0
$\begingroup$

I was asked to determine the degree of the antipodal circle function $f(\theta)=\theta + \pi$. I believe the degree of the function is $1$, through pictures and illustration, but I will like to prove this using homotopy.

So given our degree $1$ map $C_{-1}(\theta):S^1 \rightarrow S^1$ is given by $$f(\cos(\theta)+i\sin(\theta))=\cos(\theta)+i\sin(\theta)$$.

Could I use as a homotopy $H(x,t)=(\cos(\theta)+i\sin(\theta))e^{t\pi i} $. If so how will prove it is continuous.

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes, you can. However, you can simplify your formula by writing $$H(z,t) = z\cdot e^{\pi i t} .$$ Note that $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. The map $H$ is continuous because complex multiplication is continuous and the exponential function is continuous.

If you want you can also write $S^1 = \{ (x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1\}$ and $$H(x,y,t) = (x\cos(\pi t) - y \sin(\pi t), x \sin(\pi t) + y \cos(\pi t)) .$$

$\endgroup$
5
  • $\begingroup$ Thank you but this does not prove our function is continuous. The space $ S^1\times[0,1]$ im not sure if it a metric space. Hence, what are you usig to prove that our function is continuous. I beloeve we need an intermediate function. $\endgroup$
    – ben huni
    May 23, 2020 at 18:11
  • 1
    $\begingroup$ It does prove it, but it takes for granted a few well-known facts. I am not going to give proofs. 1) The function $[0,1] \to \mathbb C, t \mapsto \pi i t$, is continuous. This is trivial, multiplication with the constant $c = \pi i$ is continuous. 2) The function $\mathbb C \to \mathbb C, w \to e^{iw}$, is continuous. Complex multiplication $\mathbb C \times \mathbb C \to \mathbb C, (z_1,z_2) \mapsto z_1 \cdot z_2$, is continuous. $\endgroup$
    – Paul Frost
    May 23, 2020 at 23:22
  • $\begingroup$ Wouln't we also need that that if the function $\mathbb{C} \rightarrow \mathbb{C}$ and $[0,1] \rightarrrow \mathbb{C}$ is continuous. Then $\mathbb{C} \times \rightarrow \mathbb{C} \times \mathbb{C}$ is continuous. I know a lot steps, I just need my proofs alwas to be true. $\endgroup$
    – ben huni
    May 24, 2020 at 7:10
  • 1
    $\begingroup$ @benhuni Yes, you are right. We need the fact that "product maps" (here $\mathbb C \times [0,1] \to \mathbb C \times \mathbb C$) are continuous. This is true for all pairs of maps between metric (or more generally topological) spaces. By the way, you can also an $\epsilon$-$\delta$-argument to show directly the $H(x,y,t)$ is continuous. $\endgroup$
    – Paul Frost
    May 24, 2020 at 8:24
  • $\begingroup$ ahh! I see now thanks! $\endgroup$
    – ben huni
    May 24, 2020 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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