# $R^2$ with $n$ points removed is a bouquet of $n$ circles

It seems almost obvious that $$\mathbb{R}^2$$ with $$n$$ points removed is homotopy equivalent to the wedge sum of $$n$$ circles. Take sufficiently small circles around the removed points, connect each circle to the next one using non-intersecting paths, and take a deformation retraction which radially projects the points inside each circle out from the removed point, and deformation retracts everything outside onto the circles and paths connecting them. Then the paths can be deformation retracted to points and the points connecting circles pairwise can be brought to a single point by deformation retracting arcs of the circles between two such points.

While the visual intuition is clear, I have yet to encounter a rigorous proof. Constructing an explicit deformation retraction seems daunting. A stronger claim which could help is the claim for any finite sets $$F_1,F_2\subset\mathbb{R}^2$$, if $$\vert F_1\vert=\vert F_2\vert$$, then $$\mathbb{R}^2\setminus F_1$$ is homeomorphic to $$\mathbb{R}^2\setminus F_2$$. If I had this stronger claim, then I think I could prove explicitly that $$\mathbb{R}^2$$ with $$n$$ points removed is homotopy equivalent to a wedge sum of $$n$$ circles. However, the stronger claim seems even more daunting. Does anyone know of a rigorous proof?

Let $$F \subset \mathbb R^2$$ be a finite set. We want to construct a homeomorphism $$f: \mathbb R^2 \rightarrow \mathbb R^2$$ mapping $$F$$ to a fixed finite set of size $$n$$. We will adapt the convention that "applying a map $$f$$ to a point $$p$$" means that the letter $$p$$ now denotes what was previously $$f(p)$$. This gives the following procedure an algorithmic feel, which is easier to follow I hope.
First, note that we can ensure that one of the elements of $$F$$ is zero by just applying a translation. Once we are in this situation, let $$p \in F$$. Then applying a rotation (around zero) arranges that $$p$$ gets mapped to the $$x$$-axis. Applying a transformation of the form $$(x,y) \mapsto (x, \lambda y)$$ for big enough $$\lambda$$ ensures that all $$p \in F \setminus x\text{-axis}$$ have modulus strictly greater than all $$q \in x\text{-axis}$$.
Now comes the key step. Let $$R_{\alpha}: \mathbb R^2 \rightarrow \mathbb R^2$$ be a rotation (around zero) of angle $$\alpha$$ mapping a point $$p \in F \setminus x\text{-axis}$$ of minimal modulus to the $$x$$-axis. Then let $$f: \mathbb R^2 \rightarrow \mathbb R^2$$ be given as $$f(s) = \begin{cases} s & |s| \leq \max_{q \in x\text{-axis}}|q| \\ R_{(L(|s|))\alpha}(s) & \max_{q \in x\text{-axis}}|q|< |s|<\min_{q \in F \setminus x\text{-axis}}|q| \\ R_\alpha & |s| \geq \min_{q \in F \setminus x\text{-axis}} |q|, \end{cases}$$ where $$L: \mathbb R \rightarrow R$$ is a linear function with $$L(\max_{q \in x\text{-axis}} |q|)= 0$$ and $$L(\min_{q \in F\setminus x\text{-axis}}|q|) = 1$$. This $$f$$ can be visualised as some kind of rotation twisting only outside of a ball around zero. Applying this function and again some $$(x,y) \mapsto (x, \lambda y)$$ produces the same situation but with strictly more points on the $$x$$-axis.
Proceeding by induction, we obtain that there exists an $$f: \mathbb R^2 \rightarrow \mathbb R^2$$ mapping $$F$$ to the $$x$$-axis. All that is left to show is that there exists a homeomorphism $$h: \mathbb R^2 \rightarrow \mathbb R^2$$ mapping $$F \subset \mathbb R$$ to some fixed set, say $$\{0,1, \dots, |F|-1\}$$. But this is easy, just consider some piecewise linear function $$h$$ that does this, and then apply $$(x,y) \mapsto (h(x), y)$$ to obtain our desired final function mapping $$F$$ to a specific set, namely $$\{(0,0), (0,1), \dots, (0, |F|-1)\}$$.
• This is great! Thank you! I imagine this could be generalized to $\mathbb{R}^m$ inductively by using the same method to reduce it to $\mathbb{R}^{m-1}$. This definitely puts my nerves at ease. Commented Nov 9, 2019 at 21:48