How to prove a function is a covering map?

Question

I am learning differential geometry. The problem is I can't find some solved exercises so it is a bit hard to understand how can I solve them.

I found this exercise on the web: Let $B = \{(u,v)\ |\ u^2+v^2 \leq 1\}$ and $S^2 = \{(x,y,z)\ |\ x^2+y^2+z^2=1\}$. Let $f:S^2\rightarrow B,\ f(x,y,z) = (x,y)$. Check if $f$ is a covering map.

Starting with the definition in my mind (http://mathworld.wolfram.com/CoveringMap.html), I really don't know how to start or what should I do to check that.

Thank you very much!

Answer 1

I recommend you to study any book on this topic (e.g. http://www.maths.ed.ac.uk/~v1ranick/papers/maybook.pdf). But if you take the wolfram-article and believe what it says (proofs are not hard, but not given there), then you will see that the cardinal number $f^{-1}(y)$ must be independent of $y$ for a covering $f$. Now consider your example and compute $f^{-1}(y)$ for $y = (0,1)$ and $y = (0,0)$.

Answer 2

$f$ is not a covering map

$f$ is continuous and surjective. The issue for $f$ to be a covering map arises on the points that are at the border of the disk $B$.

Take for example $(1,0) \in B$. Any open subset of $B$ containing $(1,0)$ will contain an open subset $U = \{p \in B ; \Vert p-(1,0) \Vert < r\}$ with $r < 1/2$.

If you denote $C$ the circle $x^2+y^2=1$, then for $p \in U \cap C$, $f^{-1}(p)$ has only one point. However $f^{-1}(p)$ has two points for $p \in U \setminus C$. Hence $f$ cannot be a covering map.