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We are given a nonempty fibrant simplicial set $X$ (a Kan complex) and $v\in X_0$ is any vertex. We are $\alpha\in X_n$ and the map $\alpha:\Delta^n\to X$ is homotopic to $v:\Delta^n\overset{!}{\longrightarrow}\Delta^0\overset{v}{\longrightarrow}X$ relative to $\partial\Delta^n$ (throughout, I will abuse notation as Goerss and Jardine do, by letting $v$ denote any degenerate image of $v$ in any $X_n$).

Let $h:v\simeq\alpha:\Delta^n\times\Delta^1\to X$ be such a homotopy. We have $h\big|_{\partial\Delta^n\times\Delta^1}=v\circ\pi:\partial\Delta^n\times\Delta^1\to\partial\Delta^n\to X$ and $d_i\alpha=v$ for all $0\le i\le n$ by definition of relative homotopy.

The task is to show there exists $\omega\in X_{n+1}$ with boundary $\partial\omega=(d_0\omega,d_1\omega,\cdots,d_n\omega,d_{n+1}\omega)=(v,v,\cdots,v,\alpha)$.

I'm having surprising difficulty with this. I know that we can exploit anodyne extensions to 'fill horns' or similar, e.g. I considered in particular the following: $$(\Delta^1\times\Lambda^n_i)\cup(\partial\Delta^1\times\Delta^n)\hookrightarrow\Delta^1\times\Delta^n\\(\Delta^1\times\partial\Delta^n)\cup(\partial\Delta^1\times\Delta^n)\hookrightarrow\Delta^1\times\Delta^n\\(\Delta^1\times\partial\Delta^n)\cup(\Lambda^1_0\times\Delta^n)\hookrightarrow\Delta^1\times\Delta^n$$To generate new maps $\Delta^1\times\Delta^n\to X$. This was to no avail, because the only sensible maps to put on the components would have to use $h$ somewhere, but $h$ acts trivially or uninterestingly on each such one (e.g. $h$'s restriction to $\partial\Delta^1\times\Delta^n$ is the map $(v,\alpha)$ and this map exists regardless of whether or not $h$ does). In short, I did not see a way to exploit $h$ here.

I might consider similar extensions but with $n+1$ instead of $n$. However, a resulting map $\Delta^1\times\Delta^{n+1}\to X$ does not induce an interesting $\omega:\Delta^{n+1}\to X$, unless I'm mistaken.

Another approach might be to directly use $h$ in some composite $\omega:\Delta^{n+1}\overset{p}{\longrightarrow}\Delta^1\times\Delta^n\overset{h}{\longrightarrow}X$. In order for this to have $d_{n+1}\omega=\alpha$, we need $p(\delta_{n+1})$ equal to $(1,\mathrm{id}_n)$, and for $d_i\omega=v$ for all $i\le n$ we need to guarantee $p(\delta_i)$ is either in $\{0\}\times\Delta^n$ or in $\Delta^1\times\partial\Delta^n$. To make the two compatible, it can only be that $p(\delta_i)$ is in $\Delta^1\times\partial\Delta^n$.

But, examining all possible $(n+1)$-simplices of $\Delta^1\times\Delta^n$, I am fairly sure no such $p$ exists. I don't know what other tools I have at my disposal, so I'd really appreciate any hints or answers.

EDIT: Here is an explicit version of the below answer.

let $h:v\simeq\alpha$ rel. $\partial\Delta^n$. The maps $v\pi:\Delta^1\times\Delta^n\to\Delta^n\to\Delta^0\to X$ and $h:\Delta^1\times\Delta^n$ determine, by a known coequaliser, a unique map $(v\pi,v\pi,\cdots,h):\Delta^1\times\partial\Delta^{n+1}\to X$ since $h$ restricts to $v$ on its boundaries (relative homotopy condition). $v\pi:\{0\}\times\Delta^{n+1}\to\Delta^0\to X$ is compatible with $(v,\cdots,h)$ on the common intersection $\{0\}\times\partial\Delta^n$ since $h$ restricts to $v$ on $\{0\}\times(-)$, so we have a well defined map: $$(\Delta^1\times\partial\Delta^{n+1})\cup(\{0\}\times\Delta^{n+1})\overset{((v\pi,\cdots,v\pi,h),v\pi)}{\longrightarrow}X$$As $X$ is fibrant and the (obvious) inclusion $(\Delta^1\times\partial\Delta^{n+1})\cup(\{0\}\times\Delta^{n+1})\hookrightarrow\Delta^1\times\Delta^{n+1}$ is anodyne, there is an extension of this map to $\psi:\Delta^1\times\Delta^{n+1}\to X$.

