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We have two spaces $X=\{(x,1/n):n\neq 0, n\in\mathbb{Z}, x\in\mathbb{R}\}$ and $Y=\{(x,n):n\neq 0, n\in\mathbb{Z}, x\in\mathbb{R}\}$. On both spaces we introduce the equivalent relation $(x,y)\sim (x',y')$ if $x=x'$ and $y=y'$ or $x=x'=0$. That is, all points on the $y$ axis are collapsed to the same point.

We are asked whether $X/\sim$ and $Y/\sim$ are homeomorphic in quotient topologies.

It is easy to show that the original spaces are homeomorphic. However, I don't know how to answer the question about the quotient spaces.

My guess is that they might not be homeomorphic and some problem might occur at the $origin$ but I am not sure.

Any hint would be helpful! Thanks!

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1 Answer 1

up vote 3 down vote accepted

Once you've shown that $X$ and $Y$ are homeomorphic, it's almost immediate to show that $X / \sim $ and $ Y / \sim $ are homeomorphic :

if $f$ is the natural homeomorphism between $X$ and $Y$ just notice that $x \sim y $ if and only if $f(x) \sim f(y)$ . Then $g$ the induced fonction between $X / \sim $ and $ Y / \sim $ is continuous(by definition of the quotient topology) and one to one. Do the same with $f^{-1}$ and then $g$ is a homeomorphism.

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