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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 9th 2011
   • (edited Feb 9th 2011)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexReposted from [MO](http://mathoverflow.net/questions/54866/the-pushout-of-a-mapping-simplex-along-products-is-a-mapping-simplex): Background ---- ---- The mapping simplex: We define a functor $M^n:Fun([n],sSet)\to sSet$ sending the functor $\phi:[n]\to sSet$ classifying the sequence $$A^0\leftarrow A^1\leftarrow\dots\leftarrow A^n$$ (notice that $A^k=\phi(n-k)$.) to the simplicial set $M^n(\phi)$ defined as a representable functor as follows: $$Hom_{sSet}(\Delta^k,M^n(\phi))=\coprod_{\{g:[k]\to[n]\}}Hom_{sSet}(\Delta^k,A^{g(k)})$$ The identity map $M^n(\phi)\to M^n(\phi)$ then determines a unique map $M^n(\phi)\to \Delta^n$ by the fact that simplicial sets are colimits of their simplices. Also, given a map $\Delta^m\to \Delta^n$ corresponding to a functor $F:[m]\to [n]$ (by full and faithfulness of the Yoneda embedding), it's not hard to see that $$M^n(\phi)\times_{\Delta^n} \Delta^m\cong M^m(F^*\phi)$$ Let $\phi:[n]\to sSet$ be a functor classifying a sequence of composable maps $$A^0\leftarrow A^1\leftarrow\dots\leftarrow A^n$$ Then let $c_{A^n}$ be the constant functor $[n]\to sSet$ at $A^n$. Then by the functoriality of $M^n$, we have a map $M^n(c_{A^n})\to M^n(\phi)$ induced by the obvious natural transformation $\alpha:c_{A^n}\to \phi$, which is defined componentwise as the composite map $\alpha_i:A^n\to A^i$ for each $i$ (the naturality of this map is immediate). Notice that $M^n(c_A)$ for any simplicial set $A$ is canonically isomorphic to the product $A\times \Delta^n$ since $$Hom(\Delta^m,A\times \Delta^n)\cong Hom(\Delta^m,A)\times Hom(\Delta^m,\Delta^n)=\coprod_{\{[m]\to [n]\}}Hom(\Delta^m,A).$$ Problem ----- ----- Let $\iota:[n-1]\to [n]$ be the obvious inclusion on the last $n-1$ objects of $[n]$, and let $\phi:[n]\to sSet$ be a functor parameterizing a sequence of composable maps $$A^0\leftarrow A^1\leftarrow\dots\leftarrow A^n.$$ Let $\phi'=\iota^*\phi$, which parameterizes the sequence: $$A^0\leftarrow A^1\leftarrow\dots\leftarrow A^{n-1}$$ (yes, the indexing is annoying, since $A^k=\phi(n-k)$). Also, let $c_{A^n}:[n]\to sSet$ be the constant functor at $A^n$, and let $c'_{A^n}=\iota^*c_{A^n}$. We also define a third sequence $d_{A^n}:[n]\to sSet$ to be the sequence $$A^n\leftarrow A^n\leftarrow\dots\leftarrow A^n\leftarrow \emptyset$$ which extends $c'_{A^n}$ by the empty object on the right. It is immediate from the definition that we have a canonical isomorphism $M^{n-1}(c'_{A^n})\cong M^n(d_{A^n})$. This gives us a natural inclusion $M^{n-1}(c'_{A^n})\cong M^n(d_{A^n})\hookrightarrow M^n(c_{A^n})$ Then we apparently have the following pushout square: $$\begin{matrix}M^{n-1}(c'_{A^n})&\hookrightarrow &M^n(c_{A^n})\\ \downarrow&&\downarrow\\ M^{n-1}(\phi')&\hookrightarrow& P\end{matrix}$$ Claim: $P\cong M^n(\phi)$ Then why does the claim hold?

   Reposted from MO:

   Background

   ---

   The mapping simplex: We define a functor sending the functor classifying the sequence

   (notice that .) to the simplicial set defined as a representable functor as follows:

   The identity map then determines a unique map by the fact that simplicial sets are colimits of their simplices.

   Also, given a map corresponding to a functor (by full and faithfulness of the Yoneda embedding), it’s not hard to see that

   Let be a functor classifying a sequence of composable maps

   Then let be the constant functor at . Then by the functoriality of , we have a map induced by the obvious natural transformation , which is defined componentwise as the composite map for each (the naturality of this map is immediate).

   Notice that for any simplicial set is canonically isomorphic to the product since

   Problem

   ---

   Let be the obvious inclusion on the last objects of , and let be a functor parameterizing a sequence of composable maps

   Let , which parameterizes the sequence:

   (yes, the indexing is annoying, since ).

   Also, let be the constant functor at , and let . We also define a third sequence to be the sequence

   which extends by the empty object on the right. It is immediate from the definition that we have a canonical isomorphism . This gives us a natural inclusion

   Then we apparently have the following pushout square:

   Claim:

   Then why does the claim hold?

2. • CommentRowNumber2.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 10th 2011
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   Author: Harry Gindi
   Format: MarkdownItexDoes anyone know how to show this? It's geometrically obvious, but I'm having a rough time showing it to be true. The mapping properties are in exactly the wrong directions =(.

   Does anyone know how to show this? It’s geometrically obvious, but I’m having a rough time showing it to be true. The mapping properties are in exactly the wrong directions =(.

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 11th 2011
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   Author: Harry Gindi
   Format: MarkdownItexAlright, I think I answered my own question on MO. Any chance anyone can tell me if the proof works?

   Alright, I think I answered my own question on MO. Any chance anyone can tell me if the proof works?