# lifting of automorphisms of Riemann surface

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"?)

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Yes, since biholomorphy is an entirely local condition, it will be preserved under lifts to a cover. – Jim Belk Feb 12 '11 at 8:32
@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. – Gunnar Þór Magnússon Feb 12 '11 at 12:40
but the truly global part is the bijective part, which is already known topologically. – Soarer Feb 12 '11 at 17:29

It's a standard result from topology that the homeomorphism $f: X\to X$ can be lifted, and that the lifted map $g: Y \to Y$ is a homeomorphism. The only thing left to check is that $g$ (and its inverse) are actually holomorphic.
Topological coverings have a nice property, that is, the covering topological manifold inherits the structure of the base manifold. In your specific case, if $\pi: Y \to X$ is a (universal) cover, and $X$ is a Riemann Surface, then there is a unique analytical structure on $Y$ such that $\pi$ is a local biholomorphism. (Thus, if $Y$ is connected, $Y$ is a Riemann Surface too)
By using this fact, and that $g$ is the lift of $f$ (i.e. $\pi \circ g = f \circ \pi$), it is trivial to check that $g$ is actually a biholomorphism. The crucial point is that holomorphy is a local condition, and locally $\pi$ is a biholomorphism. Then, $f$ is a biholomorphism, both $g$ and its inverse are.