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The usual way to prove that two model categories are equivalent is to construct a zigzag of Quillen equivalences between them, but is it always possible? We can ask a more general question.

Question: Suppose that objects $X$ and $Y$ of a category with weak equivalences $(\mathcal{C},\mathcal{W})$ are isomorphic in its localization. Under which conditions on $(\mathcal{C},\mathcal{W})$ is there a zigzag of weak equivalences between $X$ and $Y$?

This is true if $(\mathcal{C},\mathcal{W})$ is a model category, but is there a weaker condition? It is not true in general. The counterexample is the category $A \to B \to C \to D$ in which $A \to C$ and $B \to D$ are weak equivalences. It is also not enough to assume that $(\mathcal{C},\mathcal{W})$ satisfies 2-out-of-6. The counterexample is the simplest category with a retract of a weak equivalence. So, here are some natural conditions on $(\mathcal{C},\mathcal{W})$ which might be sufficient:

  • $(\mathcal{C},\mathcal{W})$ satisfies 2-out-of-6 and is closed under weak retracts (a weak retract is like an ordinary retract, but identity maps are replaced with weak equivalences).
  • $\mathcal{W}$ is saturated (this implies the previous condition).
  • $(\mathcal{C},\mathcal{W})$ has the structure of a category of fibrant objects.
  • $\mathcal{C}$ is the category of (small) model categories and Quillen adjunctions between them and $\mathcal{W}$ is the class of Quillen equivalences.
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  • $\begingroup$ The introductory sentence seems to presuppose that the right notion of two model categories being "equivalent" is that they become isomorphic after localizing at the Quillen equivalences; is this clear? $\endgroup$
    – Reid Barton
    Commented Jul 28, 2019 at 3:06
  • $\begingroup$ @ReidBarton I would say that the "correct" notion of an equivalence between two model categories is an equivalence between their simplicial localizations. Since the simplicial localization functor maps Quillen equivalences to equivalences of simplicial categories, we have the following implications: (there is a zigzag of Quillen equivalences between M and N) $\implies$ (M and N become isomorphic after localizing Quillen equivalences) $\implies$ (M and N have equivalent simplicial localizations). So, the suggested in the post notion of an equivalence is closer to the "correct" one. $\endgroup$
    – Valery Isaev
    Commented Jul 28, 2019 at 4:00
  • $\begingroup$ For combinatorial model categories it is also known that these implication are all equivalences. For general model categories I don't know. $\endgroup$
    – Simon Henry
    Commented Jul 28, 2019 at 12:11
  • $\begingroup$ A necessary and sufficient condition is the existence of a functor $F\colon \mathcal{C}\rightarrow \mathcal{D}$ such that $\mathcal{W}$ is the inverse image of isomorphisms in $\mathcal{D}$ along $F$. Probably this is not very helpful for you since, a fortiori, $F$ can be taken to be the localization functor. Nevertheless it is useful in specific examples. $\endgroup$
    – Fernando Muro
    Commented Jul 28, 2019 at 15:01
  • $\begingroup$ @FernandoMuro Your condition is equivalent to saturation of $\mathcal{W}$, which I mentioned in the question. I don't see why it is sufficient and it is certainly not necessary. Take my example $A \to B \to C \to D$ and add a long zigzag of weak equivalences between $A$ and $B$. Then there is a zigzag of weak equivalences between every pair of objects, but $\mathcal{W}$ is not saturated. $\endgroup$
    – Valery Isaev
    Commented Jul 28, 2019 at 15:49

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