# How do I prove that a map is not a covering map?

I'm thinking how to prove that a map is not a covering map. For example let $p:\mathbb R_+\to S^1$ be a map defined by $p(\theta)=(\cos(2\pi\theta),\sin(2\pi\theta))$. I'm trying to find a point which doesn't have a neighborhood evenly covered by $p$. I'm thinking about the point $(-1,0)$, am I in correct way?

I need a hand here

Thanks

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What do you mean by $\Bbb R_+$? Do you mean $[0,+\infty)$ or $(0,+\infty)$? –  Olivier Bégassat Dec 11 '12 at 4:34
@OlivierBégassat $(0,+\infty)$. –  user42912 Dec 11 '12 at 4:41
Do you mean $(1,0)$? One way to think about it is that $\mathbb{R} \to S^1$ with the same map is a covering that wraps the real axis around the circle - what you do in the question wraps "half" of the real axis around, so you can imagine that the starting point, which is the origin, may be problematic. –  Sanchez Dec 11 '12 at 5:22
Look at the point $(0,1) \in S^1$. Take some open neighbourhood about it (I like to think of $S^1$ as a subset of $\mathbb{C}$ when thinking in this way), and ask yourself if it is evenly covered. –  andybenji Dec 11 '12 at 5:25
You may identify the fiber as the natural numbers, and then see how the fundamental group of the circle could possibly act on this set, and derive a contradiction. –  user17786 Dec 11 '12 at 8:15
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