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?

  • $\begingroup$ 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

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 $B_{\epsilon/2}(0))$.

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

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