# 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?

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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
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In addition to what Damiano says, characteristic zero resolution is canonical in a stronger sense: a readable account is given in a paper by Hauser. In particular, resolution commutes with smooth morphisms and commutes with group actions. Recently, this has also been shown to be true for all excellent reduced schemes of dimension at most two by Cossart, Jannsen, and Saito

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Thank you mdeland! The Hauser paper containes exactly what I want! –  Colin Tan Apr 2 '10 at 3:55
The paper by Hauser really looks great: thanks for the link! –  damiano Apr 2 '10 at 8:15
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.