Example 3 (Plane deprived of $n$ points)

The fundamental group of $\mathbf{R}^2$ $\{0\}$ is isomorphic to $\mathbf{Z}$, and the class of the path $t\mapsto e^{2\pi it}$ is a generator of it (IV, p. 347, corollary). More generally, let $n$ be a natural number and let $A =\{z_1, . . . , z_n\}$ be a set of $n$ points of $\mathbf{R}^2$. Let Y be the space $\mathbf{R}^2-$ A; we shall prove that the fundamental group of Y is isomorphic to the free group $F_n$ on $n$ generators. For every $i$, put $z_i= (u_i, v_i)$. Replacing $z_i$ if necessary by $f(z_i)$, where $f:\mathbf{R}^2\rightarrow \mathbf{R}^2$ is a homeomorphism of the form $(u, v)\mapsto (u+\alpha v, v)$, one may suppose that the abscissae of the $z_i$ are pairwise distinct. There is likewise no loss in supposing that one has $u_1<\cdots < u_n$.

Put $V_1= ]-\infty , u_2[\times \mathbf{R},V_i= ]u_{i-1}, u_{i+1}[\times \mathbf{R}$ for $2\leqslant i\leqslant n-1$, $V_n= ]u_{n-1},+\infty [\times \mathbf{R}$. For $1\leqslant i\leqslant n$, the set $U_i= V_i-\{z_i\}$ is open in the plane and homeomorphic to $\mathbf{R}^2-\{0\}$. The family $(U_i)_{1\leqslant i\leqslant n}$ is an open covering of the space $Y =\mathbf{R}^2-$ A. The intersection $U_i\cap U_j$ is empty for $|i-j|\geqslant 2$, homeomorphic to $\mathbf{R}^2$ for $|i-j|= 1$. By Corollary 3, the fundamental group of Y is isomorphic to the free group $F_n$.

Let $a$ be a point of Y. For every integer $i\in \{1, . . . , n\}$, let $r_i$ be a real number strictly positive and strictly less than the distances from $z_i$ to the points $z_j$, for $j=\not i$. Let us denote by $v_i$ the class of the loop $t\mapsto$ $z_i+r_ie^{2\pi it}$ at the point $z_i+r_i$ of Y. Let $\theta_i$ be the class of a path $\gamma_i$ joining the point $a$ to the point $z_i+r_i$ in Y. If the paths $\gamma_i$ are injective and if their images meet only at $a$, one can prove that the unique homomorphism of the free group $F(t_1, . . . , t_n)$ into $\pi_1(Y, a)$ such that $\varphi (t_i) =\theta_iv_i\theta^{-1}_i$ for every $i\in \{1, . . . , n\}$ is a group isomorphism.

An analogous reasoning makes it possible to prove that for every discrete closed subset A of the plane, the fundamental group of $\mathbf{R}^2-$ A is isomorphic to F(A) (IV, p. 463, Exercise 1).