It is said that the one point compactification of R is a circle. But how do i show it? I know it suffices to show R is homeomorphic to a punctured circle but how can i prove it?
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$\begingroup$ You can search: Alexandroff compactification: en.wikipedia.org/wiki/Alexandroff_extension $\endgroup$– Emilio NovatiJun 16, 2016 at 15:04
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1$\begingroup$ SImply consider $\Bbb{R} \to S^1$ defined by $t \mapsto e^{2i \arctan t}$. $\endgroup$– CrostulJun 16, 2016 at 15:14
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$\begingroup$ @Novati: Thanks i will check it up $\endgroup$– MathchoJun 16, 2016 at 15:44
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$\begingroup$ @Crostul: That is not even a surjection $\endgroup$– MathchoJun 16, 2016 at 15:55
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1$\begingroup$ It's a surjection onto $S^1 \setminus \{ -1 \}$. $\endgroup$– CrostulJun 16, 2016 at 16:43
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1 Answer
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Pictorially, you can imagine $\mathbb R$ to be a tangent of the circle at point $x \in S^1$. Now for each point $y \in \mathbb R$ imagine a segment joining $y$ and $-x$; $z$ be the point of intersection of this segment and the circle. The correspondence $y \mapsto z$ is a homeomorphism between $\mathbb R$ and $S^1 - \{-x\}$.
The nice thing about this construction is that it is generalizable!
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$\begingroup$ this is a quite old answer, but nonetheless; shouldn't it read, for any point $y\in\mathbb R$ instead of $y\in S^1$? $\endgroup$– ChaosMar 11, 2020 at 9:50
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