nLab pullback in a derivator

Idea

A pullback in a derivator is the generalization to the context of a derivator of the notion of pullback in ordinary category theory. Viewing a derivator as the “shadow” of an (∞,1)-category, the notion of pullback therein coincides with the notion of homotopy pullback in an $(\infty,1)$-category.

Definition

Let $\square$ denote the category

$$\array{a & \to & b \\ \downarrow & & \downarrow \\ c & \to & d}$$

that is the “free-living commutative square”, and let $L$ be the full subcategory of $\square$ on $b,c,d$, with inclusion $u\colon L\to \square$.

Let $D$ be a derivator and let $X\in D(\square)$ be a commutative square in $D$. We say that $X$ is a pullback square, or is cartesian, if the unit $X\to u_* u^* X$ of the adjunction $u^*\dashv u_*$ is an isomorphism. Since $u$ is fully faithful, so is $u_*$, so this is equivalent to saying that there exists some $Y\in D(L)$ such that $X\cong u_* Y$.

If $D$ is merely a prederivator, then we can phrase the same definition by saying that $X$ has the universal property that $u_* u^* X$ would, if the whole functor $u_*$ existed.

The dual notion, of course, is a pushout or cocartesian square.

Properties

Pasting law

Using properties of homotopy exact squares, we can prove the “pasting law” for pullback squares in a derivator:

Lemma

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.

The following proof should be compared and contrasted with the standard proof for pullbacks in 1-categories, and the quasi-categorical proof for pullbacks in $(\infty,1)$-categories. In particular, note that the statement for derivators is a generalization of both, since both 1-categories and $(\infty,1)$-categories give rise to derivators.

Proof

First of all, by “a diagram” in a derivator, we mean an object of $D(X)$ for some suitable category $X$. In the above case, $X$ is the category consisting of two commutative squares, as pictured above. We’ll write $abcdef$ for this $X$, and similarly $cdef$ for its lower-right L-shaped subcategory, and so on. We leave the verification of homotopy exactness of all squares to the reader.

Firstly, 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 $u_\ast u^\ast F$.

Detection Lemma

The following lemma, which detects when squares occurring in a Kan extension are pullbacks or pushouts, is due to Jens Franke; see also Groth. We state it in terms of pushouts.

Lemma

Let $f\colon K\to J$ be any functor and let $i\colon \square \to J$ be injective on objects, with lower vertex $i(1,1) = z$. Suppose that $z$ is not in the image of $f$, and that the induced functor $\Gamma \to (J \setminus z)/z$ is a nerve equivalence (such as if it has an adjoint). Then for any derivator $D$ and any $Y\in D(K)$, the square $i^* f_! Y$ is cocartesian.

Proof

Since $f_! = j_! \bar{f}_!$ where $\bar{f}\colon K\to (J\setminus z)$ is induced by $f$ and $j\colon (J\setminus z) \to J$ is the inclusion, it suffices to suppose that $K = (J\setminus z)$. Now what we want is to prove that the following square is homotopy exact:

$$\array{ \Gamma &\to & (J\setminus z) \\ \downarrow && \downarrow\\ \square &\to & J}$$

Exactness is trivial at all objects of $\square$ except $(1,1)$. In that case, we paste with another square:

$$\array{ \Gamma &\to& \Gamma &\to & (J\setminus z) \\ \downarrow & \swArrow & \downarrow && \downarrow\\ \ast &\underset{(1,1)}{\to} &\square &\to & J}$$

The left-hand square is a comma square, hence homotopy exact, so it suffices to show that the composite square is homotopy exact. But the comma object associated to the cospan $\ast \to J \leftarrow (J\setminus z)$ is $(J \setminus z)/z$, and of course this comma square is also exact. And the composite square factors through this comma square by the functor $\Gamma \to (J \setminus z)/z$ which is assumed a nerve equivalence; hence it is also homotopy exact.

References

See all references at derivator. Referred to particularly above are:

• Jens Franke, Uniqueness theorems for certain triangulated categories with an Adams spectral sequence, K-theory archive

• Moritz Groth, Derivators, pointed derivators, and stable derivators pdf