# Fundamental group of the complement of $k$ points in $\mathbb{R}^2$

Let $$S = \{p_1, \ldots, p_k\}$$ be a set of $$k$$ points in $$\mathbb{R}^2$$ ($$1 \leq k < \infty$$). My goal is to calculate the fundamental group of $$\mathbb{R}^2 \setminus S$$ using van Kampen's theorem.

Clearly, $$\mathbb{R}^2 \setminus \{p\}$$ deformation retracts onto $$S^1$$ for any point $$p \in \mathbb{R}^2$$, and thus $$\pi_1(\mathbb{R}^2 \setminus \{p\})$$ is isomorphic to $$\mathbb{Z}$$.

Since we can decompose $$\mathbb{R}^2 \setminus S$$ into

$$$$\mathbb{R}^2 \setminus \{p_1\},\, \mathbb{R}^2 \setminus \{p_2\},\, \ldots,\, \mathbb{R}^2 \setminus \{p_k\}$$$$

and all intersections between any number of these sets is clearly path connected, we can apply van Kampen's theorem to first obtain a surjection $$\varphi: *_i \pi_1(\mathbb{R}^2 \setminus \{p_i\}) \to \pi_1(\mathbb{R}^2 \setminus S)$$, and then make this surjection into an isomorphism by calculating $$\operatorname{Ker}(\varphi)$$ and forming $$\phi: *_i \pi_1(\mathbb{R}^2 \setminus \{p_i\})\left/\operatorname{Ker}(\varphi)\right. \to \pi_1(\mathbb{R}^2 \setminus S)$$.

This is where I'm stuck. Intuitively, I would expect the fundamental group of $$\mathbb{R}^2 \setminus S$$ to be isomorphic to $$*_{i = 1}^k \mathbb{Z}$$, and this seems to agree with other solutions found online. However, in order for this to be the case, the kernel should be trivial, since we already have $$*_i \pi_1(\mathbb{R}^2 \setminus \{p_i\}) \cong *_{i = 1}^k \mathbb{Z}$$. But according to Theorem 1.20 in Hatcher's "Algebraic Topology", the kernel of $$\varphi$$ is generated by elements of the form $$i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$$ for $$\omega \in \pi_1(\mathbb{R}^2 \setminus \{p_\alpha, p_\beta\})$$, where $$i_{xy}: \pi_1(\mathbb{R}^2 \setminus \{p_x, p_y\}) \to \pi_1(\mathbb{R}^2 \setminus \{p_x\})$$ is the group homomorphism induced via the inclusion $$\mathbb{R}^2 \setminus \{p_x, p_y\} \to \mathbb{R}^2 \setminus \{p_x\}$$.

But if $$\omega$$ is an element of $$\pi_1(\mathbb{R}^2 \setminus \{p_x, p_y\})$$, we can write $$\omega$$ as $$\omega = [\gamma_1][\delta_1]\cdots[\gamma_n][\delta_n]$$, treating $$\pi_1(\mathbb{R}^2 \setminus \{p_x, p_y\})$$ as a subgroup of $$\mathbb{Z} * \mathbb{Z}$$ since we know that $$\pi_1(\mathbb{R}^2 \setminus \{p_x\}) * \pi_1(\mathbb{R}^2 \setminus \{p_y\}) \to \pi_1(\mathbb{R}^2 \setminus \{p_x, p_y\})$$ is surjective. Assuming that $$[\gamma_i]$$ is a homotopy class in $$\pi_1(\mathbb{R}^2 \setminus \{p_x\})$$ and $$[\delta_i]$$ is a homotopy class in $$\pi_1(\mathbb{R}^2 \setminus \{p_y\})$$ for all $$i$$, we have that

\begin{align} i_{xy}(\omega) &= [\gamma_1][\gamma_2]\cdots[\gamma_n]\\ i_{yx}(\omega) &= [\delta_1][\delta_2]\cdots[\delta_n] \end{align}

since any loop around the "hole" $$p_y$$ is homotopic to the constant loop in $$\mathbb{R}^2 \setminus \{p_x\}$$ and vice-versa. But then we have

$$$$i_{xy}(\omega)i_{yx}(\omega)^{-1} = [\gamma_1][\gamma_2]\cdots[\gamma_n][\delta_n]^{-1}[\delta_{n - 1}]^{-1}\cdots[\delta_1]^{-1}$$$$

which is generally not the identity element of $$*_i \pi_1(\mathbb{R}^2 \setminus \{p_i\})$$, and so $$\operatorname{Ker}(\varphi)$$ can't be trivial.

What am I doing wrong here? I suspect that I'm doing the kernel computation incorrectly, but I haven't been able to spot my mistake.

You are working to hard. The problem that the sets $$\mathbb{R}^{2} \backslash\left\{p_{1}\right\}, \mathbb{R}^{2} \backslash\left\{p_{2}\right\}, \ldots, \mathbb{R}^{2} \backslash\left\{p_{k}\right\}$$ are not subsets of $$\mathbb{R}\backslash S$$.
Usually the trick with van kampfen is to choose the sets wisely so the calculations will turn out easy. Let's do the case for $$S=\{p_1,p_2\}$$ and I will let you generalize it.
Let $$p_1=(x_1,y_1)$$ and $$p_2=(x_2,y_2)$$ assume without loss of generality that $$x_1. Define $$\varepsilon = x_2-x_1$$ and write: $$U=\{(x,y) \in \mathbb{R} \backslash S : x Write: $$V=\{(x,y) \in \mathbb{R} \backslash S : x>x_1 + \varepsilon/3 \}$$ Now we get $$U \cap V=\{(x,y) \in \mathbb{R} \backslash S :x_1 + \varepsilon/3 Now $$U \cap V$$ is contractible hence $$\pi(U \cap V)$$ is the trivial group. The kernel of any homomorphism from a trivial group is trivial so the kernel of $$\varphi$$ is trivial(It is generated by trivial elements Theorem 1.20 in Hatcher's "Algebraic Topology"). Its easy to see that $$\pi(V),\pi(U)\cong\pi(S^1)$$ for example by showing that the map $$x \mapsto \frac{x-p_i}{||x-p_i||}$$ is a homotopy equivalence. So we get $$\pi(\mathbb{R}\backslash S)= \mathbb{Z} * \mathbb{Z}$$.