### Exact Squares

#### Posted by Mike Shulman

One of the slogans of $(\infty,1)$-category theory is that most theorems of ordinary 1-category theory generalize to the $(\infty,1)$-world if you’re careful enough with how you phrase them. One might hope that that the *proofs* of these theorems could likewise be generalized, and this is sometimes the case, but many of the more complex tools of $(\infty,1)$-category theory don’t have much of a counterpart in the 1-categorical world, such as the study of various types of anodyne morphisms of simplicial sets.

Recently I’ve been talking, reading, and writing about derivators, which I view as a way of working with $(\infty,1)$-categories that is less dependent on technical details of any particular definition of $(\infty,1)$-categories, is more closely connected to familiar techniques of 1-category theory, and may hopefully be easier, in some absolute sense, than working directly with $(\infty,1)$-categories. However, generalizing an argument from 1-categories to derivators does usually still require rephrasing it a bit, because derivators are based around an approach to 1-categorical limits and colimits that isn’t that well known in many circles – the theory of *exact squares* of categories. So in this post I want to talk about the calculus of exact squares, mainly from a purely 1-categorical viewpoint, but I’ll add some remarks at the end about how easy it is to generalize the arguments to derivators.

In order to get used to the calculus of exact squares, the first thing we have to change is to start thinking more in terms of *Kan extensions* rather than *limits and colimits*. For instance, it’s usual to think of a *pullback* as the limit of a diagram of shape
$B \to C \leftarrow A$
(a cospan) i.e. an object $P$ equipped with a natural transformation of cospans:
$\array{ P & \to & P & \leftarrow & P\\
\downarrow & & \downarrow & & \downarrow \\
B & \to & C & \leftarrow & A }$
which is universal. From this point of view, a *pullback square* is just a square
$\array { P & \to & A \\ \downarrow & & \downarrow \\ B & \to & C}$
that we put together from some of the arrows drawn above; thus the presence of that square is “hidden” inside the natural transformation of cospans.

A different approach, however, is to give primacy to this square, rather than to the object $P$. In this approach, we start instead by defining a *pullback square* of a cospan $B \to C \leftarrow A$ to be a commutative square
$\array { P & \to & A' \\ \downarrow & & \downarrow \\ B' & \to & C'}$
equipped with a natural transformation of cospans
$\array{ B' & \to & C' & \leftarrow & A'\\
\downarrow & & \downarrow & & \downarrow \\
B & \to & C & \leftarrow & A }$
which is universal. In other words, we ask for a (pointwise) *right Kan extension* along the inclusion of the “walking cospan” into the “walking commutative square.”

In order to be sure that this gives the same notion of pullback that we’re used to, we need to know two things:

- The underlying cospan $B' \to C' \leftarrow A'$ of the pullback square should be the same as the cospan we started with, and
- The resulting universal property of the object $P$ and its projections to $A$ and $B$ should be the same.

Both of these facts are naturally expressed in the language of exact squares.

What is an exact square? Suppose we have a square in $Cat$ inhabited by a natural transformation:
$\array{ X & \overset{u}{\to} & Y \\ ^v\downarrow & ^\alpha \swArrow & \downarrow^g \\ Z & \underset{f}{\to} & W }$
Then if $M$ is some ambient category, we write $f^\ast\colon M^W \to M^Z$ for restriction along $f$, and $f_! \colon M^Z \to M^W$ for left Kan extension along $f$, and $f_\ast\colon M^Z \to M^W$ for right Kan extension along $f$, insofar as these extensions exist. The given transformation $\alpha$ induces a transformation $u^\ast g^\ast \to v^\ast f^\ast$, which has mates
$v_! u^\ast \to v_! u^\ast g^\ast g_! \to v_! v^\ast f^\ast g_! \to f^\ast g_!$
and
$g^\ast f_\ast \to u_\ast u^\ast g^\ast f_\ast \to u_\ast v^\ast f^\ast f_\ast \to u_\ast v^\ast.$
We say that the square $\alpha$ is **exact** if these two transformations $v_! u^\ast \to f^\ast g_!$ and $g^\ast f_\ast \to u_\ast v^\ast$ are isomorphisms for any $M$, insofar as the relevant pointwise Kan extensions exist. (In fact, by the general calculus of mates, if one is an isomorphism, so must the other be.)

