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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeOct 16th 2016
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   Author: Mike Shulman
   Format: MarkdownItexFor an $(\infty,1)$-topos to be cohesive, we require that $\Delta$ have a left adjoint $ʃ$ that preserves products. However, even if no such adjoint exists, we can still define the "shape" of an object $X\in H$ to be the hom-functor $H(X,\Delta(-)) : \infty Gpd \to \infty Gpd$, which is left-exact and accessible. (This is just the [[shape of an (infinity,1)-topos|shape]] of the over-topos $H/X$.) Is there a condition on $H$ under which this non-representable shape preserves products? That is, such that $H(X,\Delta(-)) \times H(Y,\Delta(-)) \simeq H(X\times Y,\Delta(-))$? Here the product on the left should probably be taken in the category $LexAcc(\infty Gpd, \infty Gpd)$, since that's what's true if $\Delta$ does have a left adjoint preserving products.

   For an -topos to be cohesive, we require that have a left adjoint that preserves products. However, even if no such adjoint exists, we can still define the “shape” of an object to be the hom-functor , which is left-exact and accessible. (This is just the shape of the over-topos .) Is there a condition on under which this non-representable shape preserves products? That is, such that ? Here the product on the left should probably be taken in the category , since that’s what’s true if does have a left adjoint preserving products.

2. • CommentRowNumber2.
   • CommentAuthorCharles Rezk
   • CommentTimeOct 16th 2016
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   Author: Charles Rezk
   Format: MarkdownItexI don't think that's formulated right. Even if $\int$ exists, I don't think we expect that $H(\int X, -)\times H(\int Y,-)=H(\int(X\times Y),-)$, do we?

   I don’t think that’s formulated right. Even if exists, I don’t think we expect that , do we?

3. • CommentRowNumber3.
   • CommentAuthorMarc Hoyois
   • CommentTimeOct 16th 2016
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   Author: Marc Hoyois
   Format: MarkdownItexThis is probably not quite what you’re looking for, but $\int:\infty Top \to \infty Grpd$ preserves finite products of proper ∞-topoi. More generally, if $X$ is proper then $\int(X\times Y)\simeq \int Y\circ\int X$ for any $Y$, by proper base change, and $\int X$ preserves filtered colimits. If $F,G\in LexAcc$, there's always a canonical map $F\times G\to F\circ G$, and it's an equivalence if $F$ preserves filtered colimits.

   This is probably not quite what you’re looking for, but preserves finite products of proper ∞-topoi. More generally, if is proper then for any , by proper base change, and preserves filtered colimits. If , there’s always a canonical map , and it’s an equivalence if preserves filtered colimits.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeOct 16th 2016
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   Author: Mike Shulman
   Format: MarkdownItex@Charles: Oh, oops, I meant the product should be taken in $LexAcc(\infty Gpd,\infty Gpd)^{op}$. That's necessary to get the variance right, and I believe the dual Yoneda embedding landing in there does preserve products, just like the ordinary Yoneda embedding landing in a category of colimit-preserving functions preserves those colimits. @Marc: What do you mean by $ʃ : \infty Top \to \infty Gpd$? Only locally contractible $\infty$-topoi have a representable shape, right?

   @Charles: Oh, oops, I meant the product should be taken in . That’s necessary to get the variance right, and I believe the dual Yoneda embedding landing in there does preserve products, just like the ordinary Yoneda embedding landing in a category of colimit-preserving functions preserves those colimits.

   @Marc: What do you mean by ? Only locally contractible -topoi have a representable shape, right?

5. • CommentRowNumber5.
   • CommentAuthorMarc Hoyois
   • CommentTimeOct 17th 2016
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   Author: Marc Hoyois
   Format: MarkdownItexMy bad, I meant $\int: \infty Top \to LexAcc(\infty Gpd,\infty Gpd)^{op}$.

   My bad, I meant .

