TITLE: Existence of a simple closed curve which is not null-homotopic
QUESTION [8 upvotes]: Problem. Assume that $U$ is an open and connected subset of $\mathbb R^2$, and $\gamma :[0,1]\to U$ is a closed curve, which is not null-homotopic in $U$ and not necessarily simple closed. Show that there exists a simple closed curve, which is not null-homotopic in $U$.
I am sure that this is a standard problem, and I am looking for an elegant, if possible, solution or a reference. I have in mind the following sketch of proof:
a. Construct a polygonal closed curve $\tilde\gamma$ which is homotopic with $\gamma$ in $U$.
b. Construct $\tilde\gamma$ so that it intersects itself in finitely many points, and not whole segments.
c. Use induction on the number of self–intersections of $\tilde\gamma$, in order to show that $\tilde\gamma$ is written as a product (in the fundamental group's sense) of simple closed polygonal curves.

REPLY [2 votes]: The answer of D. Fischer is very elegant and of course correct. I am providing a modification of a part of his answer which:
a. Avoids the use of the Uniformisation Theorem of Riemann Surfaces, and
b. It applies even in the case $U$ is a topological manifold of any dimension.
Assume that $U$ is a connected but topological manifold, which is not simply connected, and hence there exists a non null-homotopic closed curve $\gamma: [0,1]\to U$. Let $X$ be the universal covering space of $U$, and $\pi: X\to U$ the corresponding covering map. (Such $X$ and $\pi$ do exist, as $U$ is a connected manifold.) Let $\tilde\gamma :[0,1]\to X$ be a lifting of $\gamma$, i.e., $\pi\circ\tilde\gamma=\gamma$. We shall show that there exists a curve $\hat\gamma : [0,1]\to X$, homotopic to $\tilde\gamma$, which is injective. Once we achieve that, the rest is obtained by D. Fischer's proof.
In fact, it suffices to show the following: 
Fact. If $X$ is a connected topological manifold, and $x,y\in X$, $x\ne y$, then there exists an injective curve $\gamma : [0,1]\to X$, connecting $x$ and $y$.
Proof of the Fact. Fix $x_0\in X$. It suffices to show that the set
$$
W=\{x\in X: x\ne x_0\,\,\,\text{and $x_0$ is connected to $x$ by an injective curve}\}\cup\{x_0\},
$$
is both open and closed. 
a. $W$ is open: Let $x_1\in W$, and $V$ an open neighbourhood of $x_1$ which is homeomorphic to the unit ball $B_1(0)$, with $p: V\to B_1(0)$ a homeomorphism, $p(x_1)=0$, and $x_2\in V$. If $x_2$ belongs to the curve $\gamma : [0,1]\to X$, which connects $x_0$ and $x_1$ we are done. Assume that $x_2$ does not belong to the curve, and let $r=\|p(x_2)\|<1$.  Let
$$
d=\mathrm{dist}\big(p(x_2),\overline{B}(0,r)\cap p(\gamma[I])\big)>0.
$$
Let $y_3\in \overline{B}(0,r)$, such that $\|y_3-p(x_2)\|=d$, (such $y_3$ exists since $\overline{B}(0,r)\cap p(\gamma[I])$ is compact)
and $J$ the segment connecting $y_3$ and $p(x_2)$. Then the union of $p^{-1}(J)$ and the part of $\gamma$ which connects $x_0$ with $x_2$ is an injective curve connecting $x_0$ and $x_1$.
b. $W$ is closed: Using a similar approach to the one we just used, it is readily obtained that a limit point of $W$ also belong to $W$.