Suppose we have two categories $\mathcal{C}$ and $\mathcal{D}$, and equivalence between them (in the sense of a strong equivalence of categories) and a model structure on $\mathcal{C}$. Then it is pretty obvious that you can transfer the model structure along the adjunction giving the equivalence to obtain a model structure on $\mathcal{D}$ which makes the equivalence a Quillen equivalence between the two resulting model categories. I need this fact for my research, but I don't want to have to write down the proof in the article I'm writing as it looks like a standard fact. Does anyone have a reference for it?

I looked a bit in the nlab and in Hovey's Model Categories, but I couldn't find it.

Update: The answer of @KyleFerendo is helpful, but as we are in the case of an equivalence of categories, and not merely of an adjunction, I would expect that one doesn't need that many assumptions in order to be able to transfer the model structure. (In fact, I would expect that you can always transfer model structures in this situation, but feel free to prove me wrong!)

Update: You can indeed always transfer the model structure in the case of an equivalence of categories. I provided a full proof in an answer below.


This is probably helpful:


Which points to, among other places, page 20 of https://arxiv.org/pdf/math/0609537.pdf

The theorem in that paper uses the following assumptions: that your model structure on $\mathcal{C}$ be cofibrantly generated, that the right adjoint $G:\mathcal{D}\to\mathcal{C}$ commute with sequential colimits (this shouldn't be a problem for you since you're using an equivalence of categories), and apparently you still have to check that when a cofibration in $\mathcal{D}$ has the left lifting property with respect to all fibrations, it is an acyclic cofibration. This second requirement was unexpected to me, but hopefully it is straightforward for you to verify, again since you are using an (adjoint) equivalence of categories.

  • $\begingroup$ Thank you! This can probably work in my case. However, it looks a bit of an overkill, as I have an equivalence of categories and not just an adjunction. Also, I would expect that in the case of an equivalence of categories some assumptions can be dropped (e.g. the requirement that the model structure is cofibrantly generated...) $\endgroup$ – Daniel Robert-Nicoud Mar 5 '17 at 11:33
  • $\begingroup$ @DanielRobert-Nicoud it is a well known result that every equivalence is part of an adjoint equivalence. For a reference take a look at this. $\endgroup$ – Giorgio Mossa Mar 6 '17 at 16:02
  • $\begingroup$ @GiorgioMossa I am aware of this fact. Both functors are at the same time left and right adjoint one to the other, so that the commuting with limits and colimits is not a problem. The condition on the lifting property is part of my definition of model structure, so no problem there. The thing that bothers me is the assumption that the model structure is cofibrantly generated! Is it really necessary? $\endgroup$ – Daniel Robert-Nicoud Mar 6 '17 at 16:43
  • $\begingroup$ @DanielRobert-Nicoud, oh I see now. I should think about it. $\endgroup$ – Giorgio Mossa Mar 7 '17 at 20:29

I have checked the details and what I stated in the OP is indeed true. I will present a full proof here, but I would be really surprised if it isn't already present in the literature, and I am still looking for a standard reference.

Any comment about the correctness of the proof, as well as advice on how to simplify or ameliorate it are more than welcome!

The statement we are going to prove is the following.

Proposition: Let $(\mathcal{C},W,C,F)$ be a closed model category, let $\mathcal{D}$ be another category and suppose we have a strong equivalence of categories $$G:\mathcal{D}\longrightarrow\mathcal{C}\ .$$ Then the classes $W':=G^{-1}(W)$, $C':=G^{-1}(C)$ and $F':=G^{-1}(F)$ make $\mathcal{D}$ into a closed model category.

Remark: I will prove this for the definition given in the book Simplicial Homotopy Theory by Goerss and Jardine (pp. 71-72), but you can easily adapt the demonstration to the other, slightly different definitions found throughout the literature (e.g. demanding the splitting of arrows into cofibrations and fibrations to be functorial).

We start by fixing a functor $$F:\mathcal{C}\longrightarrow\mathcal{D}$$ which is both left and right adjoint to $G$ with unit $\epsilon:FG\cong 1$ and counit $\eta:1\cong GF$. This always exists (as remarked e.g. by @GiorgioMossa in the comments to the other answer).

CM1: Suppose $D:I\to\mathcal{D}$ is a diagram in $\mathcal{D}$. Then $GD:I\to\mathcal{C}$ admits a (co)limit $\lim GD$ because $\mathcal{C}$ is complete and cocomplete. Now we apply $F$ and use the fact that it commutes to (co)limits (since it is left and right adjoint): $$F(\lim GD) \cong\lim FGD\ .$$ But $\epsilon$ provides a natural isomorphism of diagrams $FGD\cong D$, and thus $F(\lim GD)$ is a (co)limit for $D$.

CM2: We prove the $2$-out-of-$3$ property for $W'$. Let $f,g$ be two composable arrows in $\mathcal{D}$ such that two of $f,g,gf$ are in $W'$. Then two of $G(f),G(g),G(gf)$ are in $W$, so that the third is. Thus, by definition of $W'$, all of $f,g,gf$ are in $W'$.

CM3: Suppose $w\in W'$ and $v$ is a retract of $w$, i.e. we have a diagram $$\require{AMScd} \begin{CD} @>>> @>>> \\ @V{v}VV @V{w}VV @V{v}VV \\ @>>> @>>> \end{CD}$$ with both horizontal compositions giving the identity. Applying $G$, we see that $G(v)$ is a retract of $G(w)$, so that $G(v)\in W$. Therefore, $v\in W$. The same proof works replacing $W$ by $C$ or $F$.

CM4: Let $(i,p)\in(C',F'\cap W')$ or $(i,p)\in(C'\cap W',F')$ and suppose we have a commutative diagram $$\begin{CD} @>>> \\ @V{i}VV @VV{p}V \\ @>>> \end{CD}$$ By applying $G$, we get a diagonal filler $\alpha$ for the diagram $$\begin{CD} @>>> \\ @V{G(i)}VV @VV{G(p)}V \\ @>>> \end{CD}$$ (sorry, no diagonal arrows with the amsCD package!) Then the composition of $F(\alpha)$ with the appropriate unit $\epsilon$ (and its inverse) provide a filler for the original diagram.

Before going on, we need to prove the following lemma.

Lemma: If an arrow $f$ in $\mathcal{C}$ is in $W$ (resp. in $C$ or $F$), then $F(f)$ is in $W'$ (resp. in $C'$ or $F'$).

Proof: The counit $\eta$ provides an isomorphism $f\cong GF(f)$. Since $W$ is closed under retracts, it follows immediately that $GF(f)\in W$, and thus that $F(f)\in W'$. The same works for $C$ and $F$. This concludes the proof.

CM5: Let $f$ be any arrow in $\mathcal{D}$. Then we can split $G(f)$ into $pi$ with $p\in F$ and $i\in C\cap W$ (or $p\in F\cap W$ and $i\in C$). By the Lemma above, $F(i)\in C\cap W$ and $F(p)\in F$, and they provide a splitting for $FG(f)$. By composing with $\epsilon$ where needed, these arrows give a splitting of $f$.


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