1
$\begingroup$

Consider $X = \mathbb{R}^2$ endowed with the standard topology. Define the equivalence relation $\sim$ by $(x_1,y_1) \sim (x_2,y_2)$ if $x_1 = x_2$. Let $\tau$ be the quotient topology on $X/\sim$. Show that $(X/\sim, \tau)$ is homeomorphic to $\mathbb{R}$ with the standard topology.

It looks like the equivalence classes are $E_x = \{x\} \times Y$. So I think I want to show that $\{x\} \times Y$ is homeomorphic to $Y$. I noticed that the fact is stated (without proof) in Munkres ch. 26, so it must be very obvious. But I'm not sure how to go about showing it. Any help would be appreciated.

$\endgroup$
3
$\begingroup$

HINT:

What you are looking for is a continuous bijection between $X/\sim$ and $\mathbb{R}$ which has continuous inverse. You note that the equivalence classes in $X/\sim$ look like $\{x\} \times Y$.

So, for each element $x$ in $\mathbb{R}$ you have one equivalence class: $\{ x\} \times Y$. I.e. you have a map from $\mathbb{R}$ into $X/\sim$. Is this continuous? Specifically: What do open sets in $X/\sim$ look like? Are their pre-images open?

Perhaps there exists an inverse map. Is that continuous? If so, you have a continuous map with a continuous inverse, and you have the homeomorphism you want.

$\endgroup$
  • $\begingroup$ Would it suffice to define a map $f: \mathbb{R} \rightarrow X/\sim$ given by $x \mapsto x \times Y$? The open sets in $X/\sim$ would be sets of the form $(x-\epsilon, x+\epsilon) \times Y$, so the inverse image would be $(x-\epsilon, x+ \epsilon)$, which is open in $\mathbb{R}$. So that's continuous. I don't know how to justify that the inverse map of that would be continuous though $\endgroup$ – Pawnee Oct 8 '18 at 20:25
  • $\begingroup$ Well, do you know what the inverse map might be? There is an obvious candidate, given the $f$ which you define above. $\endgroup$ – Matt Oct 8 '18 at 20:36
  • $\begingroup$ I suppose the inverse is just the projection map $g: X/\sim \: \rightarrow \mathbb{R}$ given by $x \times Y \mapsto x$. Now that I've thought about it a little longer it's pretty clearly continuous since $(x- \epsilon, x + \epsilon) \times Y$ is open in $X/\sim$. I'm really intimidated in this class but maybe that's all there is to it. Thank you for your help! $\endgroup$ – Pawnee Oct 8 '18 at 20:41
  • $\begingroup$ You're welcome. The main thing to remember in your early days doing topology is that things are nowhere near as hard as they seem. If you keep a clear head then you'll be alright. And keep working at problems. You're doing the right thing so far :) $\endgroup$ – Matt Oct 8 '18 at 23:02
0
$\begingroup$

Every value of $x$ defines (labels) an equivalence class and the quotient space is the set of equivalence classes: hence, it is the set of all $[x]$, that is the real line. Now you have to show by using the definition of quotient topology that also the induced topology is the same of the standard real line. You have to consider the sets of the plane with an open preimage under the surjective map $(x,y) \rightarrow x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.