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>)



Lifting a morphism
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. 


