I encountered this problem while reading the proof of Lemma 1.2.4.17 in Jacob Lurie's Higher Algebra
Recall that a topological simplex $|\Delta^n|$ can be identified with $\{0\leq x_1\leq\cdots\leq x_n\leq1\}\subset[0,1]^n$
Let $|X|\subset|\Delta^m|\times|\Delta^n|$ be the following space:
$$|X|=\big\{(0\leq x_1\leq\cdots\leq x_m\leq1,0\leq y_1\leq\cdots\leq y_n\leq1):\{x_1,\cdots,x_m\}\subset\{0,y_1,\cdots,y_n,1\}\big\}$$
Equivalent description: $|X|=\bigcup_\alpha im(|\alpha|,id)$, where $[n]=\{0<1<\cdots<n\}$ finite partially ordered set,
$\alpha:[n]\to[m]$ runs through all order preserving maps, $(|\alpha|,id):|\Delta^n|\to|\Delta^m|\times|\Delta^n|$ the induced map.
We need to prove that $|X|$ is a contractible space.
Lurie claimed that it suffices to prove that each fiber of the projection $X\to|\Delta^m|$ is contractible, which was showed in HA Lemma 1.2.4.16.
However, as can be seen in the case for small $m,n$, the projection $X\to|\Delta^m|$ is never fibration. It seems not sufficient to just prove that every fiber is contractible.
$\require{AMScd}$
The case $m=n=1$: first coordinate for $\Delta^m$, second coordinate for $\Delta^n$:
\begin{CD}&&(0,0)@>>>(0,1)\\&@VVV\\(1,0)@>>>(1,1)\end{CD}
The case $m=1,n=2$ is also easy to draw, but I can't type it here use the poorly behaved “AMScd”.
Both the above two graph can be firstly contracted to the diagonal segment $(0,0)\to(1,1)$, and then contracted to a point.
The case $m=n=2$ is already quite complicated, and I didn't find a good way to deal with it, let alone more general case.
Update:
(1) Let $X=\bigcup_\alpha im(\alpha,id)$ the underlying simplicial set,
where $(\alpha,id):\Delta^n\to\Delta^m\times\Delta^n$ is defined as above.
It suffices to show that $X\subset\Delta^m\times\Delta^n$ is anodyne.
Perhaps there's a combinatorial proof?
(2) We should assume in addition that $m\leq n$
For the case $m=2,n=1$, $|X|$ is a 1-dim complex, and there is a nontrivial loop in $|X|$,
$(0,0)\to(1,1)$, $(0,0)\to(1,2)$, $(0,1)\to(1,1)$, $(0,1)\to(1,2)$\
Oh, I've worked it out.
Just note that the projection $p:X\to\Delta^m$ admits a section.
For example, let $\tau:[n]\to[m]$, $\tau(i)=i,\forall i\leq m$, $\tau(i)=m,\forall i>m$, let $\iota:[m]\subset[n]$, $\iota(j)=j$
Then $\Delta^m\simeq im\big((id,\iota):\Delta^m\to\Delta^m\times\Delta^n\big)\subset im\big((\tau,id):\Delta^n\to\Delta^m\times\Delta^n\big)$ is a section of $p$
This induced a section $|s|:|\Delta^m|\hookrightarrow|X|$ of $|p|:|X|\to|\Delta^m|$
As proved in HA 1.2.4.16 (by unwind the definitions), every fiber of $|p|$ is a topological simplex.
More precisely, for $(0\leq x_1\leq\cdots\leq x_m\leq1)=x\in|\Delta^m|$,
let $r=\#\{x_1,\cdots,x_m\}$, then $|p|^{-1}(x)\simeq|\Delta^{n-r}|$.
In particular, $\forall z\in|p|^{-1}(x)$, there is a line segment joining $z$ to $|s|(x)$
Now there is a deformation retract $|X|\to|\Delta^m|$ by shrinking each fiber $|p|^{-1}(x)$ to the point $|s|(x)$
Hence $|X|$ is homotopy equivalent to $|\Delta^m|$, which is contractible.
The condition $n\geq m$ is necessary, to ensure the existence of a surjection $\tau:[n]\to[m]$