If $\iota:\Delta^{n+1}\hookrightarrow\Delta^1\times\Delta^{n+1}$ maps $\sigma\mapsto(1,\sigma)$, then let $\omega:=\psi\circ\iota:\Delta^{n+1}\to X$ (considered also as an $(n+1)$-simplex). If $i\le n$, then $d_i\omega=\omega(\delta_i)=v\pi((1,1))=v$ and $d_n\omega=\omega(\delta_n)=h(1,1)=\alpha(1)=\alpha$ as required.

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2 Answers 2

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The first approach should work.

Map one side of the prism $\Delta^1 \times \Delta^{n+1}$ via $h$ and all others via the constant map equal to $v$. Map the button to $v$ as well. This produces a well-defined map from a "cup" to $X$. Extend it to a map $\Delta^1 \times \Delta^{n+1} \to X.$ Restrict to the top of the prism for the desired simplex.

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  • $\begingroup$ I see I was too quick to dismiss the idea of filling in a map $\Delta^1\times\Delta^{n+1}\to X$ (you should replace $n$ with $(n+1)$ for the benefit of others). $\endgroup$
    – FShrike
    Nov 23, 2022 at 9:49
  • $\begingroup$ @FShrike Oh, thanks $\endgroup$ Nov 23, 2022 at 19:32
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This can be usefully generalised, and I think this generalisation is what Goerss-Jardine wanted us to use (for the later proof, also left as an exercise).

Lemma:

Let $X$ be any nonempty fibrant simplicial set, $v\in X_0$ any distinguished vertex and $n\in\Bbb N$, $0\le i\le n+1$ arbitrary integers. For $\alpha\in X_n$ an $n$-simplex with boundary $(v,v,\cdots,v)$, $\alpha$ is homotopic to $v$ rel. $\partial\Delta^n$ iff. there exists an $\omega\in X_{n+1}$ with boundary $\underset{(0,1,\cdots,i-1,i,i+1,\cdots,n+1)}{\underbrace{(v,\cdots,v,\alpha,v,\cdots,v)}}$.

Proof:

Let $X,\alpha,i,n$ be given and assume existence of such an $\omega$. There is a map: $$\large(\Delta^{n+1}\times\partial\Delta^1)\cup(\Lambda^{n+1}_i\times\Delta^1)\overset{((\omega,v),(v,\cdots,v,-,v,\cdots,v))}{\longrightarrow}X$$

Since $d_j\omega=v$ for $j\neq i$. Since $X$ is fibrant, there is a lift of this map along the anodyne extension $(\Delta^{n+1}\times\partial\Delta^1)\cup(\Lambda^{n+1}_i\times\Delta^1)\hookrightarrow\Delta^{n+1}\times\Delta^1$ given by some $\psi:\Delta^{n+1}\times\Delta^1\to X$.

In fact, $\psi$ is a homotopy $\omega\simeq v$, but this homotopy is not necessarily relative to $\partial\Delta^{n+1}$. Anyway, define $h:\Delta^n\times\Delta^1\to X$ as the composite $\psi\circ((\delta_i\circ(-))\times1)$.

We have that $h$ restricts to $v$ on $\partial\Delta^n\times\Delta^1$ since $\delta_i\circ f\in\Lambda^{n+1}_i$ for any non-surjective $f$ and $h$ restricts to $(d_i\omega,v)$ on $\Delta^n\times\partial\Delta^1$, so $h:\alpha\simeq v$ is a relative homotopy.

Conversely suppose such an $h$ exists, $\alpha\simeq v$ rel. $\partial\Delta^n$. There is a map: $$\large(\Delta^{n+1}\times\{1\})\cup(\partial\Delta^{n+1}\times\Delta^1)\overset{\overset{(0,\cdots,i-1,i,i+1,\cdots,n+1)}{\overbrace{(v,(v,\cdots,v,h,v,\cdots,v))}}}{\longrightarrow}X$$

Valid for reasons described in my question. There is a lift of this map along the anodyne extension $(\Delta^{n+1}\times\{1\})\cup(\partial\Delta^{n+1}\times\Delta^1)\hookrightarrow\Delta^{n+1}\times\Delta^1$ given by some $\psi:\Delta^{n+1}\times\Delta^1\to X$. Define $\omega:=\psi\circ(-,0):\Delta^{n+1}\hookrightarrow\Delta^{n+1}\times\Delta^1\to X$.

The boundary of this simplex $\omega$ is precisely $(v,\cdots,v,\alpha,v,\cdots,v)$ as required.

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