Let $E\subseteq\mathbb{R}^2$ consist of $k$ points. Show that $\mathbb{R}^2\setminus E\simeq\bigvee_1^kS^1$ (homotopy equivalent).
I can see this geometrically, but is there a rigorous way to prove that? At the bottom of page 54 from Tammo tom Dieck's Algebraic Topology, the author draws a picture for the case $k=2$ and remarks:
In order to give a formal proof, without writing down explicit formulas, it is advisable to wait for the method of cofibrations.
Although I have now learned a bit about cofibrations, I cannot see how to do it…
Best Answer
The hard part of this is to show that given $2n$ distinct points of $\Bbb R^{2}$ $x_1,\ldots,x_n,y_1,\ldots,y_n$ there exists a homeomorphism from $\Bbb R^{2}$ onto itself which sends $x_i$ to $y_i$ for all $i=1,\ldots,n$. This can be done using tubular neighborhoods as follows:
First, consider the set of all ellipses whose endpoints of their mayor axes are $x_1$ and $y_1$. Notice that if we take two distinct such ellipses $C_1$ and $C_2$, then $C_1\setminus\{x_1,y_1\}$ and $C_2\setminus\{x_1,y_1\}$ are disjoint, and thus as there infinitely many such ellipses there must exist one which does not contain any of the points $x_2,\ldots,x_n,y_2,\ldots,y_n$.
Let $C$ be such an ellipse. The tubular neighborhood theorem says that there exists some $\epsilon>0$ and a homeomorphism $\varphi:[-1,1]\times C\rightarrow \overline{D_\epsilon(C)}$ such that $\varphi(\{0\}\times C)=C$, and this also holds for any other positive $\delta<\epsilon$, where $D_\epsilon(C):=\{x \in\Bbb R^2:d(C,x)<\epsilon\}$.You can try to prove this by taking any $\epsilon>0$ less than half the length of the minor axis of the ellipse. Pick $\epsilon$ small enough so that $\overline{D_\epsilon(C)}$ does not contain any of the points $x_2,\ldots,x_n,y_2,\ldots,y_n$.
As $C$ is homeomorphic to a circle, it is easy to see that there is a homeomorphism $[-1,1]\times C\rightarrow [-1,1]\times C$ sending $(0,x_1)$ to $(0,y_1)$ and which is the identity on the boundary of $[-1,1]\times C$(Can you see why?). Thus there is a homeomorphism of $\overline{D_\epsilon(C)}$ onto itself which sends $x_1$ to $y_1$ and is the identity on $\partial \overline{D_\epsilon(C)}$. Hence this homeomorphism extends to a homeomorphism of $\Bbb R^2$.
Continuing this process by induction we get the desired homeomorphism of $\Bbb R^2$; at intermediate steps we pick the tubular nhoods so that they don't touch the other previous tubular nhoods.
By what we did above we know $\Bbb R^2\setminus\{n$ points $\}$ is homeomorphic to $\Bbb R^2\setminus\{(1,0),\ldots,(n,0)\}$.
There is a strong deformation retract of $\overline{D_1(0)}$ onto its boundary. Thus $\Bbb R^2\setminus\{(1,0),\ldots,(n,0)\}$ is homotopic to $\Bbb R^2\setminus(D_{1/2}((1,0))\cup\cdots\cup D_{1/2}((n,0)))$.
Now $\Bbb R^2\setminus(D_{1/2}((1,0))\cup\cdots\cup D_{1/2}((n,0)))$ is clearly homotopic to $X=S_{1/2}((1,0))\cup\cdots\cup S_{1/2}((n,0))\cup\{(x,0):x\in (-\infty,1/2]\cup[n+1/2,\infty)\}$, where $S_i(x)=\{y\in\Bbb R^2:||x-y||=i\}$. These last topological space is in turn clearly homotopic to the wedge product of $n$-circles.