Let $X= \{(x, y) \in \mathbb{R^{2}} \mid |(x, y)| \leq 2\}$ be the closed unit disc in $\mathbb{R^{2}}$ with vectors of length at most $2$. Also let $A = \{(x, y) \in \mathbb{R^{2}} \mid 1 ≤ |(x, y)| ≤ 2\} ⊆ X$.
Then we have that $X/A$ is the quotient space.
(a) Is the quotient space Hausdorff? Compact? Connected?
(b) What is the homotopy type of X, and
what is the homotopy type of X/A with one point removed?
I think for (a) that the quotient space is Hausdorff and compact but not connected but I'm not sure how to construct an argument for this that's concise. If my intuition on it being Hausdorff and compact is even correct. Does part (a) have anything to so with my assumption that the quotient space is homeomorphic to $S^{2}$, if that's right? Then for (b) I'm just kinda lost.
Any hints would be greatly appreciated!
Best Answer
Quotient space of a connected space is always connected, because the quotient map $\pi:X\to X/\sim$ is continuous and surjective. By similar argument it is also compact.
Whether $X/A$ is Hausdorff depends on what it means exactly. Typically it is defined as the collapse, i.e. $x\sim y$ iff $x=y$ or $x,y\in A$.
Now, it is well known that a quotient $X/\sim$ is Hausdorff if and only if $\sim$ is a closed subset in $X\times X$ (see this). With that note that our relationship $\sim$ is actually equal to $(A\times A)\cup\Delta$ where $\Delta=\{(x,x)\ |\ x\in X\}$ is the diagonal. The diagonal $\Delta$ is compact, since $X$ is and $x\mapsto (x,x)$ is a homeomorphism. $A\times A$ is also compact as a product of compact spaces. Therefore our collapse is compact, as a union of compact subsets of $X\times X$, and so it is closed as well. This implies $X/A$ is Hausdorff.
$X$ is contractible, via $(t,v)\mapsto tv$ homotopy.
$(X/A)\backslash\{v\}$ is contractible as well. To see this we can consider two cases:
It can actually be shown that $X/A$ is homeomorphic to $S^2$, but this requires a bit more work.