# Are strict pushout squares in Cat exact squares?

Let $C,C',D\in \operatorname{Cat}$, and consider the following strict pushout square

$$\begin{matrix} C&\overset{f}{\to} &D^{op}\\ \downarrow^\pi&\swarrow&\downarrow^{\iota_2}\\ C'&\underset{\iota_1}{\to}&D'^{op} \end{matrix}$$

where $D':=C'^{op}\coprod_{C^{op}} D$ and the 2-cell (denoted by $\swarrow$) is the identity.

Is this square exact (or at least $Set$-exact) in the sense that the canonical Beck-Chevalley transformation

$$\pi_!f^*\to \pi_!f^*\iota_2^*\iota_{2!} = \pi_!\pi^*\iota_1^*\iota_{2!}\to \iota_1^*\iota_{2!}$$

is an isomorphism?

Note that $(-)_!$ denotes the left kan extension (that is, the left adjoint of the canonical pullback $(-)^*$ (looking at diagram categories over an arbitrary base category $M$ (resp. looking at diagram categories into $Set$))).

If it is true, does it hold for simplicial categories (resp. for the diagram categories targeting $sSet$)?

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Does strict pushout mean what I think it means? If so, then no. Take C' and D to each be the category with one morphism, and take C to have two objects and two morphisms. – Tom Goodwillie Jan 23 '11 at 14:32
Yes, and correct!. – Harry Gindi Jan 23 '11 at 17:17

No, it's false for general categories; it's true for groupoids, however.

A crucial test case for Beck-Chevalley is often called (by categorical logicians especially) Frobenius reciprocity; it's the case where one forms the pullback

$$\begin{matrix} C & \overset{\Delta}{\to} & C \times C \\ \downarrow^\Delta & & \downarrow^{1 \times \Delta} \\ C \times C & \underset{\Delta \times 1}{\to} & C \times C \times C \end{matrix}$$

and asks whether the canonical map $\Delta_! \Delta^\ast \to (\Delta \times 1)^\ast (1 \times \Delta)_!$ is invertible. You probably recognize this sort of thing from other contexts; in the monoidal category as opposed to monoidal bicategory context, where this arrow is replaced by an equation, this is the defining equation for a Frobenius monoid, the one which relates a monoid structure to a comonoid structure. It also crops up in representation theory, and in any context where one has ambidextrous adjunctions (where one functor is simultaneously a left and right adjoint to another). Most famously, it arises in the monoidal category of 2-cobordisms between oriented compact 1-dimensional manifolds. (You can see the cobordisms by drawing the string diagrams for both sides of the equation, where you get the famous "I = H" equation, and then thickening it by taking the boundary of small $\varepsilon$-neighborhoods of these string diagrams in 3-space.)

Anyway, it's fun to calculate the canonical transformation directly and see what happens. My preferred formalism is composition of profunctors, because it basically behaves like matrix multiplication. On the "I" side of the equation, one is composing the profunctor

$$C^{op} \times C \times C \to Set: (d, c_1, c_2) \mapsto (C \times C)(\Delta d, \langle c_1, c_2 \rangle)$$

with the profunctor

$$(C \times C)^{op} \times C \to Set: (c_3, c_4, d') \mapsto (C \times C)(\langle c_3, c_4 \rangle, \Delta d')$$

(This should be liked to multiplying a $1 \times 2$ matrix by a $2 \times 1$ matrix.) Officially, the result is a profunctor

$$(C \times C)^{op} \times (C \times C) \to Set$$

$$(\langle c_3, c_4 \rangle, \langle c_1, c_2 \rangle) \mapsto \int^d (C \times C)(\langle c_3, c_4 \rangle, \Delta d) \times (C \times C)(\Delta d, \langle c_1, c_2 \rangle)$$

where the coend can be rewritten

$$\int^d C(c_3, d) \times c_4, d) \times C(d, c_1) \times C(d, c_2)$$

This may look complicated, but it's not. The elements of the coend at a particular 4-tuple are equivalence classes of diagrams which look like this:

$$\begin{matrix} c_3 & & & & c_4 \\ & \searrow^{\phi_3} & & \swarrow^{\phi_4} & \\ & & d & & \\ & \swarrow^{\phi_1} & & \searrow^{\phi_2} & \\ c_1 & & & & c_2 \end{matrix}$$

where the equivalence relation is generated by stipulating that two 4-tuples (one factoring through $d$, another through $d'$) are equivalent if both arise from a diagram of shape

$$\begin{matrix} c_3 & & & & c_4 \\ & \searrow & & \swarrow \\ & & d & & \\ & & \downarrow & & \\ & & d' & & \\ & \swarrow & & \searrow & \\ c_1 & & & & c_2 \end{matrix}$$

(which looks like an "I"). Now for the "H" side; it's actually easier because there are various Yoneda lemma reductions (which I'll skip) that allow one to boil down the profunctor composite to

$$(C \times C)^{op} \times (C \times C) \to Set$$

$$(\langle c_3, c_4 \rangle, \langle c_1, c_2 \rangle) \mapsto C(c_3, c_1) \times C(c_4, c_1) \times C(c_4, c_2)$$

whose elements are diagrams in a kind of "H" shape:

$$\begin{matrix} c_3 & & c_4 \\ \downarrow & \swarrow & \downarrow \\ c_1 & & c_2 \end{matrix}$$

And now you should be able to see that the two profunctor composites are definitely distinct in general. For example, in the I-composite, $c_3$ is connected by a morphism to $c_2$, but not so in the H-composite. Of course, in groupoids they are connected because we can travel along inverses, and the isomorphism between the diagram classes is easy to establish.

There are probably less elaborate calculations which arrive at the same point, but it is often the case (for example in certain simple type-theoretic contexts) where verification of the Beck-Chevalley condition boils down to Frobenius reciprocity. I may come back to edit if I think of a clearer way to say it, but anyway Frobenius is often a crucial test case, as I said at the beginning.

Finally: Harry, please don't delete this question. Very often I find that you have deleted questions that I was considering or preparing an answer for. In the present case, the question is good, and I hope you find my answer helpful.

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Oh for God's sake, I wrote the answer for pullbacks, not pushouts! I'm very sorry. I'll leave the answer here for a while, but may come back to delete it. – Todd Trimble Jan 23 '11 at 15:37
Dear Todd, it's false anyway, and this answer is really good, so I'm accepting it (therefore precluding you from deleting it). How do you like that? – Harry Gindi Jan 23 '11 at 16:09
Um, I like it fine (thanks)! If nothing else, this answer could be transported to the nLab at some appropriate spot regarding Beck-Chevalley or Frobenius. – Todd Trimble Jan 23 '11 at 16:22