Let $f\colon X\to Y$ be a birational map of complex algebraic varieties. Are there necessarily open immersions of $X$ and $Y$ into varieties $X’$ and $Y’$, resp., which admit a proper morphism $f’\colon X’\to Y’$ extending $f$?

4$\begingroup$ This follows from Nagata's compactification theorem and its relative version (also due to Nagata). Indeed, compactify $Y$ to $Y \hookrightarrow Y'$, wlog integral, and then compactify the morphism $X \to Y'$ to $X' \to Y'$, again wlog $X'$ integral. $\endgroup$– R. van Dobben de BruynJul 4, 2019 at 20:42

$\begingroup$ @R.vanDobbendeBruyn can you make this an answer? $\endgroup$– Avi SteinerJul 4, 2019 at 21:41

$\begingroup$ @R.vanDobbendeBruyn Do you know if in addition you can require that $f’^{1}(Y)=X$? I realized that this is what I meant to ask $\endgroup$– Avi SteinerJul 4, 2019 at 21:50

1$\begingroup$ I'm not sure this is possible. Take for instance $X':=\mathrm{Bl}_p \mathbb P^2$, $Y'=Y=\mathbb P^2$ and $f':X'\to Y'$ the natural blow up map. Then defined $X:=X'\setminus \{x\}$ for some point $x$ on the exceptional divisor. Then any extension of $f$ will send the closure of the exceptional divisor to the point $p$. $\endgroup$– HenriJul 4, 2019 at 22:15

$\begingroup$ You can assert $f'^{1}(Y) = X$ if and only if the map $f$ you start with is proper. $\endgroup$– R. van Dobben de BruynJul 5, 2019 at 12:23
1 Answer
This follows from Nagata's compactification theorem [Nag62] and its relative version [Nag63]. Indeed, one may compactify $Y$ to get an open immersion $Y \hookrightarrow Y'$ with $Y'$ proper [Nag62]. Replacing $Y'$ by the reduced structure on the closure of $Y$, we may assume $Y'$ is integral.
Now apply relative compactification to $X \to Y'$ to get an open immersion $X \hookrightarrow X'$ of $Y'$schemes [Nag63]. Again, we may assume $X'$ is integral. This gives a commutative diagram $$\begin{array}{ccc}X & \stackrel{f}\longrightarrow & Y \\ \downarrow & & \downarrow \\ X' & \stackrel{f'}\longrightarrow & Y',\!\!\!\end{array}$$ and we have $f'^{1}(Y) = X$ if and only if $f$ is proper. Indeed, if $f'^{1}(Y) = X$ then $f = f' \times_{Y'} Y$ is proper. Conversely, if $f$ is proper, then $X \to f'^{1}(Y)$ is an open immersion of proper $Y$schemes, hence a closed immersion. It is also dense since $X \subseteq X'$ is dense, so we must have $X = f'^{1}(Y)$. $\square$
This appears as Lemma 5.1 in a preprint of mine [vDdB], but surely must be written up somewhere else as well. (Does anybody know a canonical reference?)
References.
[vDdB] R. van Dobben de Bruyn, The equivalence of several conjectures on independence of $\ell$. arXiv: 1808.00119
[Nag62] M. Nagata, Imbedding of an abstract variety in a complete variety. J. Math. Kyoto Univ. 2, p. 110 (1962). ZBL0109.39503.
[Nag63] M. Nagata, A generalization of the imbedding problem of an abstract variety in a complete variety. J. Math. Kyoto Univ. 3, p. 89102 (1963). ZBL0223.14011.