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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?

JAEMTO
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4 Answers4

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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 $n=m$).

Watson
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  • is a good catch phrase: "if you can learn something new in one Field, what you cant in the other, then its not homeomorphic"? – SAJW Oct 26 '16 at 11:49
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    @saturatedexpo : in some sense yes, but you have to check that this "something new" is preserved under homeomorphisms. – Watson Oct 26 '16 at 11:56
  • @Watson Well, I am reading this repeatedly, but I cannot understand second paragraph. Can you explain more specifically? Thank you... – JAEMTO Oct 26 '16 at 11:58
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    @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. – Watson Oct 26 '16 at 12:00
  • @Watson After you edited, Now that's okay. Thx. :D – JAEMTO Oct 26 '16 at 12:05
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    @JAEMTO : you can also see this question or this one or that one. – Watson Oct 26 '16 at 12:06
  • @Watson But I got one thing: How can I sure X->Y homeomorphic in first paragraph, without origin of $\mathbb {R}^2,\mathbb{R}^3 $ – JAEMTO Oct 26 '16 at 12:07
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    @JAEMTO : I hope this is more clear now. – Watson Oct 26 '16 at 12:12
  • Your final parenthetical remark says that a non-empty open set of $\mathbb R^n$ is not homotopy equivalent to the same in $\mathbb R^m$, but open balls in any dimension are homotopic to a point. Can your clarify what you meant? – Mark S. Oct 26 '16 at 17:35
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    @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. – Carmeister Oct 26 '16 at 18:46
  • Yes, thank you, and sorry about this mistake. This is because I was thinking to homotopy invariance of homology…! – Watson Oct 26 '16 at 18:59
  • The fact that (a non-empty open subset of) $\mathbb{R}^n$ is not homeomorphic to (any non-empty open subset of) $\mathbb{R}^m$ for $n \ne m$ is often called ___invariance of dimension___, see for example Terence Tao's blog post Brouwer’s fixed point and invariance of domain theorems, and Hilbert’s fifth problem (Corollary 3 near the top of the page), and many other sources. – Jeppe Stig Nielsen Oct 26 '16 at 21:03
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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.

Tsemo Aristide
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https://en.wikipedia.org/wiki/Planar_graph

We can paint K5 in R3, but not in R2

kotomord
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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 $\mathbb R^3$.

Nitin Uniyal
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    To avoid confusions, you should maybe have said "deleting something homeomorphic to a straight line doesn't disturb the connectedness in $\Bbb R^3$." – Watson Oct 26 '16 at 13:41
  • yes, you are right. – Nitin Uniyal Oct 26 '16 at 17:19
  • @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. – Ethan Bolker Oct 26 '16 at 18:58
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    @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. – Watson Oct 26 '16 at 19:04
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    @Watson You're right. I was thinking only about the line in the domain. – Ethan Bolker Oct 26 '16 at 19:13