Let $\Delta$ be the category of finite nonempty ordinals, known as the simplex category. Its objects are finite nonempty ordinals and its morphisms are order-preserving maps between those. I want to know if every epimorphism in $\Delta$ is surjective.
There are well-known proofs of the fact that monomorphisms (resp., epimorphisms) in $\mathsf{Set}$ are precisely injective (resp., surjective) functions. The proof of the monomorphism case carry out line-by-line as the one-element set $\{0\}$ is a finite ordinal and any function $f\colon\{0\}\to n$ is trivially order-preserving. The same cannot be said about epimorphisms. If $f\colon m\to n$ is an epimorphism in $\Delta$ which is not surjective (a start of a proof by contradiction) and $k \in n\setminus\mathrm{im}(f)$, then a function $g\colon n\to n\cup\{n\} = n + 1$ defined by $g(k') = k$ if $k' \neq k$ and $g(k) = n$ is not necessarily order-preserving (this is a step in the proof that every epimorphism in $\mathsf{Set}$ is surjective where instead of $n$ we take any $a$ such that $a \notin n$).
Best Answer
All maps here will be (weakly) order-preserving.
Let $f:[m]\to[n]$ not be surjective. We claim there are distinct maps $g_1,g_2 :[n]\to[n]$ with $g_1\circ f\ne g_2\circ f$. Suppose $j$ is omitted from the image of $f$. Let $g_1$ be the identity, and $g_2$ be the identity on all elements other than $j$ and also $g_2(j)=j-1$ or $j+1$ (one of which is in $[n]$).
This works unless $n=1$, and then $m$ must be $0$, in which case we need some ad hoc argument.