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Riemann surfaces

Let F: X -> Y be a surjective map between two compact connected Riemann surfaces which has the property that for every x in X, there is some open U containing x such that F restricted to U is a homeomorphism onto its image.

Prove that F is also a covering in the sense of topology (for every V open in Y, the pre-image of V by F is homeomorphic to V x W for W discrete).

have no idea where to even begin. i have some fact in my notes that given y in Y, let x be a preimage in X, then F a topological covering if every path starting at y can be lifted to a path starting at X.
Reply 1
Avatar for Zhen Lin
Zhen Lin
12
Original post by around
Let F: X -> Y be a surjective map between two compact connected Riemann surfaces which has the property that for every x in X, there is some open U containing x such that F restricted to U is a homeomorphism onto its image.

Prove that F is also a covering in the sense of topology (for every V open in Y, the pre-image of V by F is homeomorphic to V x W for W discrete).

have no idea where to even begin. i have some fact in my notes that given y in Y, let x be a preimage in X, then F a topological covering if every path starting at y can be lifted to a path starting at X.


This is a really easy question and has nothing to do with Riemann surfaces and everything to do with compactness. (Also, don't use the path lifting property. Use the definition of covering space given.)
Reply 2
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around
OP
13
Original post by Zhen Lin
This is a really easy question and has nothing to do with Riemann surfaces and everything to do with compactness. (Also, don't use the path lifting property. Use the definition of covering space given.)


Yeah, I did it in the end using the stack-of-records theorem from Differential Geometry.
Reply 3
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Hathlan
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Original post by around

Prove that F is also a covering in the sense of topology (for every V open in Y, the pre-image of V by F is homeomorphic to V x W for W discrete).


Hey this definition is a bit wrong.
Reply 4
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around
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I've already been told this, but it's the 'definition' given by our lecturer. I think if you stare hard enough the definitions are 'equivalent' on the context we're working in, but yeah, that's not quite the right one.
Reply 5
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Hathlan
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(What exactly is the context?)

C\mathbb{C} and the torus are Riemann surfaces, and the standard covering of the torus is a topological covering, but not one of your coverings.
Reply 6
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around
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Original post by Hathlan
(What exactly is the context?)

C\mathbb{C} and the torus are Riemann surfaces, and the standard covering of the torus is a topological covering, but not one of your coverings.


Yeah, the lecturer's got it wrong (the preimage of any open set U on the torus is certainly of the form W x D for W open in C, but W isn't necessarily biholomorphic to U)
Reply 7
Avatar for Zhen Lin
Zhen Lin
12
To clarify:

Definition. A covering map is a continuous map f:E→Bf : E \to B such that for each point p in B there is an open neighbourhood U such that f−1Uf^{-1} U is homeomorphic to U×DU \times D for some discrete set D.

So to fix the given definition, all we have to do is replace "for every open set" with "for every small enough open set".

Proposition. Let f:E→Bf : E \to B be a local homeomorphism with E quasicompact and B satisfying the T1T_1 separation axiom. Then f is a covering map.

Proof. It is easy to check that the inverse image of every point in B is a closed discrete subset of E, and a compact discrete set is automatically finite. So f has finite fibres. But f is a local homeomorphism, so there is a neighbourhood of each point such that f restricts to a homeomorphism on that neighbourhood, and by finiteness, we can just take the intersection of all of these to obtain the desired evenly-covered neighbourhood of B.

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