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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 an -category (model category) with products and its simplicial objects, let be the localization (Bousfield localization) at the morphisms .

   Is ?

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 be regarded as degreewise discrete simplicial “space”. First, let us show that the projections get inverted. But these are simplicial homotopy equivalences, so they become invertible if you invert the projection for every simplicial “space” . 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 -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 -cube spaces all being equivalent it follows that the -simplex spaces 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 localized at ?

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 in the slice category is right orthogonal to if and only if is right orthogonal to , so is right orthogonal to every in the slice category if and only if every is right orthogonal to , but I can’t really make heads or tails of that.

   The special case where is itself “constant” should be more straightforward. Assuming “geometric realisation” 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 and in the slice over , I believe that (which I assume is what you mean in #6 by ) should be the same as . So is right orthogonal to every in the slice over iff is right orthogonal to in the slice over for every . And it’s probably enough to consider the universal cases …