Consider $\Delta$ the simplex category, with objects $[n]=\{0,\dots,n\}$ and morphisms $f:[n]\to [m]$ such that $i<j\implies f(i)\leq f(j)$ (my definition is with $i<j$). I have shown that epimorphisms are exactly surjective maps.
Now I want to show that epimorphisms are split, that is if $f:[n]\to [m]$ is surjective, then $\exists s:[m]\to [n]$ such that $f\circ s=id_{[m]}$. To do so I'm told to use cosimplicial identities, that is $\delta^j \delta^i=\delta^i \delta^{j-1}$ and $\sigma^j \sigma^i=\sigma^i \sigma^{j+1}$ which are defined for $n\geq 1$ and $0\leq j \leq n$ by:
$\delta^j:[n-1]\to[n]$, $\delta^j(x)=x$ if $x<j$, $\delta^j(x)=x+1$ if $x\geq j$
$\sigma^j:[n+1]\to[n]$, $\sigma^j(x)=x$ if $x\leq j$, $\sigma^j(x)=x-1$ if $x>j$
However ignoring this and defining $s:[m]\to [n]$ by $s(x)=\text{min}\{y\in[n]|f(y)=x\}$ (exists since $f$ surjective) we get a section of $f$ (at least I think).
We have that $f(s(x))=f(\text{min}\{y\in[n]|f(y)=x\})=x$ If $x<y$, suppose for contradiction that $s(x)>s(y)$, then $f(s(y))\leq f(s(x))\implies y\leq x$ absurd so $f$ is split.
I have two questions, is it correct ? How do we do this using cosimplicial identities ?
Your definition of $s$ is good, perfectly fine.
How to do it with cosimplicial identities? Well, we have a canonical factorisation of any $f$ as $f=\delta\delta\delta\cdots\sigma\sigma\sigma$ where I've omitted whatever subscripts or superscripts are necessary. If $f$ is epi, that means there are no $\delta$ terms in this canonical factorisation. So $f$ is just the composite of some string of $\sigma_i$'s.
The cosimplicial identities give you an explicit way to find a right inverse to such a string. Well, the identities give you that canonical factorisation of $f$ (proof found in CWM, e.g.) and the easier cosimplicial identity: $\sigma_i\delta_i=\sigma_i\delta_{i+1}=\mathrm{Id}$ allows you to find a right inverse.