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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeJun 1st 2017
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItex[[differential cohesive (infinity,1)-topos]] says > where $i_!$ is a [[full and faithful (∞,1)-functor|full and faithful]] and preserves [[finite products]]. > Equivalently this means that $(i_! \dashv i^\ast) \colon \mathbf{H}_{th} \longrightarrow \mathbf{H}$ is a [[local geometric morphism]] with a further [[right adjoint]] to the right adjoint to the [[direct image]]. But for $(i_! \dashv i^\ast)$ to be a geometric morphism, $i_!$ would need to preserve all finite *limits*?

   differential cohesive (infinity,1)-topos says

   where is a full and faithful and preserves finite products.

   Equivalently this means that is a local geometric morphism with a further right adjoint to the right adjoint to the direct image.

   But for to be a geometric morphism, would need to preserve all finite limits?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJun 2nd 2017
   • (edited Jun 2nd 2017)
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIf we write $p_! \dashv p^* \dashv p_* \dashv p^!$ for the adjoint string exhibiting cohesion of $\mathbf{H}$, then the composite adjunction $(i_* p^*, p_* i^!) : H_{th} \to \infty Gpd$ is a geometric morphism, hence the unique global sections geometric morphism. If $i_!$ were left exact, then the composite adjunction $(i_! p^*, p_* i^*): H_{th} \to \infty Gpd$ would also be a geometric morphism, hence also the unique global sections geometric morphism, and so we would have $p_* i^! = p_* i^*$ and $i_! p^* = i_* p^*$. The first would mean that the global sections of the infinitesimal shape and flat of something coincide (internally, something like $\sharp\Im X = \sharp 🙰X$), and the second would mean that the reduced and coreduced versions of a discrete object coincide (internally, something like $\Re\flat X = \Im\flat X$). I'm guessing this is not the case in general, but I don't have any intuition for why not. How can discrete cohesion determine nonconstant infinitesimal directions?

   If we write for the adjoint string exhibiting cohesion of , then the composite adjunction is a geometric morphism, hence the unique global sections geometric morphism. If were left exact, then the composite adjunction would also be a geometric morphism, hence also the unique global sections geometric morphism, and so we would have and . The first would mean that the global sections of the infinitesimal shape and flat of something coincide (internally, something like ), and the second would mean that the reduced and coreduced versions of a discrete object coincide (internally, something like ). I’m guessing this is not the case in general, but I don’t have any intuition for why not. How can discrete cohesion determine nonconstant infinitesimal directions?

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeJun 2nd 2017
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI guess related to this is the question of how the notions of "discrete" and "codiscrete", and their associated modalities, are related in $\mathbf{H}$ and $\mathbf{H}_{th}$. Since $(i_* p^*, p_* i^!)$ is the global sections of $\mathbf{H}_{th}$, it follows that the $\flat$ of $\mathbf{H}_{th}$ is $i_* p^* p_* i^!$, which is the $\flat$ of $\mathbf{H}$ bracketed by infinitesimal flat and coreduction. But the only manifestation I see in $\mathbf{H}_{th}$ of the $\sharp$ of $\mathbf{H}$ is the monad $i_* p^! p_* i^*$, which I think is right adjoint to $\Re \flat\Im$ on $\mathbf{H}_{th}$, but which I don't see any way to express in terms of the six modalities $ʃ,\flat,\sharp,\Re,\Im,\&$ of $\mathbf{H}_{th}$ (since none of them involve $p^!$). And the $\sharp$ of $\mathbf{H}_{th}$ doesn't seem to be related to $\mathbf{H}$ at all. (This is sort of a relaying of a question from Max New, with my own musings around it.)

   I guess related to this is the question of how the notions of “discrete” and “codiscrete”, and their associated modalities, are related in and . Since is the global sections of , it follows that the of is , which is the of bracketed by infinitesimal flat and coreduction. But the only manifestation I see in of the of is the monad , which I think is right adjoint to on , but which I don’t see any way to express in terms of the six modalities of (since none of them involve ). And the of doesn’t seem to be related to at all. (This is sort of a relaying of a question from Max New, with my own musings around it.)

