If a deck-transformation has a fixed-point, then it is the identity (trivial) transformation and all points are fixed points . So you consider only transformations without fixed points. Sorry for misunderstandings.

That's what was implied by the no-fixed point rule, I thought (if a deck has a fixed point, and we have path-connectedness, then it's the identity transformation) .

I see your point now (unique path lifting property) . I was confused.

Given that the total curvature of the sphere is 2, and the total curvature of a manifold (that might have the sphere as universal map) is an integer, this argument (if made rigorous) simplifies things to considering only the involutory transformations (ie f(f(x)) = x) .

Thanks :) .I sketched something similar to 2.2 upon understanding the uniformization theorem (however I wasn't confident enough it's correct).I modeled H^2 as the Poncaire disc, then showed that the Poincare metric commutes with all possible transformations on the unit disk that preserve the conformal class (the complex automorphisms of the unit disk, and their reflections).I tried to figure out a similar trick with the metric induced by stereographicly by the Riemann sphere, but that turns out to be more difficult (the extended complex plane has the most possible mobius transformations).

How does one prove that given D. a double cover of a manifold S, and a constant curvature metric for D, you can 'push' the metric trough the covering map to obtain a constant-curvature metric for S?

Uniformization applies to Riemann surfaces .I'm asking for 2d manifolds in general. As far as I know, you can assign a complex structure to a manifold if it's orientable. It would need an extension to non-orientable manifolds.