# $\mathbb{R}^2/{\sim}$ is not homeomorphic to $\mathbb{R}^2$

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.

• @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. $\qquad$ – Michael Hardy Aug 7 '16 at 6:19
• @MichaelHardy: I wish I understood why, Latex is a recipe based thing for me, I am missing the 'big picture'. – copper.hat Aug 7 '16 at 6:24
• @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 $\mathbb R/\sim$. $\qquad$ – Michael Hardy Aug 7 '16 at 6:28
• @MichaelHardy: Thanks! Will try to remember! – copper.hat Aug 7 '16 at 6:34
• @copper.hat $(\mathbb{R}^2/{\sim})-\{(0,0)\}$ is essentially $(0,1)\times\mathbb{RP}^1\simeq \mathbb{R}\times\mathbb{S}^1$ no? – arctic tern Aug 7 '16 at 6:43

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 $0\in{\mathbb C}$.
• @Ravi: copper hat is wrong. If you remove the origin from ${\mathbb R}^2/{\sim}$ the resulting space is homeomorphic to a punctured disk. – Christian Blatter Aug 15 '16 at 8:01