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeOct 17th 2016
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   Author: Mike Shulman
   Format: MarkdownItexAh, okay. That seems at least related. Perhaps asking whether $ʃ:H\to LexAcc^{op}$ preserves products isn't quite the right question. When $H$ is cohesive, we have $ʃ H = 1$, so that the product $ʃ X \times ʃ Y$ is equivalently the pullback $ʃ X \times_{ʃ 1} ʃ Y$, i.e. the pullback $ʃ(H/X) \times_{ʃ H} ʃ(H/Y)$. So maybe in general the place to start is by asking what *pullbacks* $ʃ : Top \to LexAcc^{op}$ preserves, and then in what cases $(H/X) \times_H (H/Y)$ is such, and then perhaps add the condition that $H$ is connected (so that $ʃH = 1$). By the way, let me remind everyone that $ʃ$ is not an integral sign but an esh, `ʃ`

   Ah, okay. That seems at least related.

   Perhaps asking whether preserves products isn’t quite the right question. When is cohesive, we have , so that the product is equivalently the pullback , i.e. the pullback . So maybe in general the place to start is by asking what pullbacks preserves, and then in what cases is such, and then perhaps add the condition that is connected (so that ).

   By the way, let me remind everyone that is not an integral sign but an esh, ʃ

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeOct 19th 2016
   • (edited Oct 19th 2016)
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   Author: Mike Shulman
   Format: MarkdownItexI realized there is another condition that make sense for arbitrary (connected) $H$ and is equivalent to $ʃ$ preserving products if $H$ is locally connected: namely, that the discrete objects are an [[exponential ideal]]. It seems natural to conjecture that this is equivalent to the non-representable shape preserving products, and I can see part of one direction of a proof, but I'm stuck at > If $F,G\in LexAcc$, there's always a canonical map $F\times G\to F\circ G$, and it's an equivalence if $F$ preserves filtered colimits. What is this canonical map, and why is it an equivalence if $F$ preserves filtered colimits?

   I realized there is another condition that make sense for arbitrary (connected) and is equivalent to preserving products if is locally connected: namely, that the discrete objects are an exponential ideal. It seems natural to conjecture that this is equivalent to the non-representable shape preserving products, and I can see part of one direction of a proof, but I’m stuck at

   If , there’s always a canonical map , and it’s an equivalence if preserves filtered colimits.

   What is this canonical map, and why is it an equivalence if preserves filtered colimits?

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeOct 19th 2016
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   Author: Mike Shulman
   Format: MarkdownItexOh, oops -- if the discrete objects are an exponential ideal, then in particular they are closed under exponentials, and hence under arbitrary products; thus $\Delta$ is continuous, so by the adjoint functor theorem it has a left adjoint. So that condition is actually no weaker.

   Oh, oops – if the discrete objects are an exponential ideal, then in particular they are closed under exponentials, and hence under arbitrary products; thus is continuous, so by the adjoint functor theorem it has a left adjoint. So that condition is actually no weaker.

9. • CommentRowNumber9.
   • CommentAuthorMarc Hoyois
   • CommentTimeOct 20th 2016
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   Author: Marc Hoyois
   Format: MarkdownItex@Mike Suppose $F$ is co-represented by the cofiltered diagram $(X_i)$, and similarly $G$ by $(Y_j)$. Then $F\times G$ is co-represented by $(X_i\times Y_j)$. So we have a map $$ (F \times G)(U) = colim_{i,j} Map(X_i\times Y_j,U) = colim_{i,j}Map(X_i, Map(Y_j, U)) \to colim_i Map(X_i, colim_j Map(Y_j, U)) = (F\circ G)(U). $$ The arrow can also be written as $colim_j F(Map(Y_j, U)) \to F(colim_j Map(Y_j, U))$.

   @Mike Suppose is co-represented by the cofiltered diagram , and similarly by . Then is co-represented by . So we have a map

   The arrow can also be written as .

10. • CommentRowNumber10.
    • CommentAuthorMarc Hoyois
    • CommentTimeOct 20th 2016
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    Author: Marc Hoyois
    Format: MarkdownItexAh, but of course the map goes the other way in Pro(∞Gpd).

    Ah, but of course the map goes the other way in Pro(∞Gpd).

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeOct 20th 2016
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    Author: Mike Shulman
    Format: MarkdownItexOk, thanks.

    Ok, thanks.