## November 7, 2011

### Productive Homotopy Pullbacks

#### Posted by Mike Shulman

The last time I talked about derivators, we ended up making an interesting comparison between derivator techniques and quasicategory techniques. In particular, we looked at a proof of the pasting law for homotopy pullbacks using derivators (presented in that post) and the analogous proof for quasicategories from Higher Topos Theory (partially described here). We thought at the time it would be nice to continue with further comparisons, but got sidetracked.

Along those lines, here’s another little fact that was easy for me to prove with derivators. Anyone care to try their hand at a proof using quasicategories, so we can compare them?

Here’s the setup. In my last post I mentioned that there are some squares built out of cartesian products which are always pullback squares. Todd Trimble has a nice discussion of these here; he calls them productive pullbacks. I also mentioned that some, but not all, of these squares are actually always homotopy pullbacks. Let’s prove that.

The first one is $\array{ A\times C & \xrightarrow{f\times 1}& B\times C \\ ^{1\times g} \downarrow & & \downarrow^{1\times g}\\ A\times D & \xrightarrow{f\times 1} & B\times D}$ For this, it suffices to prove that

1. The pointwise product of pullback squares is a pullback square;
2. For any $f\colon A\to B$, the square $\array{ A & \xrightarrow{f} & B \\ \downarrow & & \downarrow \\ A & \xrightarrow{f} & B }$ is a pullback; and
3. The transpose of a pullback square is a pullback square.

I’m going to say “pullback” instead of “homotopy pullback” from now on. Also, in case you’ve forgotten, this page contains the characterization of homotopy exact squares, the main tool we use for computing in a derivator, as well as some important examples.

Let $\Box$ denote the free-living commutative square, and $R$ its lower-right corner (the free-living cospan), with inclusion functor $r\colon R \to \Box$. In a derivator, a pullback square is a $\Box$-diagram which is isomorphic to $r_\ast$ of something.

Now since $r$ commutes with the “transpose” automorphisms of $R$ and $\Box$, the third point above follows immediately. For the first, we observe that the following square of functors commutes: $\array{R\sqcup R & \overset{}{\to} &R \\ \downarrow && \downarrow\\ \Box\sqcup \Box& \underset{}{\to} & \Box}$ where the rightward-pointing arrows are the fold maps. But right Kan extension along a fold map is precisely how we calculate products of diagrams (using the fact that a derivator $D\colon Cat^{op}\to CAT$ takes coproducts to products). Thus, it suffices to observe that right Kan extension along $r\sqcup r\colon R\sqcup R \to \Box\sqcup \Box$, when restricted to each factor, computes the right Kan extension along $r$. This is because the following square is homotopy exact: $\array{R & \overset{}{\to} & R\sqcup R\\ \downarrow && \downarrow\\ \Box& \underset{}{\to} & \Box\sqcup\Box}$ Finally, for the second point above, consider the following square, where $I$ is the interval category: $\array{R & \overset{r}{\to} & \Box\\ ^a\downarrow && \downarrow^b\\ I& \underset{1}{\to} & I}$ The downward-pointing arrows $a$ and $b$ project a commutative square to one of its axes (say, the horizontal one). This square is homotopy exact, so $r_\ast a^\ast \cong b^\ast$. But $a^\ast$ takes an arrow $f\colon A\to B$ to the cospan $\array{ & & B\\ && \downarrow^{1}\\ A& \underset{f}{\to} & B}$ while $b^\ast$ takes it to the square $\array{ A & \xrightarrow{f} & B \\ \downarrow & & \downarrow \\ A & \xrightarrow{f} & B }$ so this says that the latter is the pullback of the former, which is what we want.

The second productive homotopy pullback is $\array{A & \overset{\Delta}{\to} & A\times A\\ ^{\Delta}\downarrow && \downarrow^{1\times \Delta}\\ A\times A& \underset{\Delta\times 1}{\to} & A\times A\times A}$ To show that this is a pullback, consider the following functor $\Box^{op} \to Set$. $\array{\{0\} & \overset{}{\leftarrow} & \{1,2\}\\ \uparrow && \uparrow\\ \{3,4\}& \underset{}{\leftarrow} & \{5,6,7\}}$ The function $\{5,6,7\} \to \{1,2\}$ takes 5 to 1 and takes 6 and 7 to 2. The function $\{5,6,7\} \to \{3,4\}$ takes 5 and 6 to 3 and takes 7 to 4. Let $p\colon X \to \Box$ be the discrete fibration corresponding to this presheaf, and let $\array{Y & \overset{s}{\to} & X\\ ^q\downarrow && \downarrow^p\\ R & \underset{r}{\to} & \Box}$ be a pullback. Thus, $p_\ast s_\ast \cong r_\ast q_\ast$.

Now since $p$ and $q$ are fibrations, right Kan extension along them is computed by taking limits over the fibers. This is because pullback squares such as $\array{p^{-1}(x) & \overset{}{\to} & 1\\ \downarrow && \downarrow^x\\ X & \underset{p}{\to} & \Box}$ are homotopy exact when $p$ is a fibration. Therefore, if $F$ is an $X$-shaped diagram in a derivator, then $p_\ast(F)$ looks like $\array{F_0 & \overset{}{\to} & F_1\times F_2\\ \downarrow && \downarrow\\ F_3\times F_4& \underset{}{\to} & F_5\times F_6\times F_7.}$ The morphisms, of course, are induced from those in $X$, which contains maps $1\to 5$, $2\to 6$, $2\to 7$, $3\to 5$, $3\to 6$, and $4\to 7$. Therefore, the square we want to show to be a pullback is $p_\ast$ of the constant diagram on $A$, which is to say $p_\ast X^\ast (A)$ (where $X^\ast$ is the restriction along $X\to 1$).

