Simple closed curve in $\mathbb{R}^2\backslash \{x_1,...,x_n\}$ as product of generators of fundamental group Consider $U=\mathbb{R}^2\backslash \{x_1,...,x_n\}$, by Seifert-Van Kampen theorem $\pi_1(U,p)\cong \underbrace{\mathbb{Z}*\mathbb{Z}*...*\mathbb{Z}}_{n-times} \ \forall \ p\in U$ and we can choose as generators some simple closed curves $\gamma_1,...,\gamma_n:[0,1]\to U$ such that, $\forall \ j\in\{1,...,n\}$, the internal region enclosed by $\gamma_j$ intersects the set $\{x_1,...,x_n\}$ only in $x_j$.
I am trying to prove that a simple closed curve $\gamma:[0,1]\to U$ can be written as product of generators in the form $\prod\limits_{i\in I}\gamma_i^{\epsilon_i}$ (for some $I\subseteq \{1,...,n\}$) where $\epsilon_i=1 \ \forall i\in I$ or $\epsilon_i=-1 \ \forall i\in I$.
It looks obvious but I don't see how to proove it rigorously
 A: I believe the main idea is to use the Jordan curve theorem. For any closed curve $\gamma$ let $n_\gamma:\mathbb{R}^2\setminus \gamma([0,1]) \to \mathbb{Z}$ map a point to the winding number of $\gamma$ around the point. For a simple closed curve $\gamma:[0,1] \to \mathbb{R}^2$, the image of this map is either $\{0,1\}$ or $\{-1,0\}$.
Lets focus on the first case. The curve divides the plane in two disjoint path connected sets
$$ \mathbb{R}^2 \setminus \gamma([0,1]) = n_\gamma^{-1}(0) \cup n_\gamma^{-1}( 1)$$
Now let $U_n = \mathbb{R}^2 \setminus \{x_1,\ldots, x_n\}$. Pick your generators in a way, such that $n_{\gamma_i}(x_j) = \delta_{ij}$. The claim is
$$ \gamma = \prod\limits_{i=1}^n \gamma_i^{n_\gamma(x_i)} \in \pi_1(U_n).$$
For $n=1$ we have $\gamma = \gamma_1^a$ for some $a \in \mathbb{Z}$ and I assume we already know $a = n_\gamma(x_1) = 1$.
Without loss of generality assume $n_\gamma(x_i) = 1$ for all $i \in \{ 1,\ldots, n\}$. 
Since $n_\gamma^{-1}(1)$ is path connected, we can choose a path $\alpha:[0,1] \to n_\gamma^{-1}(1)$ connecting all $x_i$ and with $\alpha(\frac{i-1}{n-1}) = x_i$. Consider the inlcusions $ \mathbb{R}^2\setminus \alpha([0,1]) \hookrightarrow U_n \hookrightarrow  \mathbb{R}^2\setminus{x_i}$.
The composition is an isomorphism on $\pi_1$ and we know (from the $n=1$ case) that the second inclusion maps $\gamma$ to  $\gamma_i$. Therefore $\gamma$ represents a generator of $\pi_1(\mathbb{R}^2\setminus \alpha([0,1]))$. Now its left to show, that an explicit representant of this generator is mapped to the desired $\gamma_1\cdot \ldots \cdot \gamma_n$ by the first inclusion.
