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

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?

• The normalization of a projective cuspidal curve (e.g. the homogenization of $y^2=x^3$) over an algebraically closed field is a counterexample (i.e. it is proper, birational, and bijective on the underlying topological space).
– Eoin
Commented Sep 5, 2023 at 2:35
• If the target is normal then it is an isomorphism. This is Zariaki’s main theorem. Commented Sep 5, 2023 at 13:45
• @Eoin Sorry, I forgot that I needed a stronger condition. I've corrected it now.
– Inna
Commented Sep 5, 2023 at 23:08

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. $$\square$$