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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).

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  • $\begingroup$ This is false even in 1-dimensional case (when $M$ is a unit disk: think of the Riemann mapping theorem). $\endgroup$ – Moishe Kohan Apr 29 at 14:05
  • $\begingroup$ @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. $\endgroup$ – Curiosity Apr 29 at 14:21
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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.

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