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Let $X_1$ and $X_2$ be the spaces \begin{align*} X_1&=\{(x,y)\in \mathbb{R}^2 : (x,y)\neq(0,0)\}, \\ X_2&=\{(x,y)\in \mathbb{R}^2 : (x,y)\notin [0,1]\times\{0\}\}. \end{align*} Are these spaces homeomorphic? If these spaces are homeomorphic, what is the homeomorphism map? If these spaces are not homeomorphic, why?

I suspect these spaces are homeomorphic, but I can't construct homeomorphism.

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    $\begingroup$ What have you tried so far? What do you suspect is the correct answer? Please edit your question and elaborate on these two points. $\endgroup$
    – graydad
    Feb 10, 2015 at 15:09

2 Answers 2

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You are right with your conjecture; but there is no such thing as "the" homeomorphism between these two spaces. In fact you have to put one together from your geometrical and analytical toolkit.

Here is an idea. I'm going to replace the segment $[0,1]$ in the definition of $X_2$ by $[-1,1]$, so that the formulas get simpler. We can cover $X_2$ by a family of confocal ellipses with foci $(\pm1,0)$. A typical such ellipse has parametric representation $$\gamma_b:\quad t\mapsto\bigl(\sqrt{b^2+1}\cos t,\>b\sin t\bigr)\qquad(0\leq t\leq 2\pi)$$ with $b>0$. Now set up your homeomorphism in such a way that $\gamma_b$ is mapped onto a circle of radius $b$.

As @drhab has uttered doubts I shall proceed with the details: Let $f:\>X_1\to X_2$ be such that $$(b\cos t,\>b\sin t)\mapsto\bigl(\sqrt{b^2+1}\cos t,\>b\sin t\bigr)\ .$$ Given $(x,y)=(b\cos t,\>b\sin t)$ we have $$ \cos t={x\over b},\quad \sin t={y\over b},\quad b=\sqrt{x^2+y^2}\ .$$ It follows that $$\sqrt{b^2+1}\cos t={x\sqrt{x^2+y^2+1}\over \sqrt{x^2+y^2}},\quad b\sin t= y\ .$$ This means that our homeomorphism appears in cartesian coordinates as $$f:\quad (x,y)\mapsto\left(x\sqrt{1+{1\over x^2+y^2}}, \ y\right)\ .$$

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  • $\begingroup$ Thank you very much! I understand why these two spaces are homeomorphic. $\endgroup$
    – Mizuno
    Feb 10, 2015 at 16:47
  • $\begingroup$ If this homeomorphism is prescribed by $\left(\sqrt{b^{2}+1}\cos t,b\sin t\right)\mapsto\left(b\cos t,b\sin t\right)$ then I have doubts. It seems to send open arc $\left\{ \left(\cos\varphi,\sin\varphi\right)\mid\varphi\in\left(0,\pi\right)\right\} $ (homeomorphic to $\left(0,1\right)$) to a space homeomorphic to space $S^{1}-\left\{ .\right\} $ where $\left\{ .\right\} $ denotes a one-point subset of $S^{1}$. The spaces $\left(0,1\right)$ and $S^{1}-\left\{ .\right\} $ are not homeomorphic. Actually I would rather think $X_1$ and $X_2$ are not homeomorphic. $\endgroup$
    – drhab
    Feb 10, 2015 at 19:34
  • $\begingroup$ My doubts have left me. I realized that I made a mistake concerning subspace topologies. +1 $\endgroup$
    – drhab
    Feb 11, 2015 at 7:35
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On $A=I\times([-1,1]\setminus\{0\})$ let $f:A\to\Bbb R^2$ be the map $$f(x,y)=(x|y|,\,y)$$ on $B=[1,\infty)\times[-1,1]\setminus\{(1,0)\}$ let $f:B\to\Bbb R^2$ be the map $$f(x,y)=(x-(1-|y|),\,y)$$ and on $C=\Bbb R^2\setminus\left((0,\infty)\times(-1,1)\cup\{(0,0)\}\right)$ let $f:C\to\Bbb R^2$ be defined by $$f(x,y)=(x,y)$$ Can you show that $f:X_2\to X_1$ is continuous and find its inverse?

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    $\begingroup$ $f$ is continuous because $f$ is continuous on $A, B, C$ respectively,and agree on overlaps. Inverse $g:X_1 \rightarrow X_2$ can be defined as below: on $A'=(\{(u,v)\in\mathbb{R}^2 : v\geq u,u\geq 0, v\leq 1\} \cup \{(u,v)\in\mathbb{R}^2 : v\leq -u,u\geq 0, v\geq 1\})\setminus \{(0,0)\}$, $g(u,v)=(u/|v|,v)$, on $B'=\{(u,v)\in\mathbb{R}^2 : v\leq u,v\geq -u, -1\leq v\leq 1\}\setminus \{(0,0)\}$, $g(u,v)=(u+1-|v|,v)$ and on $C'=(\{(u,v)\in\mathbb{R}^2 : v\geq 1\} \cup \{(u,v)\in\mathbb{R}^2 : v\leq -1\}\cup \{(u,v)\in\mathbb{R}^2 : u\leq 0\})\setminus \{(0,0)\}$, $g(u,v)=(u,v)$. $\endgroup$
    – Mizuno
    Feb 10, 2015 at 17:43
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    $\begingroup$ Yes, and it is essential that the sets $A,B,C$ are closed in order to conclude that the whole map is continuous. $\endgroup$ Feb 10, 2015 at 19:18
  • $\begingroup$ I see. Since $A,B,C$ are closed, we can prove that the preimage of every closed subset of $X_1$ under $f$ is closed in $X_2$. $\endgroup$
    – Mizuno
    Feb 11, 2015 at 5:19

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