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I have studied in Hatcher (Algebraic topology, p.169) an homological proof of the Jordan theorem. I would like to understand an upgrade of this theorem, the Schoenflies version.
On my way to prove it, I have studied the idea of using the Caratheodory theorem. Even if I can understand this one, I am completely unable to conclude.

The common idea (that is shared on internet), is to use the conformal mapping property, and I agree with that. But here is the problem, as everything related to the Jordan theorem, the fact that a Jordan domain is simply connected seems clear, but the demonstration is not.

So, I tried to prove this assertion. Here is my first idea:
In the Hatcher's proof of Jordan theorem, we saw that $$H_i(\mathbb{S}^2 - h(\mathbb{S}^1))=0$$ Except for $i =2-1-1=0$ (here $h$ is any continuous function). So $H_1(G)=0$ where $G$ is the bounded component of $\mathbb{R}^2-\mathbb{S}^1$. However, this is not a characterization of simple connectedness, we would like to have the save result but with $\pi_1(G)$. So there is my question, does someone know (a link to) a proof that $$ \pi_1(G)=0 ?$$ It can be a rather conceptual explanation, I would like it.

Furthermore, I have heard about a definition of simple connectedness with Jordan curves, but I don't know how to prove it either.

Thank you

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  • $\begingroup$ This was discussed at MO before: Fundamental group of a noncompact surface is free. $\endgroup$
    – Moishe Kohan
    Commented Jun 27, 2022 at 17:34
  • $\begingroup$ The idea is this 2-manifold can be understood to be built using only a single 0-handle with 1-handles attached. So that's why the $H_1$ computation suffices. $\endgroup$
    – Ryan Budney
    Commented Jun 27, 2022 at 18:01
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    $\begingroup$ In this wikipedia page you can find essentially a proof that a domain in the plane with trivial deRham cohomology is simply connected $\endgroup$
    – Denis Nardin
    Commented Jun 27, 2022 at 18:48

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