11

I'm only considering complete and cocomplete categories. A pair $(\mathfrak{X} , \mathfrak{W}) $ is, by definition, a category with weak equivalences if $ \mathfrak{X} $ is a category and $ \mathfrak{W} $ is a subcategory satisfying the $2$ out of $3$ axiom.

I was wondering if there are nice examples of categories with weak equivalences for which there is no model structure.

Thank you

David White
  • 29,779
Fernando
  • 865
  • 4
    Take a look at this question

    http://mathoverflow.net/questions/23269/non-examples-of-model-structures-that-fail-for-subtle-surprising-reasons

     – ChrisLazda Jun 07 '13 at 10:02
  • 2
    Take W to be the empty subcategory. There are also nontrivial examples. Actually, I'd say that any random choice would do. – Fernando Muro Jun 07 '13 at 10:25
  • 1
    As Fernando Muro says, examples are abundant, the generic example of a category with weak equivalences is not a model category. For a list of classes of counterexamples see http://ncatlab.org/nlab/show/category+with+weak+equivalences . For instance non-model categories with weak equivalences go by names such as "relative category", "category of fibrant objects", "cofibration category", "Waldhause category" etc. – Urs Schreiber Jun 08 '13 at 13:10
  • 1
    @Fernando Muro, I was not looking for trivial examples. Really, I forgot to restrict my question: I meant "weak equivalences" including isomorphisms. My original wondering was about a "real" localization problem which cannot be "solved" using model category theory. So, I was asking for a problem in which we have "nice" weak equivalences (including isomorphisms) which cannot be part of a model structure. Sorry any lack of precision, I was wrong in trying to avoid precision (using the word nice instead). For my lucky, professor May answered what I was looking for. – Fernando Jun 08 '13 at 16:56
  • Anyway, thank you for commenting. It was somehow helpful! – Fernando Jun 08 '13 at 16:57
  • @Urs Schreiber, thank you for the links! It seems to be very helpful for what I'm looking for (not only related to this question)! – Fernando Jun 08 '13 at 16:59
  • @ChrisLazda, Thank you for the related question. Although it's a different question, I wasn't aware of this mathoverflow question. I will take a look the examples and comments to see if there is something close to an answer of my problem. – Fernando Jun 08 '13 at 17:05
  • 2
    @Fernando, if your background interest is about localization problems, you'll like an example I first saw in Michael Weiss' "Hammock localization in Waldhausen categories". Weak equivalences are simple homotopy equivalences. They are not the weak equivalences of a model category because their saturation are all homotopy equivalences, and not all homotopy equivalences are simple. – Fernando Muro Jun 08 '13 at 20:59
  • 3
    @Fernando, yet another example: quasi-isomorphisms in the category of chain complexes over Freyd's strange abelian category without enough injectives or projectives. They fail for set-theoretical reasons: in this case the localization doesn't exist. – Fernando Muro Jun 08 '13 at 22:18
  • @Fernando Muro, Nice! Thank you very much for helping. Now, I am going to take a look carefully at your first example! Yes, my interest was to understand where model category theory fails for these kind of problem. THank you – Fernando Jun 09 '13 at 11:14

3 Answers3

20

A very interesting example: consider semi-simplicial sets (alias $\Delta$-sets). These are simplicial sets without degeneracies, and there is an ``adjoin degeneracies'' functor from semi-simplicial sets to simplicial sets that is left adjoint to the evident forget degeneracies functor. One can compose this with geometric realization or one can define geometric realization of semi-simplicial sets directly. Define a weak equivalence of semi-simplicial sets to be a map whose geometric realization is a homotopy equivalence (weak equivalence is the same since these are CW complexes). Since it is a presheaf category, the category of semi-simplicial sets is bicomplete, and I've described the obvious weak equivalences, which satisfy the two-out-of-three property. Matthew Thibault convinced me that this truly natural category with weak equivalences does not admit a model structure for any choice of cofibrations or fibrations.

EDIT: Dylan, if you believe the comments already posted, you know the proof. According to Tom, the acyclic fibrations have to be the isomorphisms. But then all maps are cofibrations, since they are the maps with the LLP wrt the acyclic fibrations. If you believe Karol that it cannot be the case that all monomorphisms are cofibrations, or just that $\ast$ cannot be cofibrant, you already have a contradiction to such a model structure.

