## The canonical model structure on Cat November 16, 2012

Posted by Chris Schommer-Pries in Algebraic Topology, Category Theory.

In this post I want to describe the following result, which I think is pretty neat and should be more widely known:

Theorem: On the category of (small) categories there is a unique model structure in which the weak equivalences are the equivalences of categories.

This unique model structure is of course the so-called “canonical” model structure of André Joyal and Myles Tierney. (The fact that it is the unique one with these weak equivalences lends credence, I think, to using the name “canonical”). It is proper, cartesian, simplicial, combinatorial, and every object is both fibrant and cofibrant. I first learned of this uniqueness result from Steve Lack’s comments on this MathOverflow question, though there were some details left to fill in. I hope to do that here.

Below I will give an elementary proof of the above theorem, partly so I have it written down somewhere for future reference. Charles Rezk has a nice write-up of this model structure, and I will start by describing it. It consists of:

• the canonical cofibrations, which are those functors of categories which are injective on objects,
• the canonical acyclic cofibrations, which are those equivalences of categories which are injective on objects (these are necessarily injective on morphisms too),
• the canonical acyclic fibrations, which are those equivalences which are surjective on objects, and
• the canonical fibrations, which are the “iso-fibrations“. These are the functors $f:X \to Y$ such that for any $x \in X$ and any isomorphism $\alpha: f(x) \cong y$ in Y, then there exists an isomorphism in X which maps to $\alpha$ under f.

Let’s fix some notation which will be useful later.

• Let E be the free walking isomorphism. This is the contractible category with two objects. (A category is contractible if it is equivalent to the terminal category pt).

There is an equivalence $pt \to E$, which includes the terminal category as one of the objects. The canonical fibrations are precisely those maps which have the right lifting property with respect to this functor.

The Proof:

Now let us suppose that we have a model category structure on the category of categories in which the weak equivalences are the equivalences of categories. Such a structure consists of certain classes of fibrations and cofibrations. Our goal is to show that these must be the canonical fibrations and canonical cofibrations above.

The proof will use some basic properties about model categories:

1. The cofibrations and acyclic cofibrations are closed under retracts, compositions, and pushouts along arbitrary maps.
2. The acyclic cofibrations are precisely those cofibrations which are also (weak) equivalences. The acyclic fibrations are also weak equivalences.
3. The fibrations have the right lifting property with respect to the acyclic cofibrations and the acyclic fibrations have the right lifting property with respect to the cofibrations.

It also rests on a

Key Fact: every equivalence class of objects contains a fibrant representative and a cofibrant representative.

(Recall that an object is cofibrant if the unique map from the initial object (the empty category, in this case) is a cofibration. Dually, and object is fibrant if the unique map to the terminal object is a fibration.)

What do these facts tell us?

Trivial Lemma: The inclusion $\emptyset \to pt$ is a cofibration.

Proof: Since each equivalence class of categories contains a cofibrant representative, we know that there exists some cofibration  $\emptyset \to A$ for a non-empty category A. The desired map is a retract of this, hence also a cofibration. ◊

The acyclic fibrations must have the right lifting property with resect to all cofibrations, hence with respect to this map  $\emptyset \to pt$. This means the acyclic fibrations must be surjective on objects. Since they are equivalences too, this implies the following consequences.

1. the acyclic fibrations are a subset of the canonical acyclic fibrations, hence
2. the cofibrations contain the canonical cofibrations, hence
3. the acyclic cofibrations contain the canonical acyclic cofibrations, and hence
4. the fibrations are a subset of the canonical fibrations.

This means we are half-way there. We must rule out the possibility that there could be more cofibrations. (This includes the case there are more acyclic cofibrations).

A Somewhat Less Trivial Lemma: If the cofibrations contain a map which is not a canonical cofibration (i.e. fails to be injective on objects), then the following map is also a cofibration (hence an acyclic cofibration as it is an equivalence):

$E \to pt$.

Proof: Suppose that we have a functor $A \to B$ which is a cofibration but not injective on objects. Then there exists at least one pair of objects in the source category which map to the same object in the target category. Call these objects x and y, and their image p.

The cofibrations are closed under pushouts along arbitrary maps and this allows us to alter this map to make a new cofibration. First note that a functor from a category to E is the same as a partition of its objects into two disjoint sets. Thus we may choose a functor $A \to E$ which separates x and y. We may form the pushout along this map to get a new cofibration:

$E \to E \cup_{A} B =: X$.

At this point we would like to form a retract onto the desired morphism. The problem is that this might not be possible as the image in X of the non-trivial isomorphism in E might fail to be an identity. If that is the case we will not be able to retract onto the desired map.

However cofibrations are also closed under composition. Let $X^\delta$ be the contractible category with the same objects as X. There is a unique functor

$X \to X^\delta$

which is the identity on objects. Since it is injective on objects it is a canonical cofibration, hence this map must also be a cofibration. Composing gives us a new cofibration:

$E \to X^\delta$

and now this retracts onto the desired map. ◊

Whew! That was the hardest part of the proof. Glad that’s over.

