TITLE: Are endomorphisms of degree one always automorphisms
QUESTION [7 upvotes]: Let $B$ be a smooth projective connected variety over $\mathbb C$. Let $\sigma:B\to B$ be an endomorphism of degree one.
Do I understand correctly that $\sigma$ is an automorphism?
I believe this should follow from Zariski's main theorem, I thought it wouldn't hurt to be sure by asking here.
More generally, what kind of assumptions do we really need on $B$ to conclude that every degree one endomorphism is an automorphism?

REPLY [2 votes]: This is true for any normal irreducible algebraic variety $B$ over any field $k$. A (rather easy) proof using Grothendieck groups of varieties over $k$ can be found in § 4, Corollary 8 of this paper.
A few words on the proof. Suppose $f$ is not finite. Then by ZMT, there exists a proper closed subset $F$ of $B$ such that $f^{-1}(B\setminus F)\to B\setminus F$ is an isomorphism, and $f^{-1}(F)\to F$ has connected fibers of positive dimension. In particular, 
$$\dim f^{-1}(F)>\dim F.$$ 
Now in the Grothendieck ring $K_0(Var_k)$ of algebraic varieties (= schemes of finite type) over $k$, we have 
$$[f^{-1}(F)]=[B]-[B\setminus f^{-1}(F)]=[B]-[B\setminus F]=[F].$$
But in the linked paper, it is proved that the dimension of a variety is determined by its class in $K_0(Var_k)$. Contradiction.