For example, suppose that $cspn$ denotes the walking cospan $\to \leftarrow$ and $\square$ denotes the walking commutative square, with $u\colon cspn \to \square$ the inclusion. Then our “thing number 1” says exactly that the square
$\array{ cspn & \to & cspn \\ \downarrow && \downarrow^u \\ cspn & \underset{u}{\to} & \square }$
(which is inhabited by the identity 2-cell) is exact: right Kan extending along $u$ and then restricting back along it is canonically isomorphic to the identity. And if we recall that *limits* in the ordinary sense can be identified with right Kan extensions along the unique functor to the terminal category 1, then our “thing number 2” says exactly that the square
$\array{ cspn & \to & 1 \\ \downarrow & \swArrow & \downarrow^p \\ cspn & \underset{u}{\to} & \square }$
is exact, where $p\colon 1\to \square$ picks out the upper-left vertex – i.e. right Kan extending a cospan along $u$ and then restricting to the upper-left vertex is canonically isomorphic to the limit of that cospan. (Exactness of this square is precisely a special case of *pointwiseness* of the Kan extension, if you know what that means.)

The reason it’s convenient to express things like this in terms of exactness of squares is that we can also give an explicit combinatorial description of when a given square is exact, allowing us to verify easily that many Kan extensions can be “computed” in terms of others. Here’s the characterization: given a square $\array{ X & \overset{u}{\to} & Y \\ ^v\downarrow & ^\alpha \swArrow & \downarrow^g \\ Z & \underset{f}{\to} & W }$ as above, fix some objects $z\in Z$ and $y\in Y$ and form the category $(y/X/z)$ whose objects are triples $(x,y\to u(x), v(x)\to z)$ and whose morphisms are morphisms in $X$ that make the two evident triangles commute. There is a functor $(y/X/z) \to W(g(y),f(z))$ (considering the latter as a discrete category) which takes $(x,y\to u(x), v(x)\to z)$ to the composite $g(y) \to g(u(x)) \to f(v(x)) \to f(z)$. The theorem is that the square $\alpha$ is exact if and only if this functor induces a bijection of connected components. By discreteness of $W(g(y),f(z))$, this is equivalent to saying that for each morphism $\xi\colon g(y) \to f(z)$ in $W$, the subcategory of $(y/X/z)$ consisting of the triples mapping to $\xi$ is connected.

Two different proofs of this characterization of exact squares can be found on the nLab. I leave it to you to verify that the two squares described above satisfy this condition.

Let’s do one more sample computation, which is maybe a bit less trivial. Consider the standard lemma on pullback squares which says that given a diagram $\array{a & \to & b & \to & c \\ \downarrow & & \downarrow & & \downarrow \\ d & \to & e & \to & f }$ in which the right-hand square $bcef$ is a pullback, then the left-hand square $abde$ is a pullback if and only if the outer rectangle $acdf$ is a pullback. Let’s prove this using exact squares. I’ll write $cdef$ for the lower-right L-shaped subcategory of the above diagram shape, and so on, and I’ll leave to you the verification that all the squares that arise are in fact exact.

