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 $(\mathrm{\infty },1)$-category.

Definition

Let $\square$ denote the category

$$\begin{array}{ccc}a& \to & b\\ \downarrow & & \downarrow \\ c& \to & d\end{array}$$\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 $u:L\to \square$.

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 $u^{*}⊣u_{*}$ 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

$$\begin{array}{ccccc}a& \to & b& \to & c\\ \downarrow & & \downarrow & & \downarrow \\ d& \to & e& \to & f\end{array}$$\array{a & \to & b & \to & c \\ \downarrow & & \downarrow & & \downarrow \\ d & \to & e & \to & f }

in which the right-hand square $\mathrm{bcef}$ is a pullback, then the left-hand square $\mathrm{abde}$ is a pullback if and only if the outer rectangle $\mathrm{acdf}$ 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 $(\mathrm{\infty },1)$-categories. In particular, note that the statement for derivators is a generalization of both, since both 1-categories and $(\mathrm{\infty },1)$-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 $\mathrm{abcdef}$ for this $X$, and similarly $\mathrm{cdef}$ 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

$$\begin{array}{ccc}\mathrm{cef}& \stackrel{}{\to }& \mathrm{bcef}\\ \downarrow & & \downarrow \\ \mathrm{cdef}& \underset{}{\to }& \mathrm{abcdef}\end{array}\mathrm{and}\begin{array}{ccc}\mathrm{cdf}& \stackrel{}{\to }& \mathrm{acdf}\\ \downarrow & & \downarrow \\ \mathrm{cdef}& \underset{}{\to }& \mathrm{abcdef}\end{array}$$\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 $\mathrm{cdef}$-diagram and right Kan extend it to a full $\mathrm{abcdef}$-diagram, then the right-hand square and outer rectangle must be pullback squares. Moreover, by composition of adjoints, right Kan extension from $\mathrm{cdef}$ to $\mathrm{abcdef}$ is equivalent to first extending to $\mathrm{bcdef}$ and then to $\mathrm{abcdef}$, and since the square

$$\begin{array}{ccc}\mathrm{bde}& \stackrel{}{\to }& \mathrm{abde}\\ \downarrow & & \downarrow \\ \mathrm{bcdef}& \underset{}{\to }& \mathrm{abcdef}\end{array}$$\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 $\mathrm{abcdef}$-diagram, say $F$, we can restrict it to a $\mathrm{cdef}$-diagram and then right Kan-extend it to a new $\mathrm{abcdef}$-diagram. If $u:\mathrm{cdef}\to \mathrm{abcdef}$ is the inclusion, then this results in $u_{*}{u}^{*}F$, and we have a canonical natural transformation $\eta :F\to u_{*}{u}^{*}F$ (the unit of the adjunction $u^{*}⊣u_{*}$). Since the square

$$\begin{array}{ccc}\mathrm{cdef}& \stackrel{}{\to }& \mathrm{cdef}\\ \downarrow & & \downarrow \\ \mathrm{cdef}& \underset{}{\to }& \mathrm{abcdef}\end{array}$$\array{cdef & \overset{}{\to} & cdef \\ \downarrow && \downarrow\\ cdef & \underset{}{\to} & abcdef}

is exact, the counit $u^{*}{u}_{*}G\to G$ is an isomorphism for any $G$, and in particular for $G=u^{*}F$, from which it follows by the triangle identities that $u^{*}F\to u^{*}{u}_{*}{u}^{*}F$ is also an isomorphism — i.e. the components of $\eta :F\to u_{*}{u}^{*}F$ 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 $u_{*}{u}^{*}F$ to $\mathrm{bcef}$ are both pullback squares; hence since the $\mathrm{cef}$-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 $\mathrm{abde}$ to show that the $a$-component of $\eta$ is an isomorphism, while if the outer rectangle is a pullback, we can play it with $\mathrm{acdf}$. 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 $u_{*}{u}^{*}F$.

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 $f:K\to J$ be any functor and let $i:\square \to J$ 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 $f_{!}=j_{!}{\overline{f}}_{!}$ where $\overline{f}:K\to (J\setminus z)$ is induced by $f$ and $j:(J\setminus z)\to J$ 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:

$$\begin{array}{ccc}\Gamma & \to & (J\setminus z)\\ \downarrow & & \downarrow \\ \square & \to & J\end{array}$$\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:

$$\begin{array}{ccccc}\Gamma & \to & \Gamma & \to & (J\setminus z)\\ \downarrow & ⇙& \downarrow & & \downarrow \\ *& \underset{(1,1)}{\to }& \square & \to & J\end{array}$$\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 $*\to J\leftarrow (J\setminus z)$ 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