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It is clear that $\mathbf{S}^1$ is a deformation retract of $\mathbf{R}^2\setminus\{0\}$ since we can consider the straight line deformation retract $H\colon (\mathbf{R}^2\setminus\{0\}) \times [0, 1] \to \mathbf{R}^2\setminus\{0\}$ defined by

$$H(x, t) = (1 - t)x + t\frac{x}{\|x\|}.$$

This satisfies the properties of a deformation retraction:

  • $H(x, 0) = x$ for all $x$,
  • $H(x, 1) = x/\|x\| \in \mathbf{S}^1$ for all $x$,
  • $H(a, t) = (1 - t)a + ta/\|a\| = a$ since $\|a\| = 1$ as $a \in \mathbf{S}^1$.

But I cannot see a way to construct such a deformation retract given the more general space $\mathbf{R} \setminus \{ x_0 \}$ for an arbitrary $x_0 \in \mathbf{R}^2$ (which is supposedly true, see for example this question), instead of the particular case where $x_0$ is the origin.

What is especially confusing is when the puncture point $x_0$ actually lies on the unit circle, what is the explicit deformation retract in this case? $\mathbf{S}^1$ is not even a subset of $\mathbf{R}^2\setminus\{x_0\}$, so how could one possibly exist!?

MrBigGoats
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    $\Bbb{R}^2-{x_0}$ is homeomorphic to $\Bbb{R}^2-{0}$ via translation. You can construct a deformation retract of $\Bbb{R}^2-{x_0}$ by using translation properly. – Hanul Jeon Sep 04 '15 at 04:57
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    This is true if and only if $|x_0| < 1$. However, for any $x_0$, $\Bbb R^2 \setminus {x_0}$ deformation retracts onto a subspace homeomorphic to the unit circle: just pick your favorite circle around $x_0$. –  Sep 04 '15 at 04:58

1 Answers1

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It isn’t true, and the answer to the linked question does not say that it is. Look again at that question: given a point $x_0\in\Bbb R^2$, how do you find a circle that is a strong deformation retract of $\Bbb R^2\setminus\{x_0\}$? The answer is that you choose a circle that surrounds $x_0$. The accepted answer shows what to do whey $x_0$ is the origin, and the OP was supposed to generalize this to arbitrary $x_0$.

Similarly, given a circle in $\Bbb R^2$, you must choose a point $x_0$ inside it if you want it to be a retract of $\Bbb R^2\setminus\{x_0\}$.

Brian M. Scott
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  • Great, thank you for clearing that up. So if we use the deformation retract $H(x, t) = (1- t)x + t\frac{x - x_0}{|x - x_0|}$ say, are the circles then related in any way? – MrBigGoats Sep 04 '15 at 06:06
  • I mean, they're pretty much the same so is there some sort of property that describes what's going on? – MrBigGoats Sep 04 '15 at 06:09
  • Oh I can see the comments above, they pretty much answer it. Another question: when Munkres says this (pp. 362), is that only true when $p$ and $q$ are (separately) in the top and bottom unit semicircles? If not, it's true that we can find a homeomorphism between a translated/scaled theta space which contains $p$ and $q$, and the "unit" theta space? – MrBigGoats Sep 04 '15 at 06:28
  • @MrBigGoats: Yes, you need to have one of the deleted points in each of the semicircular regions. – Brian M. Scott Sep 04 '15 at 06:39