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I think they are homeomorphic, intuitively.

But how can I show this intuitively, with words?

Or am I wrong? Then why?

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— No, they are not homeomorphic (for the usual topologies). Suppose that $h : \Bbb R^2 \to \Bbb R^3$ is a homeomorphism. Then $$h' : X=\Bbb R^2 \setminus \{(0,0)\} \to Y=\Bbb R^3 \setminus \{h(0,0)\}$$ would be also a homeomorphism.

[Indeed, it is well-defined: if $p \neq (0,0)$ then $h(p) \neq h(0,0)$ (because $h$ is injective), so that the image by $h$ of any point of $X$ lies in $Y$. As a restriction of a continuous map, $h'$ is also continuous. It is also injective, because $h$ is injective. Moreover, it is surjective onto $Y$ because $h$ is surjecive. Finally the inverse of $h'$ is continuous because it is the restriction to $Y$ of the inverse $h^{-1}$ of $h$, which is continuous (because $h$ is a homeomorphism).] $\\$

— However, $Y$ is simply connected, while $X$ is not. In $X$, the loop $$\gamma : t \mapsto (\cos(2\pi t), \sin(2\pi t))$$ can't be deformed continuously to get the constant path $c : t \mapsto (1,0)$. This idea is that if you want to "shrink" the circle (which is the image of $\gamma$) into a point, you would have to pass through the origin $(0,0)$... which doesn't belong to our space $X$ ! Another way would be to "cut" the circle... but then it is not a continuous deformation anymore.

— On the other hand, any path in $Y$ can be deformed continuously into a constant loop (or into any other path). This argument shows that $X$ can't be homeomorphic to $Y$ (because simply connectedness is preserved under homeomorphisms).


In order to prove in details that $X$ is not simply connected, I think that you need to compute its fundamental group, which happens to be $\Bbb Z$. More precisely, $$\pi_1(X) = \{\underbrace{\gamma * \cdots * \gamma}_{n \text{ times}} \;|\; n \in \Bbb Z\},$$ where $*$ denotes the concatenation of paths ; if $n<0$ we agree that we concatenate the opposite of $\gamma$.


You can actually show that $\Bbb R^n$ is homeomorphic to $\Bbb R^m$ if and only if $n=m$, but this is not obvious. You can do it using singular homology, for instance. (In fact, using homology you can prove that any non-empty open set of $\Bbb R^n$ is not homeomorphic to any non-empty open set of $\Bbb R^m$, unless $n=m$).

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    $\begingroup$ @saturatedexpo : in some sense yes, but you have to check that this "something new" is preserved under homeomorphisms. $\endgroup$ – Watson Oct 26 '16 at 11:56
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    $\begingroup$ @JAEMTO : I edited it. What is unclear? If you don't know the notions of simply connectedness, then I can try to give a vague explanation, but let me tell you that it is not obvious to give a formal proof here. Your question is not obvious, in some sense. $\endgroup$ – Watson Oct 26 '16 at 12:00
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    $\begingroup$ @JAEMTO : you can also see this question or this one or that one. $\endgroup$ – Watson Oct 26 '16 at 12:06
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    $\begingroup$ @JAEMTO : I hope this is more clear now. $\endgroup$ – Watson Oct 26 '16 at 12:12
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    $\begingroup$ @Watson It's true that an open set of $\mathbb{R}^n$ minus a point isn't homotopy equivalent to an open set of $\mathbb{R}^m$ minus a point (for $n\ne m$). This implies that an open set of $\mathbb{R}^n$ isn't homeomorphic to an open set of $\mathbb{R}^m$, but it doesn't imply that they aren't homotopy equivalent. Indeed, as Mark points out, they are both homotopy equivalent to a point, and thus to each other. $\endgroup$ – Carmeister Oct 26 '16 at 18:46
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No, the fundamental group of plane-point is $Z$ and the fundamental group of $R^3$-point is $\{1\}$.

You can also say that if you remove a circle to $R^2$, the space you obtain is not connected. But if you remove a circle to $R^3$ you still have a connected space.

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  • $\begingroup$ I'm sorry but it was too hard for me to understand... $\endgroup$ – JAEMTO Oct 26 '16 at 12:08
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    $\begingroup$ To avoid confusions, you should maybe have said "if you remove something homeomorphic to a circle […]". $\endgroup$ – Watson Oct 26 '16 at 12:35
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https://en.wikipedia.org/wiki/Planar_graph

We can paint K5 in R3, but not in R2

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    $\begingroup$ May be you want to say,"The complete graph $K_5$ is 5 colorable in $\mathbb R^3$ but not in $\mathbb R^2$. $\endgroup$ – Nitin Uniyal Oct 26 '16 at 17:17
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    $\begingroup$ I think you want "embed" rather than "paint". $\endgroup$ – Ethan Bolker Oct 26 '16 at 18:59
  • $\begingroup$ Yes, "embed" Sorry for my bad english. $\endgroup$ – kotomord Oct 27 '16 at 10:40
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They are not homeomorphic. Deleting a straight line from $\mathbb R^2$ leaves it disconnected but that doesn't disturb the connectedness in $\mathbb R^3$.

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    $\begingroup$ To avoid confusions, you should maybe have said "deleting something homeomorphic to a straight line doesn't disturb the connectedness in $\Bbb R^3$." $\endgroup$ – Watson Oct 26 '16 at 13:41
  • $\begingroup$ yes, you are right. $\endgroup$ – Nitin Uniyal Oct 26 '16 at 17:19
  • $\begingroup$ @Watson Since all the answer needs to provide is something to omit that destroys connectivity, I don't see any confusion that needs to be avoided. $\endgroup$ – Ethan Bolker Oct 26 '16 at 18:58
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    $\begingroup$ @EthanBolker : if I suppose that $h : \Bbb R^2 \to \Bbb R^3$ is a homeomorphism, then $h(\Bbb R^2 \setminus (\Bbb R \times \{0\})) = \Bbb R^3 \setminus h (\Bbb R \times \{0\})$, and $h(\Bbb R \times \{0\})$ is homeomorphic to a straight line, but is not a straight line itself, however. I just wanted to point out this detail.$\tag*{}$ How would you prove that $X:=\Bbb R^3 \setminus h (\Bbb R \times \{0\})$ is connected? I don't see immediately why $X$ should be homeomorphic to $Y:=\Bbb R^3 \setminus (\Bbb R \times \{(0,0)\})$, for instance. $\endgroup$ – Watson Oct 26 '16 at 19:04
  • $\begingroup$ @Watson You're right. I was thinking only about the line in the domain. $\endgroup$ – Ethan Bolker Oct 26 '16 at 19:13

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