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We want to show $\mathbb R^2 \setminus \left\{0\right\}$ and $\mathbb R^2$ are homeomorphic (Goal), but I already proved that $\mathbb R^2 \setminus \left\{0\right\}$ and $S^1$ are homeomorphic. And I already proved that $S^1$ and $B^2$ are not homeomorphic.

So I thought it only need to show that $B^2$ and $\mathbb R^2$ are homeomorphic. And I tried to construct a deformation retraction $H$ of $\mathbb R^2$ onto $B^2$, defined by $$ H(x, t)= \begin{cases} x & \;\text{ if }\; |x|\leq 1,\\ x \cdot (t/|x| + 1-t) & \; \text{ if }\; |x|>1, \end{cases}$$ (where $x$ is in $\mathbb R^2$) so that points of outer circle to be continuously collapse to the unit circle.

Is it okay to do that?

Thank you.

Gary
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girrafe
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    Isn't the first space simply connected and the second one not? – markvs Dec 14 '21 at 01:40
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    $\mathbb R^2\setminus {0}$ is not homemorphic to $ S^1.$ The two spaces are homotopy equivalent, but not homeomorphic. Take any two points out of $S^1,$ and the result is not a connected space. Not true for any two points in $\mathbb R^2\setminus{0}.$ – Thomas Andrews Dec 14 '21 at 01:43
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    $\mathbb{R}^2 - (0,0)$ and $\mathbb{R}^2$ are not homeomorphic. – David Lui Dec 14 '21 at 01:47
  • Your map $\mathbb R^2\to B^2$ is not $1-1,$ so it is not a homeomorphism. It is a homotopy equivalence. – Thomas Andrews Dec 14 '21 at 01:47

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It is not true that ${\mathbb{R}^2 - (0,0)}$ and ${\mathbb{R}^2}$ are homeomorphic. In fact, they don't even have the same homotopy type, since ${\mathbb{R}^2 - (0,0)\simeq S^1}$, we have $$ \pi_1(\mathbb{R}^2 - (0,0)) = \pi_1(S^1) = \mathbb{Z},\text{ but }\pi_1(\mathbb{R}^2) \equiv 0 $$

Riemann'sPointyNose
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