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As indicated in the title, I want to show with outmost possible rigor and without using too advanced techniques that $\Omega=\mathbb{R}^3\backslash\{(0,0,z)\ \mid\ z\in\mathbb{R}\}$ isn't simply connected (which intuitively is obvious). Here is my attempt:

Suppose $\Omega$ is simply connected, then for $\alpha,\beta:[0,\pi]\to\mathbb{R}^3$, $\alpha:t\mapsto (\cos t,\sin t,0)$, $\beta:t\mapsto(\cos t,-\sin t,0)$ there exists $F\in C^0([0,\pi]\times[0,1],\Omega)$ (we write $F=(F^1,F^2,F^3)$) such that $F(t,0)=\alpha(t)$ and $F(t,1)=\beta(t)$ for all $t\in[0,\pi]$ as well as $F(0,s)=(1,0,0)$ and $F(\pi,s)=(-1,0,0)$ for all $s\in[0,1]$.

As for $s\in[0,1]$ fix we have $F^1(0,s)=1$ and $F(\pi,s)=-1$, there exists $f(s)\in (0,\pi)$ such that $F^1(f(s),s)=0$. If we could prove that there exists such an $f:[0,1]\to(0,\pi)$ such that $f$ is continuous, then we see that $F^2(f(0),0)=1$ and $F^2(f(1),1)=-1$ and thus there would have to be an $s\in(0,1)$ such that $F^2(f(s),s)=0$ and thus $F(f(s),s)\notin\Omega$, contradiction.

However, I don't see how to prove the global existence of such an $f$ (intuitively it should exist though). By the inversion theorem, we can conclude that such an inverse exists locally around $s=0$ and $s=1$, but I fail to see how to extend this argument.

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    Does $f$ need to be surjective? Because the image of a compact set under a continuous map is compact, so that's impossible. Otherwise why not just take something like $f(x)=x+1$ – 2017-01-04
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    Does covering space theory and the homotopy lifting lemma count as "too advanced"? Because I don't think there are any shortcuts to applying the homotopy lifting lemma. – 2017-01-04
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    I just started learning homotopy so probably I won't get your answer, but if you have the time it would be interesting to see :) +1 in any case. So you don't think one could somehow finish my attempt? – 2017-01-04

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The function $f$ you seek need not exist. Here's what can go wrong:

homotopy

The picture shows a homotopy $F$, and indicates how the point at which it crosses $x=0$ can change discontinuously.

To successfully prove that $\Omega$ is not simply connected is essentially the same as proving that the circle $S^1$ is not simply connected, because $\Omega$ deformation retracts onto $S^1$. Establishing the non-simple connectedness of the circle takes some work, and I recommend the proof on pg. 29 and following of Hatcher's book Algebraic Topology.

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    Ah, I see :) Thanks for the great answer! I asked the question because it was part of an old Analysis III exam of my university, and I wondered if they would have wanted to see a rigorous proof. But as by the end of the third semester we only have some basic knowledge about general topology and haven't seen algebraic topology, I think it would have been enough to say "the curve has to cross the z-axis somewhere sometimes". :) – 2017-01-04