1
$\begingroup$

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

  • 0
    What do you mean by $\Bbb R_+$? Do you mean $[0,+\infty)$ or $(0,+\infty)$?2012-12-11
  • 0
    @OlivierBégassat $(0,+\infty)$.2012-12-11
  • 1
    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.2012-12-11
  • 1
    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.2012-12-11
  • 0
    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.2012-12-11
  • 0
    Or use that it is a local homeomorphism from a simply connected space, hence there has to be a homeomorphism of cover ing spaces with the real line, and prove that such homeomorphism cannot exist.2012-12-11

0 Answers 0