Let $f: M \rightarrow M$ be an endomorphism of a connected complex manifold $M$. Assume that $f(M)$ is a dense open set in $M$ and that $M \setminus f(M)$ is an analytic subset of $M$
Question: Can I deduce that if $f$ is injective then $f$ is surjective, thus biholomorphic?
This is true when $M$ is one dimensional (this is false,see edit) i.e. a Riemann surface but I don't know whether it is still true in higher dimension. In another category like algebraic varieties with regular morphisms, it is also true thanks to Ax-Grothendieck theorem.
Edit: This is not true thanks to the answer of Moishe. My argument is that $M$ and $f(M)$ are biholomorphic and $f(M)$ is punctured Riemann surface obtained from $M$. If $M$ is finite type then this is impossible but if $M$ is finite type (like $\mathbb{C}$ removed $\mathbb{N}$ then it is possible).
Best Answer
This is already false for Riemann surfaces: Consider $M$ equal to ${\mathbb C}$ with the set of natural numbers removed and let $f(z)=z-1$; $f(M)\subset M$ and the complement $M - f(M)$ is a singleton.