If $X$ is a compact Riemann surface, $Y$ is a universal cover of $X$, and $f\colon X\rightarrow X$ is a biholomorphic map, then, can $f$ be lifted to a $biholomorphic$ map on $Y$? (I mean, topologically, it can be lifted to a homeomorphism from $Y$ to $Y$, just by using "lifting criteria". But can the lift be "biholomorphic"?)
lifting of automorphisms of Riemann surface
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riemann-surfaces
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0Yes, since biholomorphy is an entirely local condition, it will be preserved under lifts to a cover. – 2011-02-12
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0@Jim Biholomorphy is a global condition: the map $f : \mathbb C^* \to \mathbb C^*$ which sends $z$ to $z^2$ is locally a biholomorphism, but it is not injective. – 2011-02-12
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2but the truly global part is the bijective part, which is already known topologically. – 2011-02-12