Is a proper map of varieties which is a bijection on points an isomorphism?

Question

Suppose that I have a proper morphism $f: X \to Y$ of varieties (i.e. reduced separated schemes of finite type). I am given that (a) on a dense open $U \subseteq Y$, $f$ is an isomorphism (i.e. $X\times_Y U \to U$ is an isomorphism), and (b) the pullback $(Y\backslash U)\times_Y X \to Y\backslash U$ is also an isomorphism. (Both pullbacks here are in the category of schemes.) Does it follow that $f$ must be an isomorphism?

(ETA: clarification of the condition that I need.)

It's clearly the case that this is not true if the schemes are not varieties, because the map $\operatorname{Spec}k \to \operatorname{Spec} k[\epsilon]/(\epsilon^2)$ is proper but not an isomorphism, but setting $U = \operatorname{Spec}0$ gives the above properties. But with the extra condition of varieties this seems to work, but a sanity check would be very helpful. Does anyone know of a reference for this, or a reason it's true/not true?

Answer 1

This is true. By Zariski's main theorem, we know that $f$ is finite, since it is proper and quasi-finite [Stacks, Tag 02LS]. We actually have the following:

Lemma. Let $Y$ be a reduced scheme and let $f \colon X \to Y$ be a finite morphism of schemes such that $X_y \to \{y\}$ is an isomorphism for all $y \in Y$. Then $f$ is an isomorphism.

Proof. The question is local on $Y$, so we may assume that $Y = \operatorname{Spec} A$ is affine. Then $X = \operatorname{Spec} B$ for some finite ring map $\phi \colon A \to B$. Since $\dim_{\kappa( p)}(B \otimes_A \kappa(\mathfrak p)) = 1$ for all primes $\mathfrak p \subseteq A$, we see that $B$ is locally free of rank $1$ over $A$ [Stacks, Tag 0FWG]. Localising further, we may assume that $B$ is free of rank $1$ over $A$; say $B = Ab$ for some $b \in B$. In particular, there exist $a,c \in A$ such that $1 = \phi(a)b$ and $b^2 = \phi(c)b$. The first gives $b \in B^\times$, so the second gives $b = \phi(c)$. Therefore, $\phi \colon A \to B$ is a map of free $A$-modules of rank $1$ whose image contains a generator, so it is an isomorphism. $\square$