Over $\mathbb{C}$, I am considering a map between projective curves $f: C \rightarrow C'$, where $C$ is smooth.
Suppose that $f$ is surjective as well as bijective away from the singularities of $C'$. Does it already follow that $C$ is a resolution of singularities?
The converse is clear, but I'm not sure how to formally prove this direction (if it is true at all).
We use the following fact: if $X$, $Y$ are integral schemes of finite type over $\mathbb{C}$ and $f: X \to Y$ is bijective and $Y$ is normal, then $f$ is birational.
See Akhil Matthew's post in https://mathoverflow.net/questions/73321/isomorphism-between-varieties-of-char-0 for the proof of this fact.
Now it's easy: look at $$f : C \backslash f^{-1}(\text{ Sing }(C')) \to C' \backslash \text{ Sing }(C')$$ and it satisfies the above criterion. Hence, $f$ is birational.
It's well-known that if two smooth curves are birational, they are isomorphic. Hence $C$ must be isomorphic to the normalization of $C'$.