Is this "the winding number"?

Question

Note that $p:\mathbb{C}\rightarrow \mathbb{C}\setminus\{0\}:z\mapsto e^z$ is a covering map.

Let $\alpha:[0,1]\rightarrow \mathbb{C}\setminus\{0\}$ be a closed curve.

Let $\gamma$ be any lift of $\alpha$ such that $\alpha=p\circ \gamma$.

Call $\frac{\gamma(1)-\gamma(0)}{2\pi i}$ the winding number of $\alpha$.

Is it okay to define "winding number" in this way? That is, does this definition exactly mean the winding number?

@Seirios I don't have one. This is the first time I'm studying somewhat related to that. — Rubertos, Dec 01 '14 at 10:05
@Rubertos Sorry, I deleted my initial comment, realised that I hadn't read your question properly (my fault for trying to be clever before my morning coffee). — Benjamin Alderson, Dec 01 '14 at 11:58
@Seirios and Rubertos: I understand the question to mean that ${\gamma (a) - \gamma (0) \over 2 \pi i}$ is the definition of the winding number of $\alpha$. Is it not? — , Dec 02 '14 at 03:47
@student Exactly, but now I prefer the definition here: http://math.stackexchange.com/questions/186512/definition-of-winding-number-have-doubt-in-definition — Rubertos, Dec 02 '14 at 03:50

score 2 · Accepted Answer · answered Dec 01 '14 at 09:21

This is correct, yes. If you have a closed curve $\alpha$, then from beginning to end it only differs by a phase $e^{2\pi i n}$. Using the exponential map as you have, you see that this gives you exactly the number of times that your curve has wound around the origin.

Is this "the winding number"?

1 Answers1