# morphism of smooth varieties - closure of graph smooth?

Let $X$ be a smooth projective variety (in particular irreducible) and $U\subset X$ (Zariski-)open, so again smooth as a variety. Let $f\colon U\rightarrow U$ be a smooth map of varieties and denote with $\Gamma_f\subset U\times U$ the graph subvariety (which is again a smooth variety). Consider $\Gamma_f$ as a locally closed subvariety of $X\times X$.

Is it true that the closure $\overline{\Gamma_f}$ of $\Gamma_f$ in $X\times X$ is a smooth variety? If not in gerneral, do any assumptions on $f$ or $X$ help?

If one takes for example a map $\mathbb{A}^1\rightarrow\mathbb{A^1}$, this question (is the projective closure of a smooth variety still smooth) suggests that nothing like this was to expect for closures in $\mathbb{P}^2$ - still I would hope that things turn out a bit differently for the closure in $\mathbb{P}^1\times\mathbb{P}^1$.

Thanks in advance for any hints!

• Probably not. This is why when we look at the graph of a birational map we often take a resolution of it. A first candidate to look at would be the Cremona transformation. For $X$ a smooth projective curve, however, the answer will be yes: every rational map extends to an actual morphism and so the closure of the graph of $f|_U$ will be the graph of $f\colon X\to X$. – Ben Mar 22 '17 at 15:34
• The Cremona will not be a counterexample. The graph is just $\mathbb{P}^2$ blown up in $3$ points. – Ben Mar 22 '17 at 15:58
• Won't the graph be isomorphic to $X$ blown up along the indeterminacy locus $Y$ of $f$? If this locus is not smooth, then $Bl_YX$ doesn't need to be smooth either. – Jake Levinson Mar 23 '17 at 3:31
• Thank you very much for your help! I still need a while to think things through, but I assume I'll get along pretty well with this! – user178979 Mar 23 '17 at 12:13