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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeJan 23rd 2011
   • (edited Jan 23rd 2011)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexReposted from [Math Overflow](http://mathoverflow.net/questions/52934/are-pushouts-in-cat-exact) -------------------------- Preamble ----- ----- Let $V=sSet$ with its cartesian monoidal structure. Let $f:C\to D^{op}$ be a $V$-functor, and let $\pi:C\to C'$ be another $V$-functor. Let $p\in D$ be an object, and let $h^p:=D(-,p):D^{op}\to V$. Let $\pi_!$ be the left adjoint of the pullback $V$-functor $\pi^*:V^{C'}\to V^C$ induced by $\pi$ (That is to say, for a V-functor $h:C'\to V$, the functor $\pi_!(h)$ is the left Kan extension of $h$ along $\pi$). Then $g:=h^p\circ f$ is a functor $C\to V$. Question ----- ----- Why is the following true: In terms of the pushout diagram $$D':=colim(C'^{op}\leftarrow C^{op}\rightarrow D)$$ where $\iota_1:C'^{op}\to D'$ and $\iota_2:D\to D'$ are the canonical injections: We have an isomorphism of $V$-functors $C'\to V$ $$\pi_!(g)(-)\cong D'(\iota_1^{op}(-),\iota_2(p)).$$ Idea ------ ----- I think we can do something like this: We have a diagram $$\begin{matrix} C&\overset{f}{\to} &D^{op}&\overset{h^p}{\to}& V\\ \downarrow^\pi&&\downarrow^{\iota^{op}_2}\\ C'&\underset{\iota^{op}_1}{\to}&D'^{op} \end{matrix} $$ Then we take the left Kan extension of $h^p$ by $\iota^{op}_2$, but by one of the many variations of Yoneda's lemma (computing via coends, this follows from _Yoneda reduction_), this is exactly $D'(-,\iota_2(p))$. Then it's enough to show that the composite $$k:=D'(-,\iota_2(p))\circ \iota^{op}_1=D'(\iota^{op}_1(-),\iota^{op}_2(p))$$ (with the natural map $g\to k\pi$) has the required universal property, that is, that it is initial among fillers of the Kan extension diagram: $$\begin{matrix} C&\overset{g}{\to} & V&\\ \downarrow^\pi&&\\ C'&& \end{matrix} $$ which is where I'm not sure how to proceed.

   Reposted from Math Overflow

   ---

   Preamble

   ---

   Let $V=sSet$ with its cartesian monoidal structure.

   Let $f:C\to D^{op}$ be a $V$-functor, and let $\pi:C\to C'$ be another $V$-functor. Let $p\in D$ be an object, and let $h^p:=D(-,p):D^{op}\to V$.

   Let $\pi_!$ be the left adjoint of the pullback $V$-functor $\pi^*:V^{C'}\to V^C$ induced by $\pi$ (That is to say, for a V-functor $h:C'\to V$, the functor $\pi_!(h)$ is the left Kan extension of $h$ along $\pi$).

   Then $g:=h^p\circ f$ is a functor $C\to V$.

   Question

   ---

   Why is the following true:

   In terms of the pushout diagram

   $$D':=colim(C'^{op}\leftarrow C^{op}\rightarrow D)$$

   where $\iota_1:C'^{op}\to D'$ and $\iota_2:D\to D'$ are the canonical injections:

   We have an isomorphism of $V$-functors $C'\to V$

   $$\pi_!(g)(-)\cong D'(\iota_1^{op}(-),\iota_2(p)).$$

   Idea

   ---

   I think we can do something like this:

   We have a diagram

   $$\begin{matrix} C&\overset{f}{\to} &D^{op}&\overset{h^p}{\to}& V\\ \downarrow^\pi&&\downarrow^{\iota^{op}_2}\\ C'&\underset{\iota^{op}_1}{\to}&D'^{op} \end{matrix}$$

   Then we take the left Kan extension of $h^p$ by $\iota^{op}_2$, but by one of the many variations of Yoneda’s lemma (computing via coends, this follows from Yoneda reduction), this is exactly $D'(-,\iota_2(p))$.

   Then it’s enough to show that the composite

   $$k:=D'(-,\iota_2(p))\circ \iota^{op}_1=D'(\iota^{op}_1(-),\iota^{op}_2(p))$$

   (with the natural map $g\to k\pi$) has the required universal property, that is, that it is initial among fillers of the Kan extension diagram:

   $$\begin{matrix} C&\overset{g}{\to} & V&\\ \downarrow^\pi&&\\ C'&& \end{matrix}$$

   which is where I’m not sure how to proceed.

2. • CommentRowNumber2.
   • CommentAuthorHarry Gindi
   • CommentTimeJan 23rd 2011
   • (edited Jan 23rd 2011)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexI just noticed that this is closely related to the [calculus of exact squares](http://golem.ph.utexas.edu/category/2010/06/exact_squares.html) and [[mates]]. I bet that there's some theorem that proves that pushout squares in $V$-Cat are exact, since my question is equivalent to: Is the natural Beck-Chevalley transformation $\pi_!f^*\to \iota_1^* (\iota_2)_!$ an isomorphism?

   I just noticed that this is closely related to the calculus of exact squares and mates.

   I bet that there’s some theorem that proves that pushout squares in $V$-Cat are exact, since my question is equivalent to:

   Is the natural Beck-Chevalley transformation $\pi_!f^*\to \iota_1^* (\iota_2)_!$ an isomorphism?

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeJan 23rd 2011
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexWell, *arbitrary* pushout squares aren't exact even when V=Set. Consider the pushout of discrete categories 1←2→1, which is 1. I don't think you can fix that by taking homotopy pushouts either. But *cocomma squares* are, I believe, always exact for any V, although I could be misremembering.

   Well, arbitrary pushout squares aren’t exact even when V=Set. Consider the pushout of discrete categories 1←2→1, which is 1. I don’t think you can fix that by taking homotopy pushouts either. But cocomma squares are, I believe, always exact for any V, although I could be misremembering.

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeJan 23rd 2011
   • (edited Jan 23rd 2011)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexAh, thanks! You're right. My original question was too generalized from the original source (therefore false). I e-mailed Lurie, and the answer hinges on some additional information (i.e. that the point p in my first post is a weak strict [[sink]] (or whatever the right term is) in the following sense: $D(p,p)=\Delta^0$ and $D(p,x)=\emptyset$ for all $x\neq p$ (description is evil, but the point $p$ is adjoined formally in a strict way as well (up to equality)).

   Ah, thanks! You’re right. My original question was too generalized from the original source (therefore false). I e-mailed Lurie, and the answer hinges on some additional information (i.e. that the point p in my first post is a weak strict sink (or whatever the right term is) in the following sense: $D(p,p)=\Delta^0$ and $D(p,x)=\emptyset$ for all $x\neq p$ (description is evil, but the point $p$ is adjoined formally in a strict way as well (up to equality)).