Order-preserving surjective maps between finite linear orders Suppose I have two order-preserving, surjective maps $s \colon [m] \to [n] $ and $s' \colon [m'] \to [n]$. I need to show that there exists some $l \in \mathbb{N}$ and order-preserving, surjective maps $t \colon [l] \to [m]$ and $t' \colon [l] \to [m']$ such that $s \circ t = s' \circ t'$. What's the quickest way to prove this?
I have a vague idea that it might involve the maps $\sigma_{i}^{n} \colon \left[n+1\right] \to \left[n\right]$ that are given by $$\sigma_{i}^{n}\left(j\right) = \begin{cases} j & \text{if } 0 \leq j \leq i \\ j-1 & \text{if } i+1 \leq j \leq n+1 \end{cases}$$. That is, start by writing $s$ and $s'$ as composites of the $\sigma^{n}_{i}$. But this seems way too tedious and I'm getting lost in the indices.
Another vague idea I have is to make a list of how many times each entry in $[n]$ is hit by $s$ and $s'$ and to construct an order-preserving surjective map $[l] \to [n]$ so that the hits of each entry in $[n]$ under the newly constructed map is the sum of hits under $s$ and under $s'$ (maybe we need to subtract one from the total). But then, I need to construct $t$ and $t'$ such that this newly constructed map is equal to the composites $s \circ t = s' \circ t'$. Again, getting lost in the weeds at this point.
I'd ideally like a clean, category theoretic argument (without resorting to multiple cases and indices) but I'm not sure this is possible since the maps involved are probably not unique. Any help appreciated!
 A: It helps to look at this from a different perspective. Given an order-preserving surjective map $f:[m]\to [n]$, consider the list $[f(1),f(2),\dots,f(m)]$. This will look something like
$$
[1,1,1,1,2,2,3,4,5,5,5]
$$
This can be written more succinctly as
$$
\text{four $1$'s, two $2$'s, one $3$, one $4$, three $5$'s},
$$
or even more succinctly as
$$
(\text{four, two, one, one, three})=(4,2,1,1,3)
$$
That is, order-preserving surjective maps $[m]\to[n]$ are in bijection with compositions of $m$ with $n$ parts.
You want to find $t$ and $t'$ so that $s\circ t=s'\circ t'$. Since every order-preserving surjective map can be written as a composition of the maps $\sigma^r_k$ for various choices of $r$ and $k$, it helps to understand how the composition for $s$ relates to the composition for $s\circ \sigma_{k}^{n+1}$.
Let's look at some examples, using the same $f$ from before:




map
list form
composition




$f$
$[1,1,1,1,2,2,3,4,5,5,5]$
$(4,2,1,1,3)$


$f\circ \sigma^{12}_1$
$[1,1,1,1,1,2,2,3,4,5,5,5]$
$(\color{blue}5,2,1,1,3)$


$f\circ \sigma^{12}_2$
$[1,1,1,1,1,2,2,3,4,5,5,5]$
$(\color{blue}5,2,1,1,3)$


$f\circ \sigma^{12}_6$
$[1,1,1,1,2,2,2,3,4,5,5,5]$
$(4,\color{blue}3,1,1,3)$


$f\circ \sigma^{12}_7$
$[1,1,1,1,2,2,3,3,4,5,5,5]$
$(4,2,\color{blue}2,1,3)$




If you think about these examples enough, you can see that the composition for $f\circ \sigma_k^{n+1}$ is given by incrementing one of the parts of the composition for $f$. Furthermore, by appropriately choosing $k$, you can choose to increment any of the parts in this fashion. I will refer to this operation as inflation.
Since $t$ can be written as a composition of $\sigma^r_k$ for various $r$ and $k$, finding $t$ and $t'$ for which $s\circ t=s'\circ t'$ is equivalent to finding two series of inflations, one applied to the composition for $s$, the other applied to the composition for $s'$, such that the resulting compositions are equal.  This is doable with the following very simple procedure. Suppose $s$ has composition $(a_1,a_2,\dots,a_n)$ and $s'$ has composition $(b_1,b_2,\dots,b_n)$. For each $i\in \{1,\dots,n\}$, look at $a_i$ and $b_i$. If $a_i<b_i$, then inflate $a_i$ until it equals $b_i$, and if $a_i\ge b_i$, then inflate $b_i$ until it equals $a_i$.
