Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It many sources it's stated that the winding number is invariant under homotopy, but I've yet to actually see why.

Suppose you have the formal definition of the winding number. So for a continuous loop $\gamma\colon[\alpha,\beta]\to\mathbb{C}\setminus\{a\}$ which doesn't pass through a point $a$, one has the function $\theta(t)=\text{arg}(\gamma(t)-a)\in\mathbb{R}/2\pi\mathbb{Z}$. By the lifting lemma, there exists a continuous $\tilde{\theta}\colon[\alpha,\beta]\to\mathbb{R}$, such that $[\tilde{\theta}(t)]=\theta(t)$, and the winding number of $\gamma$ around $a$ is then defined as $$n(\gamma,a)=\frac{\tilde{\theta}(\beta)-\tilde{\theta}(\alpha)}{2\pi}.$$

Is there a straightforward proof that the winding number is invariant under homotopy with this definition for continuous loops which do not pass through $a$? Thanks.

share|cite|improve this question
up vote 3 down vote accepted

If you have a homotopy $H : [\alpha, \beta] \times [0; 1] \to \mathbb{C}\setminus \{a \}$, the function $\theta : [\alpha, \beta] \times [0; 1] \to \mathbb{R}/2 \pi \mathbb{Z}$ defined by $\theta(x,t) = \arg(H(x,t)-a)$ is continuous in both variables, and can be lifted into a function $\tilde{\theta}$.

Then you can define the continuous function $n(t) = \frac {\tilde \theta (\beta,t) - \tilde \theta (\alpha,t)}{2 \pi} \in \mathbb{Z}$. Since $\mathbb{Z}$ is discrete, it has to be a constant map, so the winding numbers of $\gamma = H(.,0)$ and $H(.,1)$ are the same.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.