## September 30, 2015

### An Exact Square from a Reedy Category

#### Posted by Emily Riehl

I first learned about exact squares from a blog post written by Mike Shulman on the $n$-Category Café.

Today I want to describe a family of exact squares, which are also homotopy exact, that I had not encountered previously. These make a brief appearance in a new preprint, A necessary and sufficient condition for induced model structures, by Kathryn Hess, Magdalena Kedziorek, Brooke Shipley, and myself.

Proposition. If $R$ is any (generalized) Reedy category, with $R^+ \subset R$ the direct subcategory of degree-increasing morphisms and $R^- \subset R$ the inverse subcategory of degree-decreasing morphisms, then the pullback square: $\array{ iso(R) & \to & R^- \\ \downarrow & \swArrow id & \downarrow \\ R^+ & \to & R}$ is (homotopy) exact.

In summary, a Reedy category $(R,R^+,R^-)$ gives rise to a canonical exact square, which I’ll call the Reedy exact square.

## Exact squares and Kan extensions

Let’s recall the definition. Consider a square of functors inhabited by a natural transformation $\array{A & \overset{f}{\to} & B\\ ^u\downarrow & \swArrow\alpha & \downarrow^v\\ C& \underset{g}{\to} & D}$ For any category $M$, precomposition defines a square $\array{M^A & \overset{f^\ast}{\leftarrow} & M^B\\ ^{u^\ast}\uparrow & \swArrow \alpha^\ast & \uparrow^{v^\ast}\\ M^C& \underset{g^\ast}{\leftarrow} & M^D}$ Supposing there exist left Kan extensions $u_! \dashv u^\ast$ and $v_! \dashv v^\ast$ and right Kan extensions $f^\ast \dashv f_\ast$ and $g^\ast \dashv g_\ast$, the mates of $\alpha^*$ define canonical Beck-Chevalley transformations: $u_! f^\ast \Rightarrow g^\ast v_!\quad and \quad v^\ast g_\ast \Rightarrow f_\ast u^\ast.$ Note if either of the Beck-Chevalley transformations is an isomorphism, the other one is too by the (contravariant) correspondence between natural transformations between a pair of left adjoints and natural transformations between the corresponding right adjoints.

Definition. $\array{A & \overset{f}{\to} & B\\ ^u\downarrow & \swArrow\alpha & \downarrow^v\\ C& \underset{g}{\to} & D}$ is an exact square if, for any $M$ admitting pointwise Kan extensions, the Beck-Chevalley transformations are isomorphisms.

Comma squares provide key examples, in which case the Beck-Chevalley isomorphisms recover the limit and colimit formulas for pointwise Kan extensions.

The notion of homotopy exact square is obtained by replacing $M$ by some sort of homotopical category, the adjoints by derived functors, and “isomorphism” by “equivalence.”

## The proof

In the preprint we give a direct proof that these Reedy squares are exact by computing the Kan extensions, but exactness follows more immediately from the following characterization theorem, stated using comma categories. The natural transformation $\alpha \colon v f \Rightarrow g u$ induces a functor $B \downarrow f \times_A u \downarrow C \to v \downarrow g$ over $C \times B$ defined on objects by sending a pair $b \to f(a), u(a) \to c$ to the composite morphism $v(b) \to v f(a) \to g u(a) \to g(c)$. Fixing a pair of objects $b$ in $B$ and $c$ in $C$, this pulls back to define a functor $b \downarrow f \times_A u \downarrow c \to vb \downarrow gc.$

Theorem. A square $\array{A & \overset{f}{\to} & B\\ ^u\downarrow & \swArrow\alpha & \downarrow^v\\ C& \underset{g}{\to} & D}$ is exact if and only if each fiber of $b \downarrow f \times_A u \downarrow c \to v b \downarrow g c$ is non-empty and connected.

See the nLab for a proof. Similarly, the square is homotopy exact if and only if each fiber of this functor has a contractible nerve.

In the case of a Reedy square $\array{ iso(R) & \to & R^- \\ \downarrow & \swArrow id & \downarrow \\ R^+ & \to & R}$ these fibers are precisely the categories of Reedy factorizations of a fixed morphism. For an ordinary Reedy category $R$, Reedy factorizations are unique, and so the fibers are terminal categories. For a generalized Reedy category, Reedy factorizations are unique up to unique isomorphism, so the fibers are contractible groupoids.

