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Let $X$ be a normal projective irreducible surface over an algebraically closed field $k$. Let $\pi\colon Y\to X$ be a birational morphism, such that $Y$ is a smooth projective surface, and assume that no $(-1)$-curve of $Y$ is contracted by $\pi$. How do we prove that there is an isomorphism $\pi^{-1}(X_s)\to X_s$, where $X_s$ denotes the smooth locus of $X$, and that $\pi^{-1}(x)$ is a connected curve for each $x\in X\setminus X_s$?

I would say that it follows from the fact that there exists a unique minimal resolution and that on the smooth locus we only have blow-ups. Does someone has a reference for the statement?

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The statement that $\pi^{-1}(x)$ is (geometrically) connected for all $x \in X$ is Stein factorisation or Zariski's main theorem; see e.g. [Hartshorne, Cor. III.11.5].

For the other statement, it suffices to show that if $\pi \colon Y \to X$ is a proper birational morphism of smooth surfaces, then $\pi$ is obtained as a sequence of blow-ups in points. Indeed, applying this to the smooth locus shows that $\pi$ is an isomorphism there, as otherwise the last blowup introduces a $(-1)$-curve contracted by $\pi$.

This statement is for instance proved in [Beauville, Thm. II.11] (stated under the hypotheses that $k = \mathbf C$ and $X$ is projective, but this is not used in the proof).

References.

[Beauville] A. Beauville, Complex algebraic surfaces, second edition. London Mathematical Society Student Texts 34. Cambridge Univ. Press, 1996. ZBL0849.14014.

[Hartshorne] R. Hartshorne, Algebraic geometry. Graduate Texts in Mathematics 52. Springer-Verlag, 1977. ZBL0367.14001.

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  • $\begingroup$ Thanks for your answer. Why is $\pi^{-1}(x)$ never a point for $x\in X$ singular? [Hartshorne, Cor. III.11.5] shows that the preimage is connected, but why not only one point? Can we show that we would have a local isomorphism in this case? $\endgroup$
    – Jérémy Blanc
    May 3 at 12:39
  • $\begingroup$ If $\pi^{-1}(x) = \{y\}$, then $Y \times_X \operatorname{Spec} \mathcal O_{X,x} \cong \operatorname{Spec} \mathcal O_{Y,y}$, so $\mathcal O_{X,x} \to \mathcal O_{Y,y}$ is finite (since it is proper and quasi-finite). Since it is also birational extension, normality of $\mathcal O_{X,x}$ shows that $\mathcal O_{X,x} \stackrel\sim\to \mathcal O_{Y,y}$. $\endgroup$
    – R. van Dobben de Bruyn
    May 3 at 20:01

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