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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...

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  • $\begingroup$ There probably is. Do you have any thoughts? $\endgroup$ Dec 23 '18 at 6:23
  • $\begingroup$ @mathworker21 For me the case $k=1$ can be solved by observing that $\mathbb{R}^2\setminus\{0\}\cong S^1\times(0,\infty)$. But for the general case I don't see any way to do this rigorously, given that the $k$ points are arbitrarily placed so that an explicit formula for the deformation retraction seems impossible to me. $\endgroup$
    – Colescu
    Dec 23 '18 at 6:28
  • $\begingroup$ I think this can be solved using tubular neighborhoods. Are you familiar with this concept? $\endgroup$ Dec 23 '18 at 7:55
  • $\begingroup$ @CamiloArosemena-Serrato I've heard of this but unfortunately I'm not familiar with it. Could you please elaborate on this? $\endgroup$
    – Colescu
    Dec 23 '18 at 8:08
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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.

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Here's a nuke-proof of this, for the sake of rigor (although I honestly do not see the point of frowning at the perfectly sound geometric argument by explicitly drawing the deformation retract).

Let $z_1, \cdots, z_n$ be $n$ distinct points on $\Bbb C$. Then $X = \Bbb C \setminus \{z_1, \cdots, z_n\}$ admits a complex structure from $\Bbb C$, making it a noncompact Riemann surface. The universal cover $\widetilde{X}$ is a simply connected Riemann surface, so by uniformization theorem has to be biholomorphic to one of $\Bbb{CP}^1$, $\Bbb C$ or $\Bbb H^2$. It cannot be the first, as the only group acting freely on $\Bbb{CP}^1 \cong S^2$ is $\Bbb Z_2$, in which case the quotient is $\Bbb{RP}^2$ which is nonorientable, therefore cannot be $X$. $\widetilde{X}$ must be one of the latter two, which are both contractible. Therefore $X$ is a $K(\pi, 1)$-space. By taking an appropriate cover of $X$ (say $U$ is one side of a line in $\Bbb C$ separating $z_n$ from $z_1, \cdots, z_{n-1}$, and $V$ is the other side, both appropriately thickened to be open) and using Siefert-van Kampen theorem along with induction, you can prove that $\pi_1(X) \cong F_n$ is the free group on $n$ letters.

The above calculation implies $X$ is a $K(F_n, 1)$-space, and the same is true of $\bigvee_1^n S^1$. Homotopy type of a $K(G, 1)$-space is uniquely determined by $G$, see eg Hatcher, theorem $1$B.$8$. This proves $X$ is homotopy equivalent to the wedge of $n$ circles.


If $n \geq 2$, the universal cover is in fact biholomorphic to $\Bbb H^2$. I was trying to construct this cover explicitly for $n = 2$. First thoughts lead me to consider this picture. This is the Cayley graph of $F_2$ (with the length metric) isometrically embedded in $\Bbb H^2$. The subgroup $G = \langle \tau_1, \tau_2 \rangle < \text{Isom}(\Bbb H^2)$ generated by translation along the red $(\tau_1)$ and the blue axes $(\tau_2)$ preserves this tree. $G$ acts freely on the tree, and the quotient is precisely $S^1 \vee S^1$. Unfortunately, though, I think the quotient of a $\varepsilon$-neighborhood of the tree by $G$ is not biholomorphic to a neighborhood of $S^1 \vee S^1 \subset \Bbb C$, but rather neighborhood of the $1$-skeleton of a torus (these are not even homeomorphic, let alone biholomorphic domains). But this seemed like an interesting enough failed experiment to include it in the answer.

The proper procedure is as follows: consider the lattice $\Lambda_\tau$ in $\Bbb C$ generated by $1$ and $\tau$. Weierstrass's $\wp$-function defines a map $\Bbb C \to \Bbb{CP}^1$ which can be described as the composition of the covering map $\Bbb C \to \Bbb C/\Lambda_\tau$ with the branched $2$-fold covering $T^2 \to \Bbb{CP}^1$ given by quotienting by the $\Bbb Z_2$-reflection along an axis intersecting the torus at four points. This is because $\wp$ is doubly periodic, so factors through the universal cover, and $\wp(z) = \wp(-z)$, therefore factors through the reflection action. Note that the pole of $\wp$ occurring at the vertices of the periodic parallelogram is sent first to the point at infinity of $T^2$ and then sent to the point at infinity of $\Bbb{CP}^1$. The map $T^2 \to \Bbb{CP}^1$ has four branched points, corresponding to the three zeroes of $\wp'$ in the interior of the period parallelogram (which can be seen explicitly, as $\wp'^2$ is a cubic in $\wp$, giving three zeroes in the interior of the parallelogram) and one at infinity. The three zeroes can be explicitly seen to be $e_1 = \wp(1/2)$, $e_2 = \wp(\tau/2)$ and $e_3 = \wp((1 + \tau)/2)$, the values of $\wp$ at the midpoints of the two edges and the diagonal of the period parallelogram of $\wp$ respectively.

Define $\lambda : \Bbb H^2 \to \Bbb C$ by letting $\lambda(\tau)$ to be the cross-ratio of $e_1, e_2, e_3$ and $\infty$ - the four branch points of $T^2 \to \Bbb{CP}^1$. Explicitly, $\lambda(\tau) = (e_3 - e_2)/(e_1 - e_2)$. Note how the domain of $\lambda$ is the Teichmuller space of the torus, as any $\tau$ with $\text{Im}(\tau)> 0$ determines a complex structure on the torus coming from the quotient $\Bbb C/\Lambda_\tau$. It can be checked that $\lambda$ is invariant under $\tau \mapsto \tau + 2$ and $\tau \mapsto \tau/(1 + 2\tau)$ which correspond to the fractional linear transformation given by the matrices $(1, 2|0, 1)$ and $(1, 0|2, 1)$, which generate the congruence subgroup $\Gamma(2) < \text{SL}_2(\Bbb Z)$, kernel of the map $\text{SL}_2(\Bbb Z) \to \text{SL}_2(\Bbb Z_2)$ given by reducing modulo $2$. As $e_1, e_2, e_3$ are all distinct, $\lambda$ misses the points $0$ and $1$, and elsehwere $\lambda : \Bbb H^2 \to \Bbb C \setminus \{0, 1\}$ defines a holomorphic universal cover. In retrospect, I should have seen this picture as apparently the fundamental domain for the action of $\Gamma(2)$ on $\Bbb H^2$ is as follows. The quotient identifies the two sides, and the two arcs of the domain, so that the quotient is $\Bbb{CP}^1 \setminus \{0, 1, \infty\}$.

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