Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
Take a look at this question mathoverflow.net/questions/23269/… –  ChrisLazda Jun 7 '13 at 10:02
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 7 '13 at 10:25
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 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 8 '13 at 13:10
@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 8 '13 at 16:56
@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 8 '13 at 22:18

2 Answers 2

up vote 9 down vote accepted

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.

share|improve this answer
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 7 '13 at 13:19
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 7 '13 at 14:34
This note uf-ias-2012.wikispaces.com/file/view/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 7 '13 at 16:47
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 7 '13 at 17:07
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 7 '13 at 17:29

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 http://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.

share|improve this answer
Do you really want to call it a "category with weak equivalences" if not every identity map is a weak equivalence? –  Tom Goodwillie Jun 7 '13 at 22:11
Inna means to include all isomorphisms in A (she said as much in my office an hour ago). –  Peter May Jun 7 '13 at 22:47
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 7 '13 at 23:05
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 8 '13 at 1:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.