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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeMay 30th 2010
   • (edited May 30th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexLemma: Any left anodyne map in SSet/S is a covariant equivalence. Proof: We can consider the case of a left horn inclusion, since these generate all left anodyne maps. Then we must show that any map $$i: LeftCone(\Lambda^n_j) \coprod_{\Lambda^n_j} S \to LeftCone(\Delta^n) \coprod_{\Delta^n} S$$ is a categorical equivalence. **However, $i$ is a pushout of of the map $\Lambda^{n+1}_{j+1} \to \Delta^{n+1}$, which is inner anodyne, so we're done.** Questions: How do we show that $i$ is a pushout as described in the bolded sentence?

   Lemma: Any left anodyne map in SSet/S is a covariant equivalence.

   Proof: We can consider the case of a left horn inclusion, since these generate all left anodyne maps.

   Then we must show that any map

   $$i: LeftCone(\Lambda^n_j) \coprod_{\Lambda^n_j} S \to LeftCone(\Delta^n) \coprod_{\Delta^n} S$$

   is a categorical equivalence. However, $i$ is a pushout of of the map $\Lambda^{n+1}_{j+1} \to \Delta^{n+1}$, which is inner anodyne, so we’re done.

   Questions:

   How do we show that $i$ is a pushout as described in the bolded sentence?

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 31st 2010
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   Author: DavidRoberts
   Format: MarkdownItexA pushout in the category of arrows? Can we replace $n+1$ and $j+1$ by $n$ and $j$ in the bold statement?

   A pushout in the category of arrows? Can we replace $n+1$ and $j+1$ by $n$ and $j$ in the bold statement?

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeMay 31st 2010
   • (edited May 31st 2010)
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   Author: Harry Gindi
   Format: MarkdownItexThe LeftCone(X) is the join $\Delta^0\star X$. Also, that definitely won't work, since if we replace n+1 and j+1 by n and j, we don't get an inner anodyne map when i=0.

   The LeftCone(X) is the join $\Delta^0\star X$. Also, that definitely won’t work, since if we replace n+1 and j+1 by n and j, we don’t get an inner anodyne map when i=0.

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeMay 31st 2010
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   Author: Harry Gindi
   Format: MarkdownItexAlso, I'm not sure if the pushout is in the category of arrows or not, or if it means that it's a pushout by some morphism (although the second one sounds more plausible).

   Also, I’m not sure if the pushout is in the category of arrows or not, or if it means that it’s a pushout by some morphism (although the second one sounds more plausible).

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMay 31st 2010
   • (edited May 31st 2010)
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   Author: Urs
   Format: MarkdownItex(HTT, lemma 2.1.4.6) The input data is a morphism $\sigma : \Delta^n \to S$. The pushout diagram in question is $$ \array{ \Lambda^{n+1}_{i+1} &\to& (\Lambda^n_i)^{\triangleleft} \coprod_{\Lambda^n_i} S \\ \downarrow && \downarrow \\ \Delta^{n+1} &\to& (\Delta^n)^{\triangleleft} \coprod_{\Delta^n} S } \,. $$ Writing out the cells here is, as usual, obvious but tedious. Think about it in low dimensions, where you can visualize the simplices: Set $n = 2$. Start with a 2-simplex $\sigma$ in $S$. Then $(\Delta^2)^{\triangleleft} \coprod_{\Delta^2} S$ is the original simplicial set $S$ together with a tetrahedron $\Delta^3$ built over $\sigma$. One face of the tetrahedron is the original 2-simplex $\sigma$ in $S$, the three others "stick out" of $S$: The simplicial set $(\Lambda^2_1)^{\triangleleft} \coprod_{\Lambda^2_1} S$ is accordingly the simplicial set $S$ with only two of the three faces of this tetrahedron over $\sigma$ erected. The map $(\Lambda^3_2) \to (\Delta^2)^{\triangleleft} \coprod_{\Delta^2} S $ identifies the horn of this tetrahedron given by these two new faces and the original face $\sigma$. The pushout therefore glues in the remaining face of the tetrahedron and fills it with a 3-cell.

   (HTT, lemma 2.1.4.6)

   The input data is a morphism $\sigma : \Delta^n \to S$.

   The pushout diagram in question is

   $$\array{ \Lambda^{n+1}_{i+1} &\to& (\Lambda^n_i)^{\triangleleft} \coprod_{\Lambda^n_i} S \\ \downarrow && \downarrow \\ \Delta^{n+1} &\to& (\Delta^n)^{\triangleleft} \coprod_{\Delta^n} S } \,.$$

   Writing out the cells here is, as usual, obvious but tedious. Think about it in low dimensions, where you can visualize the simplices:

   Set $n = 2$. Start with a 2-simplex $\sigma$ in $S$. Then $(\Delta^2)^{\triangleleft} \coprod_{\Delta^2} S$ is the original simplicial set $S$ together with a tetrahedron $\Delta^3$ built over $\sigma$. One face of the tetrahedron is the original 2-simplex $\sigma$ in $S$, the three others “stick out” of $S$:

   The simplicial set $(\Lambda^2_1)^{\triangleleft} \coprod_{\Lambda^2_1} S$ is accordingly the simplicial set $S$ with only two of the three faces of this tetrahedron over $\sigma$ erected.

   The map $(\Lambda^3_2) \to (\Delta^2)^{\triangleleft} \coprod_{\Delta^2} S$ identifies the horn of this tetrahedron given by these two new faces and the original face $\sigma$.

   The pushout therefore glues in the remaining face of the tetrahedron and fills it with a 3-cell.

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeMay 31st 2010
   • (edited May 31st 2010)
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   Author: Harry Gindi
   Format: MarkdownItexThanks! The trick I was missing was that since we're gluing the cone over the horn back to S, we can take the original simplex as a face of the higher dimensional horn. That is, I was having a hard time seeing what the map $$\Lambda^{n+1}_{i+1}\to (\Lambda^n_i)^{\triangleleft} \coprod_{\Lambda^n_i} S$$ was. I was about to ask how to construct all of this formally, but a quick second with a pencil and paper made it clear. Thanks so much, I really appreciate it! The key point here is that $$\Lambda^{n+1}_{i+1}\cong (\Lambda^n_i)^{\triangleleft} \coprod_{\Lambda^n_i} \Delta^n$$ in a compatible way.

   Thanks! The trick I was missing was that since we’re gluing the cone over the horn back to S, we can take the original simplex as a face of the higher dimensional horn. That is, I was having a hard time seeing what the map

   $$\Lambda^{n+1}_{i+1}\to (\Lambda^n_i)^{\triangleleft} \coprod_{\Lambda^n_i} S$$

   was.

   I was about to ask how to construct all of this formally, but a quick second with a pencil and paper made it clear.

   Thanks so much, I really appreciate it!

   The key point here is that

   $$\Lambda^{n+1}_{i+1}\cong (\Lambda^n_i)^{\triangleleft} \coprod_{\Lambda^n_i} \Delta^n$$

   in a compatible way.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeMay 31st 2010
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   Author: Urs
   Format: MarkdownItexI added this, with a tiny bit of further details, to <a href="http://ncatlab.org/nlab/show/model+structure+for+left+fibrations#WeakEquivalences">model structure for left fibrations-- Properties -- Weak equivalences</a>.

   I added this, with a tiny bit of further details, to model structure for left fibrations– Properties – Weak equivalences.