There is a part of my proof of the following that I am unsure about:
Show that an arrow $1\to a$ whose domain is a terminal object must be monic.
R Goldbatt's Topoi: An object $1$ is terminal if for every $a$ there is only one arrow from $a$ to $1$.
An arrow $f:1\to a$ is monic if for any parallel arrow $g$ and $h$, if $f\circ g=f\circ h$ then $g=h$. Therefore I need to show that $g=h$.
My problem is this: if we assume $g$ and $h$ to be $g: a\to 1$ and $h: a\to 1$, then this is easy - since $1$ is terminal this must mean that there is only one unique arrow from $a$ to $1$ - meaning $g=h$.
But I am not at all sure if this assumption is legit. The definition of monic in Topoi actually gives $f:a\to b$ and $g:c\to a$ and $h:c\to a$ as an example, so the fact that I am assuming the domain/codomain of $g$ and $h$ for my own convenience seems too easy. Could anyone help please?
Your proof is fine—it really is that simple! Although you should call the domain of $g$ and $h$ something other than $a$ to avoid variable overload.
Specifically: saying $f : 1 \to a$ is monic means that, for all diagrams of the form $b \overset{g}{\underset{h}{\rightrightarrows}} 1 \xrightarrow{f} a$, if $f \circ g = f \circ h$, then $g=h$.
But the hypothesis that $f \circ g = f \circ h$ is redundant, since $g=h$ is automatic from the fact that $1$ is terminal. So you're done before you even begin.