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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

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Take a look at this question… – 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 . 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, 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 8 '13 at 20:59
@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
up vote 11 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.

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That's a very surprising example. – Omar Antolín-Camarena Jun 7 '13 at 12:51
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… 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

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 In particular, model categories can determine whether you're empty or nonempty, but they can't differentiate between different set sizes.

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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

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