In complex analysis, there is a classical result concerning simple connectivity.

A set $E \subset \mathbb{C}$ is open, bounded and connected. Then $E$ is simply connected iff $E^c$ is connected.

I know a proof by means of complex analysis. Its sketch is as follows:

"$\Leftarrow$": By the general Cauchy Theorem, for a closed curve $\gamma$ in $E$ and a function $f \in H(E)$, \begin{equation} \int_\gamma f \operatorname{d\!} z=0. \end{equation} Hence $E$ is holomorphically simply connected. Due to the Riemann Mapping Theorem, $E$ is homeomorphic to $\mathbb{D}$, providing that $E$ is simply connected.

"$\Rightarrow$": If $E^c$ is not connected, we can construct a closed curve $\gamma$ in $E$ such that $\operatorname{Ind}_\gamma(a)\ne 0$ for some $a \in E^c$, which contradicts the fact that $E$ is simply connected.

My question is whether there is a topological proof of it. Thank you very much!

  • $\begingroup$ What do you mean by a "topological proof" ? this isn't true in most metric (to make sense of bounded) spaces, so a proof has to use something specific to $\mathbb{C}$ (for instance it is blatantly false in $\mathbb{R}^3$ and higher dimensions, so not only does it have to use something specific to the reals, but it has to use dimension $2$) $\endgroup$ – Max Mar 3 at 14:00
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    $\begingroup$ It is a result just involving the structure of $\mathbb{R}^2$. But the proof by complex analysis adds some additional structures on $\mathbb{R}^2$ (such as making it become a field). I want to find a proof not using complex analysis but using, for example, algebraic topology. $\endgroup$ – Andrew Mar 3 at 14:19
  • $\begingroup$ I guess you can use Jordan's theorem, but I don't know its proof so I don't know if this would fit into what you call a topological proof $\endgroup$ – Max Mar 3 at 14:34
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    $\begingroup$ You should note that the complement in question is $\Bbb C\cup\{\infty\}\setminus E$. (Consider $\{0<y<1\}$ for example...) $\endgroup$ – David C. Ullrich Mar 3 at 15:36
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    $\begingroup$ You should state the definition of "simply connected" that you are using. The reason is that the definitions of simple connectivity given in complex analysis texts tend to be rather nonstandard. See also: math.stackexchange.com/questions/330479/… , math.stackexchange.com/questions/2415064/… $\endgroup$ – Moishe Kohan Mar 4 at 0:12

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