So we have learned that if the cofibrations contain more than just the canonical cofibrations, then they also contain $E \to pt$, which is then necessarily an acyclic cofibration. This leads us to define the following:

Definition: A category is gaunt if every isomorphism is an identity.

Gaunt categories are what you get when you take category theory and strip away the fleshy meat of topology (in this case 1-types or groupoids).  We also have this:

Another Trivial Lemma: If the acyclic cofibrations contain the map $E \to pt$, then the fibrant objects are necessarily gaunt.

The proof is just unraveling definitions.  We also have a

Trivial Observation: Not every category is equivalent to a gaunt category (e.g. non-trivial groupoids).

But now we see a contradiction emerge. For a model structure, every equivalence class of objects must contain a fibrant representative. If the cofibrations contain more that the canonical cofibrations, then $E \to pt$ is a cofibration and hence  the fibrant objects are gaunt. The equivalence class of, say, a non-trivial groupoid cannot be thus represented. We are thus led to conclude:

Theorem: There is precisely one model structure on the category of categories in which the weak equivalences are the equivalences of categories. It is the canonical model structure.

1. Charles Rezk - November 16, 2012

Nifty!

Question: If we add in $E\to pt$ to the acyclic cofibrations (and thus to the cofibrations), will we still get a model category? (Obviously, with more weak equivalences?)

2. Chris Schommer-Pries - November 16, 2012

Hi Charles!

Great question!

I think the answer is yes, but I haven’t completely thought through all the details.

I think I convinced myself that if you add the map $E \to pt$ to the canonical cofibrations, then you get that *every functor* is a cofibration. This would mean that the acyclic fibrations are just the isomorphisms.

This would also mean that the question of whether this gives a model structure is the same as whether the acyclic cofibrations are closed under the 2-out-of-3 property.

Now I know there is a “gauntification” functor $L^G$ (a localization) from categories to gaunt categories. It does not just send isomorphisms to identities; it can be very destructive.

I think the new acyclic cofibrations (= new weak equivalences) are precisely those functors $A \to B$ such that $L^G(A) \to L^G(B)$ is an isomorphism. Note that any equivalence of gaunt categories is automatically an isomorphism. If this is the case, then they clearly satisfy 2-out-of-3 and so this would form a model category.

At the homotopy theory level, this should be a sort of (left?) localization to the gaunt categories, but it is a little weird. For example it is not any kind of Bousfield localization.

This also reminds me of how there are precisely 9 model structures on the category of sets. I wonder how hard it would be to do a similar classification of all the model structures on Cat?

I have my own question for you regarding this post… in your note on the canonical model structure you mention that Joyal and Tierney constructed it first. Did you also discover it independently? What do you know of its history?

3. Charles Rezk - November 17, 2012

I used to think I discovered it independently, but I’m not sure. The reason I’m not sure is that I know that as a grad student I went through a paper by Bousfield, called “Homotopy Spectra Sequences and Obstructions”, in an appendix of which he uses the fact that Groupoid has a model category structure. Bousfield’s paper may have lodged in my mind somewhere. (The construction of the model category for Groupoid is virtually identical to that for Cat.)

Bousfield refers to a 1978 Bulletin article by Anderson, “Fibrations and Geometric Realizations”, where Anderson mentions the Groupoid model category on p.783. That is the earliest case I know.

4. Mike Shulman - November 18, 2012

Very nice! Clearly the same argument works for the category of groupoids. What about the category of 2-categories, with the biequivalences? (Maybe this was addressed in the cited MO question — I can’t get to it right now.)

5. Chris Schommer-Pries - November 19, 2012

Hi Mike,

That is also a good question.

Steve Lack would be the person to ask about the 2-categorical case, but here are a few observations. Any object equivalent to a cell retracts onto that cell. Using this you can make a similar argument as above to show any potential class of cofibrations contains the usual generating cofibrations. The above argument then certainly shows that any choice of cofibrations must be injective on objects. I am not sure that as written it can be adapted to show they are “relatively free” on 1-morphisms too. I am somewhat doubtful.

However there is a different argument which I think does the job, which is also more in line with Steve’s comments of the math overflow question. First, if I remember correctly, the Lack model structure on 2-Cat is still proper. Since this is a property of the weak equivalences alone (a non-elementary result), it means that any potential class of cofibrations must preserve arbitrary weak equivalences under pushouts.

So what you have to do is show that if you have a functor which is not relatively free, then there is an equivalence such that when you form the pushout you fail to get an equivalence. Actually after a couple of back of the envelope calculations it is not so obvious to me how to show this in general. Hmmmm….

6. Omar Antolín-Camarena - December 18, 2012

About classifying the model structures on the category of categories: it’s probably complicated since, to begin with, there should be lots of homotopy categories; among others you probably get the homotopy category of $n$-types for each $n \ge -2$ (on the category of sets you get these for $n$=-2, -1 and 0), and I would guess that you get each from lots of model structures.

(By the way, I did actually get around to writing up the nine model structures for sets after we talked about them: http://www.math.harvard.edu/~oantolin/notes/modelcatsets.html)

Sorry comments are closed for this entry