## Reedy diagrams as bialgebras

For any category $M$, the objects in the lower right-hand square $\array{ M^{iso(R)} & \leftarrow & M^{R^-} \\ \uparrow & \swArrow id & \uparrow \\ M^{R^+} & \leftarrow & M^R}$ are Reedy diagrams in $M$, and the functors restrict to various subdiagrams. Because the indexing categories all have the same objects, if $M$ is bicomplete each of these restriction functors is both monadic and comonadic. If we think of the $M^{R^-}$ as being comonadic over $M^{iso(R)}$ and $M^{R^+}$ as being monadic over $M^{iso(R)}$, then the Beck-Chevalley isomorphism exhibits $M^R$ as the category of bialgebras for the monad induced by the direct subcategory $R^+$ and the comonad induced by the inverse subcategory $R^-$.

There is a homotopy-theoretic interpretation of this, which I’ll describe in the case where $R$ is a strict Reedy category (so that $iso(R)=ob(R)$), though it works in the generalized context as well. If $M$ is a model category, then $M^{iso(R)}$ inherits a model structure, with everything defined objectwise. The Reedy model structure on $M^{R^-}$ coincides with the injective model structure, which has cofibrations and weak equivalences created by the restriction functor $M^{R^-} \to M^{iso(R)}$; we might say this model structure is “left-induced”. Dually, the Reedy model structure on $M^{R^+}$ coincides with the projective model structure, which has fibrations and weak equivalences created by $M^{R^+} \to M^{iso(R)}$; this is “right-induced”.

The Reedy model structure on $M^R$ then has two interpretations: it is right-induced along the monadic restriction functor $M^R \to M^{R^-}$ and it is left-induced along the comonadic restriction functor $M^R \to M^{R^+}$. The paper A necessary and sufficient condition for induced model structures describes a general technique for inducing model structures on categories of bialgebras, which reproduces the Reedy model structure in this special case.

Posted at September 30, 2015 8:50 PM UTC

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### Re: An exact square from a Reedy category

Hahaha! I can’t believe I didn’t notice when writing my paper on Reedy categories that the connectedness of the category of Reedy factorizations is exactly the condition for that square to be exact. Nice.

Presumably this also generalizes to what I called “c-Reedy categories”?

Posted by: Mike Shulman on October 1, 2015 12:30 AM | Permalink | Reply to this

### Re: An exact square from a Reedy category

Could you remind me what those are again? It’s been a while since I’ve looked at your paper.

Posted by: Emily Riehl on October 2, 2015 3:57 PM | Permalink | Reply to this

### Re: An exact square from a Reedy category

A c-Reedy category is a category $C$ equipped with an ordinal-valued degree function on its objects, and subcategories $\overset{\leftrightarrow}{C}$, $\overset{\to}{C}$, and $\overset{\leftarrow}{C}$ containing all the objects, such that

• $\overset{\leftrightarrow}{C} \subseteq \overset{\to}{C}\cap \overset{\leftarrow}{C}$.
• Every morphism in $\overset{\leftrightarrow}{C}$ is level (i.e. its domain and codomain have the same degree).
• Every morphism in $\overset{\to}{C}\setminus\overset{\leftrightarrow}{C}$ strictly raises degree, and every morphism in $\overset{\leftarrow}{C}\setminus\overset{\leftrightarrow}{C}$ strictly lowers degree.
• Every morphism $f$ factors as $\overset{\to}{f} \overset{\leftarrow}{f}$, where $\overset{\to}{f} \in \overset{\to}{C}$ and $\overset{\leftarrow}{f}\in\overset{\leftarrow}{C}$, and the category of such factorizations with connecting maps in $\overset{\leftrightarrow}{C}$ is connected.
• For any $x$ and any degree $\delta\lt\deg(x)$, the functor $\overset{\leftarrow}{C}(x,-):\overset{\leftrightarrow}{C} \to \Set$ is a coproduct of retracts of representables.
Posted by: Mike Shulman on October 5, 2015 7:54 PM | Permalink | Reply to this

### Re: An exact square from a Reedy category

I think the domain of the functor in your last condition is meant to be the category of level morphisms of degree $\delta$.

