Showing a space is not locally homeomorphic to $\mathbb{R}^n$

Question

I have the quotient space on $S^n$ (the n-sphere) given by the equivalence: $$x \sim y \iff x=y \text{ or }x_{n+1}=y_{n+1}=0$$ I can see that this essentially "draws" $S^{n-1}$ and then pulls that sphere to a point, splitting it into a bouquet of 2 n-spheres, joining at a point I have called $[s]$, the equivalence class of the "belt". I need to say whether this is a manifold or not, I say it is not as the space around $[s]$ doesn't look like $\mathbb{R}^n$, but I don't know how to formalise this.

Answer 1

Short answer: You can remove the special point and check how many connected components you have.

Long answer: For $n\geq 2$, if you remove the special point you will disconnect the space. Any path from $(0,...,1)\to (0,...-1)$ must pass through the equator $x_{n+1}=0$. You can see this by projecting onto the last coordinate and applying intermediate value theorem. Removing a point from a manifold of dimension $n\geq 2$ does not disconnect the space so your quotient space is not a manifold of dimension $n\geq 2$ For many reasons, it is also not now a one dimensional manifold so it is not a manifold.

For $n=1$, removing a point from a one dimensional manifold locally disconnects the space into two components. However, in your quotient space, it will locally disconnect into four. More precisely, we can find a neighbourhood of the special point which is homeomorphic to the union of two open line segments which intersect at a single point. This space is not homeomorphic to an open line segment as it can be disconnected into four components by removing a single point.

Answer 2

Every open neighborhood of $[s]$ becomes disconnected after removing $[s]$.