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Let $\sim$ be the equivalence relation on $\mathbb{R}^2$ given by $(x,y)\sim (x',y')$ iff $(x,y)=(x',y')$ or $(x,y)=(-x',-y')$. Show that $\mathbb{R}^2/{\sim}$ is not homeomorphic to $\mathbb{R}^2$.

I try to prove $\mathbb{R}^2/{\sim}$ is homeomorphic to $[0,1)\times\mathbb{R}$, but I can't construct homeomorphism between them. Please give me advice.

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  • $\begingroup$ @copper.hat : It should be typeset as $\mathbb R/{\sim}$ rather than $\mathbb R/\sim$. There's a reason why those look so different from each other. $\qquad$ $\endgroup$ – Michael Hardy Aug 7 '16 at 6:19
  • $\begingroup$ @MichaelHardy: I wish I understood why, Latex is a recipe based thing for me, I am missing the 'big picture'. $\endgroup$ – copper.hat Aug 7 '16 at 6:24
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    $\begingroup$ @copper.hat : Binary relation symbols and binary operation symbols have some space before and after them; thus in $3+5$ you see some space before and after the plus sign that you don't see in $+5$. Thus in $A=B$ or $A\sim B$ there is a thin space before and after the binary relation symbol. When you write $\mathbb R/\sim$, coded as \mathbb R/\sim, you see that space before and after the relation symbol. But with \mathbb R/{\sim}, there's nothing before or after that symbol, so those spaces are not there and you see $\mathbb R/{\sim}$ instead of $\mathbb R/\sim$. $\qquad$ $\endgroup$ – Michael Hardy Aug 7 '16 at 6:28
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    $\begingroup$ @MichaelHardy: Thanks! Will try to remember! $\endgroup$ – copper.hat Aug 7 '16 at 6:34
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    $\begingroup$ @copper.hat $(\mathbb{R}^2/{\sim})-\{(0,0)\}$ is essentially $(0,1)\times\mathbb{RP}^1\simeq \mathbb{R}\times\mathbb{S}^1$ no? $\endgroup$ – arctic tern Aug 7 '16 at 6:43
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The spaces ${\mathbb R}^2/{\sim}$ and ${\mathbb R}^2$ are in fact homeomorphic. In order to produce a homeomorphism I shall work with ${\mathbb C}/{\sim}$ and ${\mathbb C}$, and I write $\hat z$ for the equivalence class of $z$ modulo $\sim$. The homeomorphism $f$ in question is then given by $$f:\quad{\mathbb C}/{\sim}\to{\mathbb C},\qquad \hat z\mapsto w:=z^2\ .$$ This $f$ is clearly bijective. Its continuity in both directions follows from the geometrical description of the map $z\mapsto z^2$ and special considerations how circular open neighborhoods of $0\in{\mathbb C}/{\sim}$ are mapped onto circular open neighborhoods of $0\in{\mathbb C}$.

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  • $\begingroup$ As copper hat points above in the comments, removing some special points leaves one of them simply connected but not the other. How could this be the case if they are homeomorphic? $\endgroup$ – RKD Aug 15 '16 at 4:59
  • $\begingroup$ @Ravi: copper hat is wrong. If you remove the origin from ${\mathbb R}^2/{\sim}$ the resulting space is homeomorphic to a punctured disk. $\endgroup$ – Christian Blatter Aug 15 '16 at 8:01

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