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It is well-known that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^2 \setminus \{(0,0)\}$.

I have two questions.

a) Does there exist a continuous bijection $f: \mathbb{R}^2 \to \mathbb{R}^2 \setminus \{(0,0)\}$ ?

b) Does there exist a continuous bijection $g: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}^2$ ?

Thank you very much for your answers in advance!

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  • $\begingroup$ You should notice that a) implies b): make $g=f^{-1}$ $\endgroup$
    – Carlos Eugenio Thompson Pinzón
    Nov 9, 2013 at 1:31
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    $\begingroup$ @CarlosEugenioThompsonPinzón: the inverse of a continuous bijection is not necessarily continuous. $\endgroup$
    – Rob Arthan
    Nov 9, 2013 at 1:40

1 Answer 1

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No: Such maps are open by invariance of domain. They are therefore homeomorphisms. http://en.wikipedia.org/wiki/Invariance_of_domain

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  • $\begingroup$ As Tim Kimsella said, such a continuous function would be a homeomorphism. But it can't be a homeomorphism because $ X:= \mathbb{R} ^2 - \left\{ 0\right\} $ and $Y := \mathbb{R} ^2 $ have different homotopy types ( $ X\equiv S^1 $ and $ Y\equiv \ast $ ). The argument would be the same for $ \phi : X\to Y $ or $\varphi : Y\to X $. $\endgroup$
    – Fernando
    Jan 1, 2015 at 5:16

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