. $\endgroup$
    – Victor
    Commented Apr 21, 2017 at 2:11
  • $\begingroup$ The condition that $L(c) \to d$ is a weak equivalence is actually necessary. Indeed, the map $L(c_1) \to d_1$ that I constructed in the proof is a weak equivalence for every cofibration $c \to c_1$ if and only if $\hat{L} \dashv \hat{R}$ is a Quillen equivalence. But if we take $id_c : c \to c$, then this map is simply $L(c) \to d$. $\endgroup$ Commented Apr 21, 2017 at 2:19
  • $\begingroup$ Also, if $c$ is cofibrant and $c \to R(d)$ is a weak equivalence, then so is $L(c) \to d$. $\endgroup$ Commented Apr 21, 2017 at 2:32
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    $\begingroup$ $c_1$ is not cofibrant and thus the composition of the first two maps does not have to be a weak equivalence. $\endgroup$
    – Victor
    Commented Apr 21, 2017 at 3:17
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    $\begingroup$ Ah, yes, you are right. But it still seems that it is not enough to assume that $c \to R(d)$ is a weak equivalence. $\endgroup$ Commented Apr 21, 2017 at 3:28
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Actually a while ago Benoit Fresse gave me an answer to this question. I figured I should post it here.

The condition on $C$ and $D$ is that they should be left proper and $R$ should preserve weak equivalences.

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    $\begingroup$ Did he provide a reference or a proof? That's what you were originally asking for, right? $\endgroup$ Commented Jan 24, 2018 at 15:14
  • $\begingroup$ I'm in a hurry now so I might have misread, but Proposition 16.2.4 of May-Ponto's "More concise algebraic topology" seems very related. $\endgroup$ Commented Jan 25, 2018 at 1:10
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    $\begingroup$ @DavidWhite He sketched the proof. We need it for our work. It is now Lemmas 6.5-6.7 in arxiv.org/abs/1708.02203. $\endgroup$
    – Victor
    Commented Jan 25, 2018 at 15:24
  • $\begingroup$ @BrunoStonek Thank you, I will check it! $\endgroup$
    – Victor
    Commented Jan 25, 2018 at 15:24

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