Why doesn't there exist a Riemann surface (except $\mathbb{P}^1$) with $\mathbb{P}^1$ as its universal cover?

Question

I can see this from the uniformization theorem but I came across a book that claims that this can be seen from the fact that Möbius transformations always have fixed points.

Answer 1

The deck transformations would have to be fixed point free bijective conformal self maps of ${\mathbb P}^1$. Now all these self maps are Moebius transformations. This is not obvious; maybe you need Liouville's theorem for a proof.But a Moebius transformation $$T(z)={az +b\over cz +d}$$ (plus exception handling rules) necessarily has one or two fixed points. These are obtained by solving the equation $T(z)=z$ for $z$.