Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users

Dashboard
Notifications
Mark all as read
Q&A

How to show if a set is simply connected?

+6
−0

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. $\left( \begin{array}{c} x\\1+\cos(\phi)\\1+\sin(\phi) \end{array} \right),\phi\in[0,2\pi[$

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comments

1 answer

+4
−0

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.

Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment

Thank you for the lecture notes. kreijstal‭ 30 days ago

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!