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• CommentRowNumber1.
• CommentAuthorHarry Gindi
• CommentTimeFeb 9th 2011
• (edited Feb 9th 2011)

Reposted from MO:

## 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?

• CommentRowNumber2.
• CommentAuthorHarry Gindi
• CommentTimeFeb 10th 2011

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 =(.

• CommentRowNumber3.
• CommentAuthorHarry Gindi
• CommentTimeFeb 11th 2011

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