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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 $(\infty,1)$-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 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.

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 $\int$ exists, I don’t think we expect that $H(\int X, -)\times H(\int Y,-)=H(\int(X\times Y),-)$, 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 $\int:\infty Top \to \infty Grpd$ 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.

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 $LexAcc(\infty Gpd,\infty Gpd)^{op}$. 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?

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 $\int: \infty Top \to LexAcc(\infty Gpd,\infty Gpd)^{op}$.

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 $ʃ:H\to LexAcc^{op}$ 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

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) $H$ 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?

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 $\Delta$ 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 $F$ 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 $colim_j F(Map(Y_j, U)) \to F(colim_j Map(Y_j, U))$.

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.