Blowup along the fundamental locus of a rational map Assume $f:X\dashrightarrow Y$ is a rational map between varieties, where $X$ is normal and $Y$ is complete. Then, the fundamental locus the $f$ (which means cannot extend the definition of $f$ on it), say $B$, is of codimension at least two. Intuitively, the map $f$ gives rise to a morphism $\hat f$ defined on $Bl_B X$ in a natural way. Is this intuition correct? And if yes, how can I prove it?
Thanks!
 A: Let me assume that $Y$ is projective. Let $\tilde{X} \subset X \times Y$ be the graph of $f$ (i.e., the closure of the graph of the restriction of $f$ to $X \setminus B$). Then $\tilde{X} \to X$ is a projective birational morphism, hence it is the blowup of an ideal $I$. This ideal is supported on the closed subset $B$, and defines a subscheme structure on it. Thus, the blowup of this subscheme is $\tilde{X}$, and after this blowup the map becomes regular.
A: This post is to document the geometry of an example suggested by Sasha in the comment (with some more details to be filled in later).
The goal is to study the relationship between the two spaces that naturally resolve a rational map $f$. One space is the closure of the graph $\overline{\Gamma(f)}$ (see Sasha's answer) and the other space is obtained via successive blowups along smooth centers (existence guaranteed by Hironaka's theorem) and there should be a minimal one, which is denoted as $M$.
Let's consider the rational map defined by Sasha:$$f:\mathbb P^2_{[x,y,z]}\dashrightarrow \mathbb P^3_{[a,b,c,d]},\  [x,y,z]\mapsto [x^2,xy,y^2,yz],$$
which is everywhere defined except at $p=[0,0,1]$. $f$ dominates the quadric cone $Q=\{ac=b^2\}\subseteq \mathbb P^3_{[a,b,c,d]}$.
Blowup the domain at $p$, the resulting space is the first Hirzebruch surface $\Sigma_1$ with exceptional divisor $E_1$, and $f$ extends to a rational map $f':\Sigma_1\dashrightarrow Q$, which is everywhere regular except for a point $p_1$ on $E_1$. Moreover, it contains a rational curve $L$, as proper transform of the line $\{y=0\}$ in $\mathbb P^2$, so $L\cdot L=0$, $E_1\cdot E_1=-1$ and $L\cdot E_1=1$.
Blowup $\Sigma_1$ along $p_1$, and denote the resulting space as $M$. One can show $M$ is a minimal resolution of $\mathbb P^2$ via successive blowups along smooth centers where $f$ extends to a regular map
$$\tilde{f}:M\to Q.$$
On $M$, the proper transform $\tilde{L}$ and $\tilde{E}_1$ becomes disjoint. Moreover, $\tilde{L}\cdot\tilde{L}=-1$ and $\tilde{E}_1\cdot\tilde{E}_1=-2$. What $f$ does is to contract the two curves $\tilde{L}$ and $\tilde{E}_1$.
We can factor $f$ in two ways depending on which curve to blow down first, namely there is a commutative diagram
$\require{AMScd}$
\begin{CD}
M @>{\tau'}>> \overline{\Gamma(f)}\\
@V{\sigma'}VV @V{\sigma}VV\\
\Sigma_2@>{\tau}>> Q 
\end{CD}
where $\overline{\Gamma(f)}$ is the closure of the graph of the rational map $f$ and $\Sigma_2$ is the second Hirzebruch surface. The vertical maps are blow down of the $(-1)$-curve $\tilde{L}$, while the horizontal maps are blow down of the $(-2)$-curve $\tilde{E}_1$.
The top row is the relationship between the two spaces that we are mainly interested in, namely, $M$ is the minimal resolution of the space $\overline{\Gamma(f)}$ by blowing up the $A_1$-singularity.
