Let $X$ and $Y$ be two complex, irreducible, normal, projective varieties (read: integral, projective, normal $\mathbb C$-schemes of finite type), projective in the sense of Hartshorne. Let $\phi:X\dashrightarrow Y$ be a birational map. I will think of $\phi$ as a morphism $\phi:U\to Y$ defined on an open set $U\subseteq X$ where we assume $U$ to be maximal, i.e. $Z:=X\setminus U$ is the closed set of points where $\phi$ is not defined.

Assume that $\phi$ is surjective, i.e. $\phi(U)=Y$. One could also say that $\phi$ is dominant and its image is closed. In this situation, is $\phi$ always proper?

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    $\begingroup$ If $\phi$ is proper, then $U$ is proper (because $Y$ is), so $U=X$. $\endgroup$ Mar 9, 2015 at 19:21
  • $\begingroup$ Yea, that's the general idea =). It would then be some blowing up of $Y$. $\endgroup$ Mar 9, 2015 at 19:44

1 Answer 1


tl;dr: No. There exists a flop $X \dashrightarrow W$ such that the flopped curve in $W$ is contained in the exceptional locus of a blowndown $W \to Y$, but isn't contracted by it. The complement of the flopping curve in $X$ still surjects onto $Y$ because the map is defined on the rest of the exceptional locus of $W \to Y$. However, the map is not a morphism since the image of the flopped curve under the contraction is not a point.

Full construction: No. Let $Y$ be your favorite variety which admits a standard flop, and let $C$ be the flopping curve. Let $f : Z \to Y$ be the blow-up along $C$, with exceptional divisor $E$. Since $C$ is a flopping curve, there's also a map $g : Z \to Y^+$ contracting $E$ along the other ruling.

Now, let $B \subset Z$ be any curve meeting the exceptional divisor $E$ once. Let $\pi : W \to Z$ be the blow-up of $Z$ along $B$. Let $\sigma$ be the section of $f\vert_E : E \to Y$ that meets the exceptional divisor of $\pi$.

The curve $\sigma$ can itself be flopped by a map $\phi : W \dashrightarrow X$. The reason is this: $W$ is constructed as a blow-up of $Y^+$: first blow up the curve $C^+$, and then blow-up $B$. We could instead blow up $B$ and then $C^+$ to get $X$. The two things are isomorphic in codim $1$, and differ by the flop of $\sigma$. Let $\sigma^+ \subset X$ be the flopped curve.

Now consider the map $X \dashrightarrow W \to Z \to Y$. The indeterminacy locus is the same as that of $\phi^{-1}$, which is a single curve $\sigma^+ \subset X$. However, $U = X \setminus \sigma^+$ dominates $Y$, because $f(E \cap U) = f(E)$ still dominates the subset that the deleted stuff maps to.

If you want to make this very concrete, you can construct $Y$ by blowing up one line in $\mathbb P^3$, and then the strict transform of another line intersecting that one. $C$ is then the fiber of the first exceptional divisor over the intersection point, which we blow up to get $W$. Then $Z$ is obtained by blowing up a fourth curve that meets the exceptional divisor. Everything in sight is obtained by blowing up these four intersecting curves in various orders.

  • $\begingroup$ Thanks a lot! I honestly do not know the term "flop", but I will look it up and for the moment, I am convinced. I expected a counterexample anyway, but could not come up with one. $\endgroup$ Mar 10, 2015 at 13:58
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    $\begingroup$ The moral is this: the map $X \dashrightarrow W$ is a map of 3-folds that's indeterminate along a single curve in both directions. You can look at a resolution $V$ of this, and the image of the exceptional divisor of $V \to W$ is a curve $\gamma$. Now, when you delete the (1-dimensional) indeterminacy from $X$ to get $U$, the image of $U$ in $Y$ loses the image of $\gamma \subset W$. But the map $W \to Y$ contracts entire divisors above $Y$, and so even after you remove the single curve $\gamma$, it's still surjective. $\endgroup$
    – user47305
    Mar 10, 2015 at 14:53
  • $\begingroup$ Thanks for that comment and +1, that actually helped me to picture it tremendously. $\endgroup$ Mar 10, 2015 at 16:36

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