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This question is succeeding a question raised in this post. Let $$f:X\dashrightarrow Y$$ be a rational map between smooth projective varieties over $\mathbb C$ with smooth fundamental locus $B$ and $\text{codim}_BX=2$. It is mentioned by Sasha that by considering the closure of graph $\overline{\Gamma(f)}$ inside the product $X\times Y$, the first projection $\pi_1:\overline{\Gamma(f)}\to X$ is a projective birational morphism, therefore the blowup along some ideal sheaf $\mathcal{I}$ supported on $B$ (Hartshorne Cha. II, 7.17) and the second projection $\pi_2:\overline{\Gamma(f)}\to Y$ becomes regular.

This is a way of extending a rational map to a morphism. Another way is to blowup smooth centers successively according to Hironaka's theorem. What I want to ask is something "converse", namely, I believe the following statement is true:

Claim: Let $\sigma:\text{Bl}_BX\to X$ be the blowup along the (reduced) subvariety $B$, and assume that the induced rational map $\tilde{f}$ on the blowup $$\tilde{f}:\text{Bl}_BX\to Y$$ is regular. Then $\text{Bl}_BX\cong \overline{\Gamma(f)}$.

In other words, what I want to prove is that if the rational map extends to the initial blowup $\text{Bl}_BX$, then the graph closure is isomorphic to blowup of the ideal sheaf $\mathcal{I}_B$ of $B$. What I have so far is that by the universality property of the blowup (Hartshorne Cha. II, 7.14), there is a unique morphism $g$ such that $\pi_1$ factors through $$\overline{\Gamma(f)}\xrightarrow{g} \text{Bl}_BX\xrightarrow{\sigma} X$$

However, to prove my claim, I also need to dominate $\overline{\Gamma(f)}$ by $\text{Bl}_BX$, somehow using the assumption that $\tilde{f}$ is a morphism, but I don't know how to go further.

Any suggestions and comments are appreciated!

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Note that $\mathrm{Bl}_B(X)$ comes with the blowup morphism to $X$ and with the morphism $\bar{f}$ to $Y$. Together, they define a morphism $$ \mathrm{Bl}_B(X) \to X \times Y. $$ Its image is an irreducible and reduced subscheme of $X \times Y$ which contains the graph of $f\vert_{X \setminus B}$ as a dense open subset, hence coincides with $\overline{\Gamma(f)}$. This gives the require morphism $\mathrm{Bl}_B(X) \to \overline{\Gamma(f)}$.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – AG learner
    Commented Sep 29, 2020 at 4:51
  • $\begingroup$ Dear Sasha, may I bother you with this problem again? I realize that the original sketch of the proof of my "claim" stated in the question is not correct: I want to show there are maps between $\overline{\Gamma(f)}$ and $Bl_B(X)$ on both sides, preserving the projection to $X$. The direction $Bl_B(X)\to \overline{\Gamma(f)}$ as you pointed out is okay (thanks). However, the other direction $\overline{\Gamma(f)}\to Bl_B(X)$ does not really come from the Universal Property of the blowup as I stated because the pullback of the ideal sheaf of $B$ via the projection $\overline{\Gamma(f)}\to X$... $\endgroup$
    – AG learner
    Commented Nov 26, 2020 at 16:30
  • $\begingroup$ is not necessarily an invertible sheaf, as by a priori, the graph closure $\overline{\Gamma(f)}$ might be arbitrary singular, but pullback being an invertible sheaf is a necessary condition according to Hartshorne II. 7.14, which I overlooked. Also according to a discussion here, in general there is no morphism between two blowups. But I still think the domination $Bl_B(X)\to \overline{\Gamma(f)}$ tells us a lot of information. Do you think my claim is still correct? $\endgroup$
    – AG learner
    Commented Nov 26, 2020 at 16:40
  • $\begingroup$ Nevermind, the answer is no. I found a simple counterexample: the rational map $f:\mathbb C^2\to \mathbb P^1, (x,y)\mapsto (x^2,y^2)$ extends to a regular map $\tilde{f}$ on $Bl_{(0,0)}\mathbb C^2$, but the graph closure $\overline{\Gamma(f)}$ is isomorphic to the blowup along the ideal $(x^2,y^2)$, which is singular along the entire exceptional divisor, and the natural map $Bl_{(0,0)}\mathbb C^2\to \overline{\Gamma(f)}$ is exactly the normalization. Thank you for your help! $\endgroup$
    – AG learner
    Commented Nov 26, 2020 at 22:20
  • $\begingroup$ @AGlearner: Yes, this seems to be right. $\endgroup$
    – Sasha
    Commented Nov 27, 2020 at 6:59

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