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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!

whatismath
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1 Answers1

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(a) Is the quotient space Hausdorff? Compact? Connected?

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.

(b) What is the homotopy type of X, and what is the homotopy type of X/A with one point removed?

$X$ is contractible, via $(t,v)\mapsto tv$ homotopy.

$(X/A)\backslash\{v\}$ is contractible as well. To see this we can consider two cases:

  1. When $v=[A]$ then $(t,[x])\mapsto [tx]$ is a well defined homotopy, because every $[x]$ has a single representative when $[A]$ is excluded. Again we simply contract everything towards origin.
  2. When $v\neq [A]$ it is a bit more complicated. Basically we want to linearly contract every point $x$ towards $[A]$ away from $v$. Or in other words by connecting $v$ to $x$ via straight line and finding the projection of that line on $A$, or more precisely on the boundary of $A$. Meaning finding an element on the line with norm $1$. This can be done by first solving $\lVert tx+(1-t)v\rVert=1$ equation (for $t>1$), say its solution is $t_x$, and let $v_x=t_xx+(1-t_x)v$. Then we put $(t,[x])\mapsto [tx+(1-t)v_x]$ and $(t,[A])\mapsto [A]$. This function is continuous, because I already know that $t_x$ is expressible in a continuous way, which I leave as an exercise.

It can actually be shown that $X/A$ is homeomorphic to $S^2$, but this requires a bit more work.

freakish
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  • So would part (a) be a similar argument then if instead of $A$ we had $B = {(x,y) \in \mathbb{R} \mid 1< |(x,y)| \leq 2 } \subseteq X$ or would, as I'm thinking, the fact that $B$ has a different equality majorly affect it? – whatismath Mar 10 '22 at 12:35
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    @whatismath the quotient would still be compact and connected, but no longer Hausdorff, not even $T_1$. That's because $T_1$ spaces have to have closed points (this can even be treated as the definition of $T_1$). But $[B]$, as a point in $X/B$ is not closed, because that is equivalent to $B$ being closed as a subset of $X$. – freakish Mar 10 '22 at 12:38
  • Thank you so much! – whatismath Mar 10 '22 at 12:41