2
$\begingroup$

Let $\{0\} \subset B \subset \mathbb{C}$ be a connected, simply connected open subset of $\mathbb{C}$ containing $0$.

I am trying to prove that $\sqrt{B}:=\{z \in \mathbb{C} \ | \ z^2 \in B\}$ is connected and simply connected without using the Riemann mapping theorem, but my attemps always stumble over the fact that I would like to use a property of $B$ which is more or less equivalent to an instance of the (Riemann mapping) theorem.

Can anybody find a more elementary solution?

$\endgroup$
  • $\begingroup$ Trying dividing $B$ into simply-connected regions on which a global square-root exists, then using the van Kampen theorem to show that $\sqrt{B}$ is still simply-connected. (Dealing with $0$ requires a bit of care. I'm not sure this would actually be any more elementary than using the Riemann mapping theorem.) $\endgroup$ – anomaly Sep 20 '16 at 20:53
  • $\begingroup$ Let alone the fact that this is not exactly elementary, I actually don't know how to divide B in such a way. $\endgroup$ – nombre Sep 20 '16 at 21:53
1
$\begingroup$

Assume you know that a loop $I\to \Bbb C$ which is surjective on $B_{\epsilon}(0)$ can be homotoped relative to $\Bbb C \setminus B_{\epsilon}(0)$ to a path which is not surjective on $B_{\epsilon}(0)$. Then this path is also homotopic to a path which image does not intersect with $B_{\varepsilon/2}(0)$ relative to $\Bbb C \setminus B_{\epsilon}(0)$.

So you can homotop any path $\gamma \colon I \to \sqrt{U}$ to a path $\tilde\gamma \colon I \to \sqrt{U}\setminus B_{\sqrt{\varepsilon/2}}(0)$. But $\tilde\gamma^2$ is homotopic to a path on the circle with radius $\varepsilon/2$. (Just look at a null homotopy of this path in polar coordinates and take the radius $r$ to be $\max(r,\varepsilon/2)$) Then since you know that $(\cdot)^2 \colon \Bbb C^{\times} \to \Bbb C^{\times}$ is a covering, by unique path lifting we can lift this homotopy to $\sqrt{U}$ and then we can easily show that this path is nullhomotopic.

How to prove the first assumption? Look at $\gamma^{-1}(B_{\sqrt{\varepsilon}}(0))$ and $\gamma^{-1}(\Bbb C \setminus B_{\epsilon/2}(0))$ which are both unions of open intervals in $I$. Since $I$ is compact you know that there is a finite number of "subpaths" of $\gamma$ in $B_{\epsilon}(0))$ which cover $B_{\epsilon/2}(0))$. Since there is only a finite number of them, you can homotope them to straight lines and a finite number of straight lines does not cover $B_{\epsilon/2}(0))$.

$\endgroup$
  • $\begingroup$ This is an interesting proof, thanks. $\endgroup$ – nombre Sep 29 '16 at 7:22

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.