If $f$ is an epimorphism of frames, it is surjective as function? If $f$ is an epimorphism of frames, then is it surjective as a function?.
The frames behaves so much like topological spaces (moreover the locales which are the opposite category), so the question is in the same same spirit. I was thinking something like the inclusion of the rationals in the reals. But neither of them are frames.
 A: No.  For instance, let $A=\mathcal{P}(\{0,1\})$ and $B=A\setminus\{\{1\}\}$ and let $f:B\to A$ be the inclusion map.  Then $f$ is a frame homomorphism, and I claim it is epic.  Indeed, suppose $g:A\to C$ is a frame homomorphism; we must show that $g(\{1\})$ is uniquely determined by the restriction of $g$ to $B$.  Let $c=g(\{0\})$; then $g(\{1\})$ is an element $d\in C$ such that $c\vee d=1$ and $c\wedge d=0$.  At most one such $d$ can exist, since if $d'$ is any other such element, then $$d'=d'\wedge(c\vee d)=(d'\wedge c)\vee (d'\wedge d)=0\vee (d'\wedge d)=d'\wedge d$$ so $d'\leq d$ and by symmetry $d\leq d'$ as well.
This should not be surprising: if we think that frames are roughly dual to topological spaces, then the dual statement about topological spaces is that for a monomorphism of topological spaces, the domain must have the subspace topology induced from the codomain.  This is of course false, since you can have a continuous injection where the domain has a strictly finer topology than the subspace topology.  Indeed, the example in frames above comes from such an example in topological spaces: a continuous bijection $X\to Y$ where $Y$ has two points and the discrete topology and $X$ has two points but only one of the points is closed.