Since $p_\ast s_\ast \cong r_\ast q_\ast$, it will therefore suffice to show that $X^\ast(A)$ is of the form $s_\ast$ applied to something, and the obvious choice for that something is $Y^\ast(A)$. Thus, we need to show the following square is homotopy exact: $\array{Y & \overset{s}{\to} & X\\ \downarrow && \downarrow\\ 1 & \underset{}{\to} & 1}$ The nontrivial bit of proving this is essentially the statement that $Y$ has a contractible nerve. But the geometric realization of the nerve of $Y$ is homeomorphic to the interval, since $Y$ looks like $1 \to 5 \leftarrow 3 \to 6 \leftarrow 2 \to 7 \leftarrow 4$ This completes the proof.

The final productive homotopy pullback square is $\array{A & \overset{(1,f)}{\to} & A\times B\\ ^f\downarrow && \downarrow^{f\times 1}\\ B & \underset{\Delta}{\to} & B\times B.}$ I’ll leave that one as an exercise for the reader.

Anyone care to try a quasicategorical proof?

Posted at November 7, 2011 12:27 AM UTC

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### Re: Productive homotopy pullbacks

I am still trying to grasp derivators but I think I am making a little bit of progress :)
Thanks to the n-Category Cafe I’m making it there!

Posted by: Jessica on November 8, 2011 12:23 AM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

I suspect that this comment was spam, since you linked to a commercial web site (now removed), but on the off-chance that it wasn’t, I’m glad to have helped. (-:

Posted by: Mike Shulman on November 8, 2011 12:41 AM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

Urs’s comment here still makes me chuckle.

Posted by: Tom Leinster on November 8, 2011 3:40 PM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

Hi Mike,

great to see you put derivators to work. I see around me something close to an explosion of interest in the notion of derivators (a relative explosion at least) so it is high time to have some actual applications, and be it just useful little facts as the ones above.

I wish I had more time to look into these exercises that you pose. My lazy knee-jerk reaction is: use the theorem that quasicatecorical limits in presentable quasicategories can be computed as homotopy limits in a presentation by a combinatorial model category. Use Dugger’s theorem to conclude that this can be taken to be a left Bousfield localization of the projective model structure on simplicial presheaves. In there, the relevant statements boil down to their 1-categorical analogs after choosing suitably good representatives of the diagrams.

But I guess that this is not what you want to hear :-)

Posted by: Urs Schreiber on November 9, 2011 9:00 PM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

But I guess that this is not what you want to hear :-)

Yes, that’s very much not the point! I could have done the same thing for locally presentable derivators. The point is to work with the homotopical structure directly.

Posted by: Mike Shulman on November 10, 2011 12:03 AM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

The point is to work with the homotopical structure directly.

Sure. So it would be fun to work with the homotopy type theory directly. (!)

I can’t really do it yet, but I would enjoy to start thinking about it from that angle.

First, I need to understand cartesian products in HoTT/Coq. I have seen Guillaume use them here in the formulation of the statement that HoTT is cartesian closed. It looks straightforward, but is there nevertheless somewhere I can read up on this?

Posted by: Urs Schreiber on November 10, 2011 9:28 AM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

You do realize, right, that the derivator context and the quasicategory context are both strictly more general than the model category context or the HoTT context? Model categories only model complete and cocomplete $(\infty,1)$-categories, while homotopy type theory additionally requires local cartesian closure in order to be much good.

Posted by: Mike Shulman on November 10, 2011 5:51 PM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

You do realize,

Sure. Where I mentioned model categories, it was, as I said, out of laziness, and where I mentioned HoTT it was because I happen to be interested in understanding these kinds of questions there.

On the other hand, a limit diagram in a quasicategory is detected by inducing under $Maps_{\mathcal{C}}(X,-)$ for each object $X$ a limit diagram in the quasicategory of Kan complexes. For the latter we can use the model category presentation to reason about it. Proceeding in this “hybrid” way it is easy to establish your first items 1.-3. for general quasicategories.

But then, when you said

The point is to work with the homotopical structure directly.

I was thinking to myself: well, derivators and quasicategories are also just models for homotopy theory, so it would be interesting think about how to approach the above questions with “pure” homotopical structure. It was maybe more a comment to myself than a contribution to your thread, I realize.

Posted by: Urs Schreiber on November 11, 2011 9:17 AM | Permalink | Reply to this

### Re: Productive homotopy pullbacks

Sorry if I sounded too patronizing. The Yoneda embedding argument is clever. (-:

I think one of the reasons I like to avoid arguments that proceed by strictification is that I believe the possibility of strictification is a “classicality” property, like excluded middle, choice, and Whitehead’s theorem, that will not survive internalization. There’s no reason that every object of an $(\infty,1)$-topos (say) should admit a surjection (by which I probably mean an effective epi) from a categorically discrete object (a “set”), but it seems to me that some property of that sort for the hom-spaces is necessary in order to be able to present an $(\infty,1)$-category (even $\infty Gpd$) using a model category.

Of course, this problem afflicts quasicategories as much as it does model categories; I think complete Segal spaces probably have the best chance of avoiding it (and thereby provide the most promising avenue to defining internal categories in HoTT). But I think derivators also avoid this problem, since we will always have discrete quotients ($\pi_0$).

Posted by: Mike Shulman on November 12, 2011 7:19 AM | Permalink | PGP Sig | Reply to this

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