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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeOct 19th 2014
   • (edited Oct 19th 2014)
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   Author: Urs
   Format: MarkdownItexI thought this would be straightforward, but now it seems I am stuck: for $\mathcal{C}$ an $\infty$-category (model category) with products and $\mathcal{C}^{\Delta^{op}}$ its simplicial objects, let $L_{\Delta^1} \mathcal{C}^{\Delta^{op}}$ be the localization (Bousfield localization) at the morphisms $(-)\times \Delta^1 \to (-)$. Is $\infty Grpd \simeq L_{\Delta^1} (\infty Grpd)^{\Delta^{op}}$?

   I thought this would be straightforward, but now it seems I am stuck: for $\mathcal{C}$ an $\infty$-category (model category) with products and $\mathcal{C}^{\Delta^{op}}$ its simplicial objects, let $L_{\Delta^1} \mathcal{C}^{\Delta^{op}}$ be the localization (Bousfield localization) at the morphisms $(-)\times \Delta^1 \to (-)$.

   Is $\infty Grpd \simeq L_{\Delta^1} (\infty Grpd)^{\Delta^{op}}$?

2. • CommentRowNumber2.
   • CommentAuthorZhen Lin
   • CommentTimeOct 19th 2014
   • (edited Oct 19th 2014)
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   Author: Zhen Lin
   Format: MarkdownItexTo avoid confusion, let $T^n$ be $\Delta^n$ regarded as degreewise discrete simplicial "space". First, let us show that the projections $T^n \to T^0$ get inverted. But these are simplicial homotopy equivalences, so they become invertible if you invert the projection $X \times T^1 \to X$ for every simplicial "space" $X$. Thus the local objects are precisely the "constant simplicial spaces", if I'm not mistaken. Left Bousfield localisation at a proper class is not known to be possible a priori, but in this case the local objects do turn out to form a reflective $(\infty, 1)$-subcategory, with reflector given by "geometric realisation" (i.e. codescent objects).

   To avoid confusion, let $T^n$ be $\Delta^n$ regarded as degreewise discrete simplicial “space”. First, let us show that the projections $T^n \to T^0$ get inverted. But these are simplicial homotopy equivalences, so they become invertible if you invert the projection $X \times T^1 \to X$ for every simplicial “space” $X$. Thus the local objects are precisely the “constant simplicial spaces”, if I’m not mistaken.

   Left Bousfield localisation at a proper class is not known to be possible a priori, but in this case the local objects do turn out to form a reflective $(\infty, 1)$-subcategory, with reflector given by “geometric realisation” (i.e. codescent objects).

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeOct 19th 2014
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   Author: Urs
   Format: MarkdownItexTrue, thanks. I was trying to see that from the $n$-cube spaces $X^{(\Delta^1)^n}$ all being equivalent it follows that the $n$-simplex spaces $X_n$ are all equivalent. But of course what you say is the right way to do it.

   True, thanks. I was trying to see that from the $n$-cube spaces $X^{(\Delta^1)^n}$ all being equivalent it follows that the $n$-simplex spaces $X_n$ are all equivalent. But of course what you say is the right way to do it.

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeOct 19th 2014
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   Author: Urs
   Format: MarkdownItexI have added that remark [here](http://ncatlab.org/nlab/show/cohesive+%28infinity%2C1%29-topos#SimplicialObjctsInACohesiveTopos) (in the Examples-section at _[[cohesive (infinity,1)-topos]]_). Feel invited to expand, if you care.

   I have added that remark here (in the Examples-section at cohesive (infinity,1)-topos). Feel invited to expand, if you care.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeOct 20th 2014
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   Author: Mike Shulman
   Format: MarkdownItexNice, thanks Zhen! Can you also characterize the objects of the slice $\infty Gpd^{\Delta^{op}}/X$ localized at $T^1$?

   Nice, thanks Zhen! Can you also characterize the objects of the slice $\infty Gpd^{\Delta^{op}}/X$ localized at $T^1$?

6. • CommentRowNumber6.
   • CommentAuthorZhen Lin
   • CommentTimeOct 20th 2014
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   Author: Zhen Lin
   Format: MarkdownItexI'm not sure. By adjunction, an object $Y$ in the slice category is right orthogonal to $X^* A \to X^* B$ if and only if $\prod_X Y$ is right orthogonal to $A \to B$, so $Y$ is right orthogonal to every $Z \times T^1 \to Z \times T^0$ in the slice category if and only if every $\prod_X (Y^Z)$ is right orthogonal to $T^1 \to T^0$, but I can't really make heads or tails of that. The special case where $X$ is itself "constant" should be more straightforward. Assuming "geometric realisation" $[\mathbf{\Delta}^{op}, \infty \mathbf{Grpd}_{/ X}] \to \infty \mathbf{Grpd}_{/ X}$ continues to send simplicial homotopy equivalences to equivalences, the same analysis should work, yielding the same conclusion.

   I’m not sure. By adjunction, an object $Y$ in the slice category is right orthogonal to $X^* A \to X^* B$ if and only if $\prod_X Y$ is right orthogonal to $A \to B$, so $Y$ is right orthogonal to every $Z \times T^1 \to Z \times T^0$ in the slice category if and only if every $\prod_X (Y^Z)$ is right orthogonal to $T^1 \to T^0$, but I can’t really make heads or tails of that.

   The special case where $X$ is itself “constant” should be more straightforward. Assuming “geometric realisation” $[\mathbf{\Delta}^{op}, \infty \mathbf{Grpd}_{/ X}] \to \infty \mathbf{Grpd}_{/ X}$ continues to send simplicial homotopy equivalences to equivalences, the same analysis should work, yielding the same conclusion.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeOct 21st 2014
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   Author: Mike Shulman
   Format: MarkdownItex> The special case where XX is itself “constant” should be more straightforward. Yes, internally: a map with modal codomain is modal iff it has modal fibers.

   The special case where XX is itself “constant” should be more straightforward.

   Yes, internally: a map with modal codomain is modal iff it has modal fibers.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeApr 7th 2015
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   Author: Mike Shulman
   Format: MarkdownItexGiven $f:Y\to X$ and $g:Z\to X$ in the slice over $X$, I believe that $\prod_X (f^g)$ (which I assume is what you mean in #6 by $\prod_X Y^Z$) should be the same as $\prod_Z g^*f$. So $Y$ is right orthogonal to every $Z\times_X T^1 \to Z$ in the slice over $X$ iff $g^*Y$ is right orthogonal to $Z\times T^1 \to Z$ in the slice over $Z$ for every $g:Z\to X$. And it's probably enough to consider the universal cases $Z = X_n \times T^n$...

   Given $f:Y\to X$ and $g:Z\to X$ in the slice over $X$, I believe that $\prod_X (f^g)$ (which I assume is what you mean in #6 by $\prod_X Y^Z$) should be the same as $\prod_Z g^*f$. So $Y$ is right orthogonal to every $Z\times_X T^1 \to Z$ in the slice over $X$ iff $g^*Y$ is right orthogonal to $Z\times T^1 \to Z$ in the slice over $Z$ for every $g:Z\to X$. And it’s probably enough to consider the universal cases $Z = X_n \times T^n$…