# Lifting a morphism

Suppose we have a morphism $\phi : S_{1} \rightarrow S_{2}$, between quasi-projective varieties of dimension $2$ over $\mathbb{C}$ with at worst quotient singularities. Suppose furthermore that $\phi$ is an isomorphism on a Zariski open subset.

Let $\tilde{S_{i}}$ denote minimal resolutions of the $S_{i}$ (exceptional divisors contain no $-1$ curves). Is it possible to lift $\phi$ uniquely to a morphism $\tilde{\phi}: \tilde{S_{1}} \rightarrow \tilde{S_{2}}$?

I asked this question already on math.stackexchange (https://math.stackexchange.com/questions/2735590/lifting-a-generically-1-1-map-to-resolutions)

## 1 Answer

Let $\phi\colon S_1\to S_2$ be a birational morphism between singular surfaces and let $f_i\colon \widetilde{S}_i \to S_i$ be the minimal resolution of singularities, for $i=1,2$.

Any birational map $\psi\colon \widetilde{S}_1 \dashrightarrow \widetilde{S}_2$ lifting $\phi$ should satisfy $\phi = f_2\circ\psi\circ f_1^{-1}$, so $\psi=f_2^{-1}\circ\phi\circ f_1$ is uniquely determined by the uniqueness of $f_1,f_2$. Clearly $\psi$ exists as a birational map.

Now suppose $\psi$ is not a morphism. Then there is a $(-1)$-curve $E\subset\widetilde{S}_2$ which is contracted to a point $P\in\widetilde{S}_1$ by $\psi^{-1}$ (since $\widetilde{S}_1$, $\widetilde{S}_2$ are smooth). But then $f_2(E)\subset S_2$ is a curve (since $f_2$ does not contract any $(-1)$-curve in $S_2$) which is contracted to a point $f_1(P)\in S_1$ by $\phi^{-1}$, so $\phi$ could not have been a morphism. Hence $\psi$ is a morphism.