6
$\begingroup$

A cofibration category is saturated if it satisfies the following equivalent conditions:

  • Every map which becomes an isomorphism in the homotopy category is already a weak equivalence.
  • The weak equivalences are closed under retracts.
  • The weak equivalences satisfy the two-out-of-six property.

The second condition is one of the axioms of a model category, and most categories with weak equivalences that arise in practice satisfy the last two properties more or less by construction. In fact, I realized I don't know any examples of cofibration categories which are not saturated. So the question is:

What is an example of a non-saturated cofibration category? And, what are its maps which become isomorphisms in the homotopy category that are not already weak equivalences?

I think that "finite spaces" (in a suitable sense) with simple homotopy equivalences as the weak equivalences is supposed to form such an example; but I don't know where to find the details.

Edit: By a "cofibration category" I mean any of

  • a "catégorie dérivable à droite" in the sense of Cisinski

  • a "precofibration category" in the sense of Radulescu-Banu

  • a "cofibration category" in the sense of Baues, without the axiom on fibrant models (though I don't see why this axiom should force the category to be saturated)

I'm not picky about exactly which axioms assume which objects are cofibrant, but I do want the weak equivalences to satisfy two-out-of-three.

$\endgroup$

1 Answer 1

6
$\begingroup$

In chapter 2 of "Spaces of PL Manifolds and Categories of Simple Maps" Waldhausen, Jahren, and Rognes show that finite simplicial sets with injective maps as cofibrations and surjective maps with contractible point inverses as weak equivalence, forms a cofibration category.

This category is not saturated. This is because all homotopy equivalences are inverted in the homotopy category since the projection $X \times I \rightarrow X$ is simple, and this forces the two inclusions $X \rightarrow X \times I$ to be equal in the homotopy category. Since there are many homotopy equivalences that are not surjective (not even taking into account Whitehead torsion), there are many maps inverted that are not weak equivalences.

$\endgroup$
4
  • 2
    $\begingroup$ ... and this notion of weak equivalence (simple maps) is a relevant one because it leads to Hatcher's Higher simple homotopy theory and Waldhausen's stable parametrized h-cobordism theorem, connecting the algebraic K-theory of spherical group rings to h-cobordism spaces, concordance/pseudoisotopy spaces and automorphism groups of manifolds $\endgroup$ Commented Feb 7, 2021 at 20:31
  • 1
    $\begingroup$ Thanks, this is an interesting example. However, I think it's not quite what I'm looking for, because the simple maps do not satisfy the two-out-of-three condition either. I'll edit the question to clarify what I mean by a cofibration category. $\endgroup$ Commented Feb 10, 2021 at 14:12
  • $\begingroup$ @ReidBarton Oh gosh -- thinking back, this has been an endless source of confusion to me. It's widely known that simple homotopy equivalences give a good example of "equivalence-like maps" which don't have all the properties you'd expect, but exactly which properties they fail to have -- 2/3? 2/6? saturation? is something I have never been 100% sure of. Perhaps part of the issue is that it depends on exactly which category / which weak equivalences of the "simple" flavor you're talking about? $\endgroup$ Commented Feb 10, 2021 at 14:35
  • $\begingroup$ @ReidBarton Good point; if you expand to allow homotopy equivalences with zero torsion, then we do get the two-out-of-three condition. I'm not sure how it affects the various pushout properties. Probably is an explicit computation using cell decompositions. $\endgroup$ Commented Feb 10, 2021 at 14:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .