The n-Category Café

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?