What role does this condition play? If I’m interpreting it correctly, to say that $\overset{\leftarrow}{C}(x,-)$ is a retract of a coproduct of representables means that there is a set of degree-decreasing maps $x \twoheadrightarrow x_i$, with $x_i$ of degree $\delta$, so that every $x \twoheadrightarrow y$, with $y$ of degree $\delta$, factors (non-uniquely?) through a unique one of these maps along some level morphism $x_i \to y$.

Returning to your original question, it seems that the fourth condition in the definition of a $c$-Reedy category is sufficient for the “Reedy square” to be exact, though not homotopy exact. Do you have any favorite examples on which to test this claim?

Posted by: Emily Riehl on October 6, 2015 9:04 PM | Permalink | Reply to this

### Re: An exact square from a Reedy category

Oops, you’re right, that subscript got lost.

That condition is a generalization of the condition on a generalized Reedy category that the isomorphisms act freely on the down-going arrows. It’s sort of a “cofibrancy” condition that makes the latching and matching object functors homotopically well-behaved.

And yes, the fourth condition does seem to give exactness but not (immediately) homotopy exactness. I don’t really have any good examples, unfortunately; the main interesting examples I know of are additionally enriched.

Posted by: Mike Shulman on October 7, 2015 12:05 AM | Permalink | Reply to this

### Re: An exact square from a Reedy category

It’s an interesting question whether the “Reedy square” must be homotopy exact for a c-Reedy category, i.e. whether the categories of factorizations in the fourth condition must actually be contractible. I think it’s not too hard to cook up examples which satisfy the first four conditions where the square is exact but not homotopy exact. For instance, take the parallel pair category $(a \rightrightarrows b)$, add a new terminal object $t$ and a new initial object $i$, and put $a$ and $b$ in degree 0 and $i$ and $t$ in degree 1. Then the category of good factorizations of the unique map $i\to t$ is the parallel pair category, hence connected but not contractible. However, this example violates the fifth “cofibrancy” condition! Is it possible that the cofibrancy condition somehow magically ensures that the exact Reedy square is in fact homotopy exact?

By the way, I should have been more precise: the fifth condition ensures good homotopical behavior of the latching objects, which is necessary because the c-Reedy model structure (like the generalized-Reedy one) uses projective model structures on each level. If we dually used injective model structures on each level, then we would instead need a dual version of the fifth condition ensuring good behavior of the matching objects.

Posted by: Mike Shulman on October 7, 2015 3:59 PM | Permalink | Reply to this

### Re: An exact square from a Reedy category

To make a foray into the introduction here, let me see if I properly understand the statement that $u_! f^\ast \implies g^\ast v_!$ is invertible if and only if $v^\ast g_\ast \implies f_\ast u^\ast$ is an isomorphism.

This does not follow by applying the mate correspondence to a square made up of $f,g,u,v$ with appropriate decorations. Instead, one applies the mate correspondence linking the squares

$\array{B & \overset{=}{\to} & B\\ ^{u_! f^\ast}\downarrow & \seArrow\overline{\alpha^\ast} & \downarrow^{g^\ast v_!}\quad \\ C& \underset{=}{\to} & C}$ and $\quad \array{B & \overset{=}{\to} & B\\ ^{f_\ast u^\ast}\downarrow & \seArrow\widetilde{\alpha^\ast} & \downarrow^{v^\ast g_\ast}\\ C& \underset{=}{\to} & C}$

The one is an isomorphism iff the other is only because two of the parallel functors are identities, so the double-categorical functoriality of the mate correspondence kicks in. The general mate correspondence does not preserve/reflect the property of being an isomorphism except in this case– this is the point of the Beck-Chevalley condition, and something we’ve discussed before.

Posted by: Tim Campion on October 2, 2015 1:44 AM | Permalink | Reply to this

### Re: An exact square from a Reedy category

Oh darn, the 2-cell in the right hand diagram should point southwest, not southeast!

Posted by: Tim Campion on October 2, 2015 1:47 AM | Permalink | Reply to this

### Re: An exact square from a Reedy category

Yes, exactly.

Posted by: Emily Riehl on October 2, 2015 3:55 PM | Permalink | Reply to this

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