# Injective holomorphic endomorphism whose image is the complement of a proper analytic subset is surjective.

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).

• This is false even in 1-dimensional case (when $M$ is a unit disk: think of the Riemann mapping theorem). – Moishe Kohan Apr 29 at 14:05
• @MoisheKohan Thank you. I edited the question. Open dense is not that strong condition. If I replace the condition is that $M \setminus f(M)$ is an analytic set then it is true in 1-dimensional by using fundamental groups since proper analytic sets in 1-dimensional are discrete sets. – Curiosity Apr 29 at 14:21

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.