No idea where to even begin. There is a hint: this requires construction of an explicit function.
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To make drawings easier, you can always use $B(0,1)$ instead of $\Bbb R^2$ and $(0,1)$ instead of $\Bbb R$. Then the homeomorphism should be clear: pull from the tiny hole in $B(0,1)$ outwards, while pulling from (nonexistent) the boundary of $B(0,1)$ to get a cylinder with no "endpoint" circles. – Pedro Nov 04 '13 at 05:05
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Can I assume that R^2 {0} and S^1 x R are continuous or is that an additional part I have to prove with U open in Y? – user105225 Nov 04 '13 at 05:16
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2Careful: functions may or not be continuous; continuity does not apply to spaces. – user99680 Nov 04 '13 at 06:46
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This question is an exercise in a well-known textbook. I wish people would stop posting questions from textbooks on sites like this. It makes it annoyingly difficult to create homework for my students, knowing that they can simply google the exact statement of the question and immediately find an answer. – David White Oct 12 '23 at 20:05
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Hint: The plane minus the origin can be written as
$$\Bbb{R}^2 \setminus \{0\} = \{(r \cos{t}, r \sin{t}) : 0 < r < \infty, 0 \le t < 2\pi\}$$
Do you see how this looks like a circle cross a line?
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3This is far more than a hint. You just can't resist giving answers rather than hints! – Ted Shifrin Nov 04 '13 at 04:35
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And would S1 x R just look like a cylinder that keeps going as the real line keeps going? – user105225 Nov 04 '13 at 04:39
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1You have a copy of circles indexed by the Reals{0}. For every r in Reals{0}, take a circle of radius r centered at the origin. – user99680 Nov 04 '13 at 04:52
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HINTS: Use polar coordinates for $\Bbb R^2\setminus\{\langle 0,0\rangle\}$ and $S^1$.
For each $\theta\in[0,2\pi)$ the ray $\{\langle r,\theta\rangle:r>0\}$ corresponds to the line $\{\langle 1,\theta\rangle\}\times\Bbb R$.
The map $$\Bbb R^+\to\Bbb R:x\mapsto\begin{cases}x-1,&\text{if }x\ge 1\\\\1-\dfrac1x,&\text{if }0<x<1\end{cases}$$ is a homeomorphism.
Brian M. Scott
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