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!?
Best Answer
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 $\Bbb R^2\setminus\{x_0\}$.