Existence of a simple closed curve which is not null-homotopic

Question

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.

Answer 1

Let $\pi \colon \mathbb{R}^2 \to U$ be the (a) universal covering. We know that the covering space is homeomorphic to $\mathbb{R}^2$ by the uniformisation theorem.

Let $\hat{\gamma} \colon [0,1] \to \mathbb{R}^2$ be a lift of $\gamma$. Since $\gamma$ is not null-homotopic, we have $\hat{\gamma}(1) \neq \hat{\gamma}(0)$. Define $\beta$ as the straight line segment connecting these points, $\beta(t) = (1-t)\hat{\gamma}(0) + t\hat{\gamma}(1)$. Next define

$$s = \inf \bigl\{ t \in (0,1] : \bigl(\exists u \in [0,t)\bigr)\bigl(\pi(\beta(t)) = \pi(\beta(u))\bigr)\bigr\}.$$

We claim that there is an $u \in [0,s)$ with $\pi(\beta(s)) = \pi(\beta(u))$: Choose a neighbourhood $V$ of $\beta(s)$ on which $\pi$ is injective. There is an $\varepsilon > 0$ such that $\beta(t) \in V$ for all $t \in [s-\varepsilon,s+\varepsilon]$. We have sequences $t_n$ and $u_n$ with $\pi(\beta(t_n)) = \pi(\beta(u_n))$ where $t_n \geqslant t_{n+1} \geqslant s$ and $u_n < t_n$. Without loss of generality, we can assume $t_n < s + \varepsilon$ for all $n$. Since $\pi$ is injective on $V$, we have $u_n < s - \varepsilon$ for all $n$. After taking a subsequence, we can assume that $u_n$ converges to some $u \leqslant s - \varepsilon$. By continuity,

$$\pi(\beta(s)) = \lim_{n\to\infty} \pi(\beta(t_n)) = \lim_{n\to \infty} \pi(\beta(u_n)) = \pi(\beta(u)).$$

Now define $\tilde{\beta}(t) = \beta((1-t)u + ts)$ and $\tilde{\gamma} = \pi \circ \tilde{\beta}$. By construction, $\tilde{\gamma}$ is a simple closed curve in $U$, and since $\beta(u) \neq \beta(s)$, it is not null-homotopic.

Answer 2

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 $W$.