The n-Category Café

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:

1. The underlying cospan $B' \to C' \leftarrow A'$ of the pullback square should be the same as the cospan we started with, and
2. 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.