# Functoriality of Hironaka's resolution of singularities

Is Hironaka's resolution of singularities functorial? I know that the resolution is not unique, there are flips etc. But if we have a rational map f:X---> Y, can we chose resolutions X'->X and Y'->Y and a map $f_*:X'\to Y'$ that makes the relevant diagram commute?

• First resolve Y, then resolve the closure of the graph of the resulting rational map: this should do the trick! – damiano Apr 1 '10 at 13:25
• I think the tag projective-resolution is intended for homological algebra use, so it is probably non appropriate here. – Andrea Ferretti Apr 1 '10 at 13:26
• @Andrea: indeed. I've taken the liberty to detag. – José Figueroa-O'Farrill Apr 1 '10 at 14:08
• Darn!! I tried to create a resolution of singularities tag, only to find that it was truncated as resolution-of-singulariti. We really need more letters for tags. – Regenbogen Apr 1 '10 at 15:00
• Regenborgen: You can create the truncated tag for now, and then later when (if?) the character limit for tags is increased, the administrators can change the tag name. – Kevin H. Lin Apr 1 '10 at 17:33

A useful (at least for me) example is given in Kollar's article/book on resolutions of singularities about how you can't expect to get a "resolution functor": take a quadric cone $$C = \{(x,y,z) \in \mathbb A^3: xy-z^2=0\}$$ in $\mathbb A^3$. Then you have the obvious map $\phi\colon \mathbb A^2 \to C$. But now suppose that $C'$ is a resolution of $C$ provided by a putative "resolution functor". Then if we let $\tilde{C}$ be the minimal resolution, $C'$ factors through $C$. If we assume that $\mathbb A^2$ is resolved by itself (as seems reasonable!) then we'd have to have $\phi$ lifting to a map $\mathbb A^2 \to \tilde{C}$ compatibly with the original morphism, which of course one cannot do.