Epimorphism and surjection

Question

I have seen on http://mathworld.wolfram.com/Epimorphism.html

A morphism $f:Y \to X$ in a category is an epimorphism if, for any two morphisms $u,v:X\to Z, uf=vf$ implies $u=v$. In the categories of sets, groups, modules, etc., an epimorphism is the same as a surjection, and is used synonymously with "surjection" outside of category theory.

My question is, if $X,Y$ are manifolds, epimorphism and surjection are the same? Because I found that, one usually uses "surjection" for a map between two non-specific manifolds $M, N$, and "epimorphism" for a map between tangent bundle $TM, TN$.

Answer 1

No. Let $X=(0,2\pi)$, $Y=S^1$, and $f(x)=(\cos x,\sin x)$. Then $f$ is not surjective, since $(1,0)\notin f\bigl((0,2\pi)\bigr)$. But $f$ is an epimorphism, because if $Z$ is another manifold, and $u,v\colon S^1\longrightarrow Z$ are differentiable maps such that $u\circ f=v\circ f$, then $u=v$, because the image of $f$ is dense in $S^1$.

Answer 2

The inclusion $i : [0,1) \hookrightarrow [0,1]$ seems to have this property. Indeed, if $u,v : [0,1] \to M$ are smooth maps which agree on $[0,1)$ then we should have $u(1) = v(1)$ by continuity.