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By "saturated class of morphisms" in a category $\mathcal{C}$, I mean a subcategory $\mathcal{W} \subset \mathcal{C}$ such that the image of $\mathcal{W}$ in $\mathcal{C}[\mathcal{W}^{-1}]$ consists of exactly the isomorphisms. By the universal property of $\mathcal{C}[\mathcal{W}^{-1}]$, this is equivalent to saying that $\mathcal{W}$ is exactly the preimage of the class of isomorphisms under some functor.

Saturated classes of morphisms are all over the place, but (in my very limited experience) I've never seen a saturated class which was not defined as the preimages of the isomorphisms under some explicit functor, so that the fact that the class is saturated follows immediately from the definition. (E.g. the weak equivalences in $\mathsf{Top_{*,\mathrm{conn}}}$ are the preimages of the isomorphisms under the functor $\prod_n \pi_n$; the quasi-isomorphisms of chain complexes are the preimages of the isomorphisms under the homology functor; I also include under this umbrella any class of morphisms defined as the saturation of some smaller class in light of the functor $\mathcal{C} \to \mathcal{C}[\mathcal{W}^{-1}]$.) I'd like an example which is naturally defined in a way that does not make the saturation obvious.

I ask this because in the definition of a model category, we hypothesize that the weak equivalences satisfy 2-out-of-3 - a consequence of saturation - and then prove that the weak equivalences are saturated. Similarly, in a homotopical category, we hypothesize that the weak equivalences satisfy 2-out-of-6 - another consequence of saturation - and then prove in light of additional hypotheses that the weak equivalences are saturated. What does this extra work buy us? If saturation is obvious in all interesting examples, why not just hypothesize that the weak equivalences are saturated from the get-go? Is the motivation purely to keep the hypotheses elementary, or to avoid reliance on the existence of a possibly non-locally-small category like $\mathcal{C}[\mathcal{W}^{-1}]$?

A couple of related questions would be: what is an example of a class of morphisms which satisfies 2-out-of-3 or 2-out-of-6 but is not saturated?

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  • $\begingroup$ I've just noticed that the basepoint-dependence of the homotopy groups means that there isn't one single functor of homotopy groups out of $\mathsf{Top}$ which can be used to make the saturation of weak equivalences in $\mathsf{Top}$ obvious in the way I described above. However, I would still maintain that saturation of the weak equivalences in $\mathsf{Top}$ follows almost immediately from the definitions, and so does not serve as an answer to my question. $\endgroup$
    – Tim Campion
    Commented Dec 30, 2014 at 19:26
  • $\begingroup$ Maybe this is not so obvious. Is this really the motivating example? If so, I find it surprising because unlike the difference between stable and unstable phenomena, basepoint issues are usually regarded as formalities: I've never seen so much hinge on basepoints before. $\endgroup$
    – Tim Campion
    Commented Dec 30, 2014 at 21:55
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    $\begingroup$ I think that simple homotopy equivalences satisfy 2-out-of-3, but are not saturated in the category of finite complexes. $\endgroup$
    – Ben Wieland
    Commented Dec 31, 2014 at 2:05
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    $\begingroup$ In my opinion, the widespread belief that basepoint issues are formalities is one of the more insidious illusions fostered by traditional algebraic topology. (-: Other contexts in which the subtlety of basepoints is noticeable include equivariant homotopy theory and fixed-point theory; in both cases a choice of consistent basepoint would have to be a fixed point, but often the existence or nonexistence of a fixed point is one of the questions we want to deploy algebraic topology to answer! $\endgroup$
    – Mike Shulman
    Commented Dec 31, 2014 at 4:45
  • $\begingroup$ I wonder if one could use the fact that at least for "many" spaces, a map $X \to Y$ is a weak homotopy equivalence iff the induced map $X \times \mathbb R^\infty \to Y \times \mathbb R^\infty$ is a homeomorphism to give a direct proof that the weak homotopy equivalences are saturated. I would find this very amusing if it worked! $\endgroup$
    – Tim Campion
    Commented Feb 9, 2021 at 22:37

1 Answer 1

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In the canonical model structure on $\omega\mathrm{Cat}$, the weak equivalences are not defined as preimages of isomorphisms under any functor, and are not even even closely related to any such preimage the way weak equivalences of unbased spaces are. And indeed, one of the hardest parts of the proof of the model structure is proving the 2-out-of-3 property.

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  • $\begingroup$ Interesting. I thought the idea of a "canonical model structure" on $n$-categories was that the weak equivalences should be "the evident generalization of equivalences of categories", and the naturalness of this choice should suggest that saturation should be obvious. As I think about it, maybe "the evident generalization" isn't so obvious when $n = \omega$. $\endgroup$
    – Tim Campion
    Commented Dec 31, 2014 at 18:16
  • $\begingroup$ Why would "naturalness" imply that saturation should be obvious? Even in the case n=1, I can't think of any obvious functor giving the equivalences of categories as the preimage of the isomorphisms. $\endgroup$
    – Mike Shulman
    Commented Jan 1, 2015 at 21:36
  • $\begingroup$ In the case $n=1$, it's obvious that when you mod out by the congruence identifying isomorphic functors, you invert exactly the equivalences of categories. I guess I assumed that a similar observation should work for every value of $n$. In some sense, this good control over equivalences is really the point of the homotopy hypothesis, isn't it? $\endgroup$
    – Tim Campion
    Commented Jan 2, 2015 at 7:54
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    $\begingroup$ Well, I think it's not quite obvious, it relies on the existence of an interval; but you're right that that's a fairly easy argument. It doesn't work for higher categories, though, because in that case in order to define an "equivalence of categories" by the existence of a quasi-inverse, you need to allow weak functors; but at least with algebraic models for higher categories, the weak functors don't generally form a well-behaved 1-category that you can put a model structure on. $\endgroup$
    – Mike Shulman
    Commented Jan 2, 2015 at 18:01

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