Peter May
  • 30,168
  • 1
    That's a very surprising example. – Omar Antolín-Camarena Jun 07 '13 at 12:51
  • 1
    Can you give a little indication of the argument for why this does not have a model structure? (This is a neat example!) – Dylan Wilson Jun 07 '13 at 13:19
  • 10
    If the class of weak equivalences is as Peter says, or smaller, then every trivial fibration $X\to Y$ is an isomorphism. Proof by induction that $X_n\to Y_n$ is bijective: Given $y:\Delta^n\to Y$, pull back the map to get a trivial fibration $y^\ast X\to \Delta^n$. An object mapped to $\Delta^n$ is at most $n$-dimensional (this is really the crux), and if the map to $\Delta^n$ is an isomorphism on the $(n-1)$-skeleton then it cannot be a weak equivalence without being an isomorphism. – Tom Goodwillie Jun 07 '13 at 14:34
  • 1
    This note http://uf-ias-2012.wikispaces.com/file/view/semisimplicialsets.pdf/421930564/semisimplicialsets.pdf claims that there is such a model structure. What's certainly true is that there is no model structure where all monomorphisms are cofibrations or even where all objects are cofibrant. – Karol Szumiło Jun 07 '13 at 16:47
  • 3
    The weak equivalences in that reference are a priori different from those I prescribed (which are the most natural ones). It uses a right adjoint rather than the left adjoint of the forgetful map to create the weak equivalences. – Peter May Jun 07 '13 at 17:07
  • 1
    Originally I thought that these weak equivalences coincide with the classical ones. But it turns out that the image of any finite dimensional semisimplicial set under the right adjoint is empty. So weak equivalences created by this functor are indeed very different. Can you give an argument for why there is no model structure with classical weak equivalences? I can see that there is no such model structure if the one-point semisimplicial set is cofibrant. – Karol Szumiło Jun 07 '13 at 17:29
  • 1
    @Karol: See my comment above. Any such model structure would be such that every trivial fibration is an isomorphism, therefore every map is a cofibration. – Tom Goodwillie Jun 07 '13 at 20:55
  • @Tom: you are absolutely right, for some reason I missed your comment. In that case perhaps the quickest way to finish off the argument is to say that if every map is a cofibration, then every pushout is a homotopy pushout which is absurd. – Karol Szumiło Jun 08 '13 at 10:09
  • Alice in Wonderland here: I edited my answer to complete the proof 20 hours ago, before the last two comments. – Peter May Jun 08 '13 at 15:42
  • Nice! Thank you! Although my question wasn't formulated precisely enough, professor Peter May answered exactly what I was looking for! Thank you very much! – Fernando Jun 08 '13 at 17:09
  • A published reference for the non-existence of this model structure is page 2 (just after Theorem 1.2) of the paper Model categories with simple homotopy categories by Jean-Marie Droz and Inna Zakharevich. Probably there are other references, but I didn't spot any in the published works of Thibault. – David White Jan 17 '24 at 01:52
8

Here is another (more combinatorial) answer. Let $\kappa$ be any (nice) cardinal, and let $\mathcal{A}$ be the union of the full subcategory of $\mathbf{Set}$ consisting of all sets of cardinality less than $\kappa$ and the subcategory of all isomorphisms in $\mathbf{Set}$. $\mathcal{A}$ is closed under retracts and 2-of-3, but it is NOT the subcategory of weak equivalences of a model structure on $\mathbf{Set}$. We can see this just by listing all of the (nine) model structures on $\mathbf{Set}$, and noting that the subcategories of weak equivalences are

  • all morphisms
  • all isomorphisms
  • all morphisms between nonempty sets (plus the identity on $\emptyset$)

(See Tom Goodwillie's answer to https://mathoverflow.net/questions/29653.) In particular, model categories can determine whether you're empty or nonempty, but they can't differentiate between different set sizes.

Community
  • 1
Inna
  • 1,025
  • 3
    Do you really want to call it a "category with weak equivalences" if not every identity map is a weak equivalence? – Tom Goodwillie Jun 07 '13 at 22:11
  • 5
    Inna means to include all isomorphisms in A (she said as much in my office an hour ago). – Peter May Jun 07 '13 at 22:47
  • 1
    Inna's example is a generalization of mine, which was the case $\kappa=0$. As Tom says, it's easy to see they are not the weak equivalences of a model structure since they do not contain all identities. Actually, as the question is phrased, there are tons of examples. Prof. May's answer above is too interesting and deserves a better question! – Fernando Muro Jun 07 '13 at 23:05
  • 3
    Yes, I meant that the subcategory of weak equivalences is the subcategory of all isomorphisms and also all morphisms in A. Obviously, without all isomorphisms it can't be the subcategory of weak equivalences, but for more trivial reasons. – Inna Jun 08 '13 at 01:17
2

This week, we learned that another example is the category of simplicial sets, and the class of weak equivalences the simplicial homotopy equivalences. All credit to Tom Goodwillie, Tim Campion, and Tyrone for figuring out why such a model structure cannot exist.

I made this answer CW, since I'm really just advertising answers written by two other people.

David White
  • 29,779