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJun 2nd 2017
   • (edited Jun 2nd 2017)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks, Mike. re #1: Right, that's an edit mistake. It puzzled me for a moment how I could have written such a blatant nonsense, but looking at the logs, here is what happened: Originally in [rev 1 on Apr 13 2011](https://ncatlab.org/nlab/revision/differential+cohesive+%28infinity%2C1%29-topos/1) I had just required that $i_!$ preserves the terminal object. Then in [rev 49 on Dec 20 2013](https://ncatlab.org/nlab/revision/diff/differential+cohesive+%28infinity%2C1%29-topos/49) something made me go and add to what had used to be "$i_!$ preserves the terminal object" the clause "... and is in fact left exact". And so I also added that comment saying that this means that it is the left adjoint of a local geometric morphism. Then in [rev 73 on Jan 6 2015](https://ncatlab.org/nlab/revision/diff/differential+cohesive+%28infinity%2C1%29-topos/73) I removed that clause again and settled for "$i_!$ preserves finite products" But, as the records show, I forgot to also remove this extra remark, which by that further edit had become obsolete and in fact nonsensical. I have removed it now. Thanks for catching this. Regarding #3: That's right, I have not considered integrating the sharp on $\mathbf{H}$ into the axiomatics in terms of endofunctors on $\mathbf{H}_{th}$. It's great that you are looking into this now. You'll probably discover various improvements. For other readers: we are talking about this diagram: $$ \array{ & && &\overset{i_!}{\hookrightarrow}& \\ & &\overset{p_!}{\longleftarrow}& &\overset{i^\ast}{\longleftarrow}& \\ \Delta \colon & \infty Grpd &\overset{p^\ast}{\hookrightarrow}& \mathbf{H} &\overset{i_\ast}{\hookrightarrow}& \mathbf{H}_{th} \\ & &\overset{p_\ast}{\longleftarrow}& &\overset{i^!}{\longleftarrow}& \\ & &\overset{p^!}{\hookrightarrow}& } $$

   Thanks, Mike.

   re #1: Right, that’s an edit mistake. It puzzled me for a moment how I could have written such a blatant nonsense, but looking at the logs, here is what happened:

   Originally in rev 1 on Apr 13 2011 I had just required that preserves the terminal object.

   Then in rev 49 on Dec 20 2013 something made me go and add to what had used to be " preserves the terminal object" the clause "… and is in fact left exact". And so I also added that comment saying that this means that it is the left adjoint of a local geometric morphism.

   Then in rev 73 on Jan 6 2015 I removed that clause again and settled for " preserves finite products" But, as the records show, I forgot to also remove this extra remark, which by that further edit had become obsolete and in fact nonsensical.

   I have removed it now. Thanks for catching this.

   Regarding #3:

   That’s right, I have not considered integrating the sharp on into the axiomatics in terms of endofunctors on .

   It’s great that you are looking into this now. You’ll probably discover various improvements.

   For other readers: we are talking about this diagram:

5. • CommentRowNumber5.
   • CommentAuthormaxsnew
   • CommentTimeJun 6th 2017
   • PermaLink
   Author: maxsnew
   Format: MarkdownItexI added a line to [[Reduction Modality]] to say that $\Re$ preserves finite products.

   I added a line to Reduction Modality to say that preserves finite products.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeJun 8th 2017
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexSo what about the composite $i_! p^*$; is that left exact? Equivalently, does $i_! p^* = i_* p^*$ and $p_*i^* = p_* i^!$?

   So what about the composite ; is that left exact? Equivalently, does and ?

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJun 9th 2017
   • PermaLink
   Author: Urs
   Format: MarkdownItexYes. $i_\ast p^\ast = \Delta$ takes an $\infty$-groupoid $X$ to the stack constant on $X$ and since $X \simeq \underset{\longrightarrow}{\lim}_X \ast$ and since $i_!(\ast) \simeq \ast$ the same is true for $i_! p^\ast$.

   Yes.

   takes an -groupoid to the stack constant on and since and since the same is true for .

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJun 9th 2017
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexAh, I see. More abstractly, $i_* p^*$ and $i_! p^*$ are both cocontinuous since they are left adjoints, and both preserve the terminal object, but $\infty Gpd$ is the free cocompletion of its terminal object, so $i_* p^* = i_! p^*$. Presumably if $\infty Gpd$ is replaced by some arbitrary base $(\infty,1)$-topos, then an analogous argument would work as long as all the adjunctions are indexed over that base. Thanks!

   Ah, I see. More abstractly, and are both cocontinuous since they are left adjoints, and both preserve the terminal object, but is the free cocompletion of its terminal object, so . Presumably if is replaced by some arbitrary base -topos, then an analogous argument would work as long as all the adjunctions are indexed over that base. Thanks!