Comments on How to show if a set is simply connected?
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How to show if a set is simply connected?
How do I show that the following set is not simply connected?
$M= \mathbb{R}^3\setminus\{(x,y,z)\in \mathbb{R}^3|y=z=1\} $
I am aware that if a set has a hole, then it isn't simply connected. But how can I show this? In the example I gave, I know that the set isn't simply connected because there is a "hole" which is the line y=z=1, basically there exists a loop on $\mathbb{R}^3$ which "contains" this hole. But I don't know how to formalize this.
An example of the loop would be something like. $\left( \begin{array}{c} x\\1+\cos(\phi)\\1+\sin(\phi) \end{array} \right),\phi\in[0,2\pi[$
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Probably one of the simpler ways of establishing this is reducing it to a case where you already know the answer. I will assume that you have proven at some point that the circle, $\mathbb S^1$, is not simply connected. Simple connectedness is a topological property so it is preserved by homeomorphism. Another important and useful fact about simply connected spaces is that a retract of a simply connected space is simply connected. The proof outline is then simply to show that the circle is a retract of $M$ and thus $M$ can't have been simply connected.
A retraction of a space $X$ onto a subspace $A \subseteq X$ is a continuous function $r : X \to A$ such that $\forall a \in A. r(a) = a$. We say that $A$ is a retract of $X$ if there exists a retraction $X \to A$.
The function $r(x,y,z) = ((y-1)^2 + (z-1)^2)^{-\frac{1}{2}}(0, y, z)$ is a retraction $M \to \mathbb S^1$ where we're identifying $\mathbb S^1$ with unit circle in the $yz$-plane centered at $(0, 1, 1)$. You should prove that this is a continuous function. (It clearly would not be continuous if we didn't omit the line $y = z = 1$.) Intuitively, this projects to the $yz$-plane and then maps each point on a ray in the $yz$-plane starting at $(0, 1, 1)$ to the corresponding point at unit distance from $(0, 1, 1)$. To be nitpicky, you should also prove that the subset of $M$ we were calling $\mathbb S^1$ is, in fact, homeomorphic to the circle. Alternatively, you can just directly prove that this subspace is not simply connected.
See these lecture notes for discussion and proofs of the subresults, though I recommend attempting to prove them yourself first if you haven't already.
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