I recently learned about simplicial sets, and now I'm studying some basic properties. I've learned that every degenerate simplex is a degeneracy of a unique non-degenerate (nice) simplex. However, I suspect that the entire representation must also be unique.
(This question started as a request for help when I was stuck, but while I was writing the post I figured it out myself, so I end up asking if you could check my proof.)
This means, given a nice simplex $x$ and
$$\large
z = s_{j_n}\cdots s_{j_1}x = s_{i_n}\cdots s_{i_1}x,
\qquad 0\le j_1 <\cdots <j_n, \;\; 0\le i_1 <\cdots< i_n
$$
does this imply that $j_k=i_k$ for all $k$? If $s_j x = s_i x,\ i<j$, for a simplex $x$, then
$$\large
x = d_i s_i x = d_i s_j x = s_{j-1} d_i x
$$
So if $x$ is nice, then $s_j x \ne s_k x$. This can be seen as the induction start.
Assume that the claim is true for $n$.
Let's first consider the case $j_{n+1} > i_{n+1}+1$. If we apply $d_{j_{n+1}}$ to both sides, we get
$$\large
d_{j_{n+1}} z = s_{j_n} \cdots s_{j_1} x =
s_{i_{n+1}} \cdots s_{i_n}d_{j_{n+1}-n-1} x
$$
which means that $d_{j_{n+1}} z$ is degeneracy of $x$ and of $d_{j_{n+1}} x$, but then $d_{j_{n+1}}x$ has to be a degeneracy of $x$. We can conclude that $j_{n+1}\in\{i_{n+1},i_{n+1}+1\}$.
Clearly, $j_{n+1} = i_{n+1}$ implies by induction that $j_k = i_k$ for all $k$. But what if $j_{n+1}=i_{n+1}+1$?
We can write
$$\large\begin{align}
s_{i_n}\cdots s_{i_1} x
& = d_{i_{n+1}} s_{i_{n+1}} \cdots s_{i_1} x \\
& = d_{i_{n+1}} s_{i_{n+1}+1} \cdots s_{j_1} x \\
& = s_{i_{n+1}} d_{i_{n+1}} s_{j_n} \cdots s_{j_1} x
\end{align}$$
If $i_{n+1} > j_n +1$, the last line would transform into an $(n+1)$-fold degeneracy of $d_{i_{n+1}-n} x$. So $d_{i_{n+1}}s_{j_n}$ must vanish, and we can use the induction hypothesis to conclude that $i_n = i_{n+1}$, which is a contradiction.
Is this proof correct? And can it be simplified?
Thanks.