Let $X,Y$ be two normal algebraic surfaces (for instance projective) and let $\varphi\colon X\dashrightarrow Y$ be a birational map which restricts to an isomorphism $(X\setminus F)\to (Y\setminus G)$ where $F\subset X,G\subset Y$ are finite subsets. Does it follow that $\varphi$ is an isomorphism?

(This is true at least when $X$ and $Y$ are smooth).

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    $\begingroup$ There are trivial (smooth) counterexamples when "projective" is dropped. I guess you mean this is true in the smooth projective (or complete) case; probably the affine smooth case follows by some kind of Hartogs phenomenon. $\endgroup$
    – YCor
    Commented Nov 10, 2016 at 7:58

1 Answer 1


I think so (even in dimension higher than 2, assuming that $F$ and $G$ are still finite sets, and not codimension two subvarieties of course). Let $H$ be an ample divisor in $X$ avoiding $F$ and let $H'$ be its strict transform in $Y$. Then $H'$ is ample in $Y$. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.


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