First of all, since the squares $\array{cef & \overset{}{\to} & bcef\\ \downarrow && \downarrow\\ cdef & \underset{}{\to} & abcdef} \qquad and \qquad \array{cdf & \overset{}{\to} & acdf\\ \downarrow && \downarrow\\ cdef& \underset{}{\to} & abcdef}$ are exact, if we start from a $cdef$-diagram and right Kan extend it to a full $abcdef$-diagram, then the right-hand square and outer rectangle must be pullback squares. Moreover, by composition of adjoints, right Kan extension from $cdef$ to $abcdef$ is equivalent to first extending to $bcdef$ and then to $abcdef$, and since the square $\array{bde & \overset{}{\to} & abde\\ \downarrow && \downarrow\\ bcdef & \underset{}{\to} & abcdef}$ is exact, the left-hand square in such an extension must also be a pullback.

Now if we start with an $abcdef$-diagram, say $F$, we can restrict it to a $cdef$-diagram and then right Kan-extend it to a new $abcdef$-diagram. If $u\colon cdef \to abcdef$ is the inclusion, then this results in $u_\ast u^\ast F$, and we have a canonical natural transformation $\eta \colon F \to u_\ast u^\ast F$ (the unit of the adjunction $u^\ast \dashv u_\ast$). Since the square $\array{cdef & \overset{}{\to} & cdef \\ \downarrow && \downarrow\\ cdef & \underset{}{\to} & abcdef}$ is exact, the counit $u^\ast u_\ast G \to G$ is an isomorphism for any $G$, and in particular for $G=u^\ast F$, from which it follows by the triangle identities that $u^\ast F \to u^\ast u_\ast u^\ast F$ is also an isomorphism — i.e. the components of $\eta\colon F \to u_\ast u^\ast F$ at $c$, $d$, $e$, and $f$ are isomorphisms. Now if the right-hand square of $F$ is a pullback, then the restrictions of $F$ and $u_\ast u^\ast F$ to $bcef$ are both pullback squares; hence since the $cef$-components of $\eta$ are isomorphisms, so is the $b$-component. And if the left-hand square of $F$ is a pullback, then we can play the same game with $abde$ to show that the $a$-component of $\eta$ is an isomorphism, while if the outer rectangle is a pullback, we can play it with $acdf$. Hence in both of these cases, $\eta$ itself is an isomorphism, since all of its components are — and thus the remaining square in $F$ is also a pullback, since we have shown that it is so in $u_\ast u^\ast F$.

This may seem like a slightly roundabout way to prove such a basic lemma. But its merit is that it generalizes directly to any other context in which we have a notion of “Kan extension” and “exact square” — in particular, to derivators. I won’t go into the definition of derivators today, since this post is already too long; you can find it on the nlab and in the references cited there. The most important change, when working with derivators, is that instead of exact squares we have to consider homotopy exact squares, which satisfy the analogous property for all *homotopy Kan extensions* (a.k.a. Kan extensions in $(\infty,1)$-categories). Being homotopy exact is a *stronger* property than being merely exact: the corresponding combinatorial characterization says that the functor $(y/X/z) \to W(g(y),f(z))$ induces a weak equivalence of nerves, rather than merely a bijection of connected components. (In fact, one might argue that *a priori* this is actually a *more natural* demand – why cut things off at level 0?) Fortunately, in practice it’s often also easy to verify this stronger condition, and all of the exact squares I drew above are actually homotopy exact; thus exactly the same proofs work in any derivator, and hence in any $(\infty,1)$-category.

## Re: Exact Squares

I tink I am following so far. What I still have only a vague feeling for – not having worked with this myself – is what this does for us in practice.

At derivator you write:

So I looked at homotopy exact square, which is very nice. But it still does not quite seem to provide me with convenient ways to compute homotopy limits. Or maybe I am missing something.

The first example listed there says that, as you amplified here, homotopy Kan extensions are pointwise. It’s nice to see how this relates to homotopy exactness of comma squares, but this is not a new technique for computing homotopy limits, or is it?

Similarly for the example of extension along fully faithful functors: it’s good to see the reformulation in terms of homotopy exact squares, but have we really gained a “convenient way to compute” the homotopy limit?

You probably have some examples where the computation of a concrete homotopy (co)limit is facilitated by appealing to exact square-reasoning. Could you spell one such example out?