Prove that a path that rotates twice around an annulus intersects itself

Let $\gamma:S^1\rightarrow S^1\times[0,1]\ $ be a closed continuous path on an annulus, with endpoints on a point in the boundary. Assume that $\gamma\ $ circles the annulus twice (For a precise definition we can integrate over the angle or use degree of a map).

How do you show that $\gamma\ $ is self intersecting (not counting the endpoints intersection obviously)?

I tried lifting $\gamma$ to a path on $\mathbb{R}\times[0,1]$, and using the intermediate value theorem somehow, but it didn't work out. Then I thought of looking at two lifts of $\gamma$ that start at a "distance of one rotation" from each other, and showing that they intersect. It would've been straight forward if I knew that a path splits the space to two connected components, but for that I believe that I need something like Jordan curve theorem which feels like a complete overkill.

Is there a nice proof of this claim?

Extras:

1. If there are easier solutions for smooth paths, I'm also interested in seeing them.
2. I'd like to see a similar proof (or a contradiction) for a path on a Torus, that rotates twice in one directions and 0 times in the second direction.

Solution 1:

Your assumption that "$\gamma$ circles the annulus twice" can be restated as the condition that $\gamma$ represents the square of a generator of $\pi_1(A)$, where $A$ is the annulus $S^1\times [0,1]$.

For some reason, you are assuming that the end-points of $\gamma$ are on the boundary of the annulus. By embedding the annulus $A=S^1\times [0,1]$ in $A'=S^1\times [-1,2]$, we achieve that the same loop $\gamma$ is disjoint from $\partial A$ and still represents the square of a generator of $\pi_1(A')$. Thus, I will be proving:

Lemma. Let $A$ be an annulus and $\gamma$ is a simple loop in $A$ disjoint from $\partial A$. Then $\gamma$ cannot represent the the square of a generator of $\pi_1(A)$.

Proof. Suppose that the image $c$ of $\gamma$ (a simple loop) does represent the square of the generator of $\pi_1(A)$.

Let $p: \tilde{A}= {\mathbb R}\times [0,1]\to A$ denote the universal covering and let $G$ denote its group of covering transformations. Set $L:=p^{-1}(c)$. Since $\gamma$ represents the square of $\alpha$, the preimage $L$ consists of exactly two connected components $L_0, L_1$, each of which is homeomorphic to ${\mathbb R}$ (the universal cover of $c$) and the generator $g$ of $G$ satisfies: $$ g: L_0\to L_1, g: L_1\to L_0. $$ Moreover, the topological line $L_0$ is invariant under the index 2 subgroup $G_0< G$. From this, it follows that $L_0$ is properly embedded in ${\mathbb R}\times [0,1]\subset {\mathbb R}^2$, i.e. every sequence in $L_0$ either contains a convergent subsequence (whose limit is in $L_0$ because it is closed in $\tilde{A}$) or diverges to $\infty$. Thus, compactifying $L_0$ in the Riemann sphere ${\mathbb R}^2\cup \{\infty\}=S^2$, we obtain a Journal curve $J\subset S^2$, a simple topological loop. Now, I use the Jordan curve theorem: Yes, this is necessary for the proof.

It follows that $L_0$ separates $\tilde{A}$ in two components $L_0^\pm$. The component $L_0^-$ contains the line $\{0\}\times {\mathbb R}$ and the other component $L_0^+$ contains the line $\{1\}\times {\mathbb R}$. The same applies to the topological line $L_1$. Since the topological lines $L_0, L_1$ are disjoint, $L_1$ is contained in one of the components $L_0^\pm$. Let's assume that it is contained in $L_0^+$ (the other case is handled by switching the roles of $L_0, L_1$).

Note that since $A$ is an annulus (rather than the Moebius band!), $G$ preserves each boundary line of the strip $\tilde{A}$. Hence, the condition that $L_1\subset L_0^+$ implies $$ g(L_0^+)=L_1^+ \subset L_0^+, $$ and $g(L_0)=L_1\subset L_0^+$. Applying $g$ again, we also obtain $$ g^2(L_0)=g(L_1)\subset g(L_0^+)\subset L_0^+. $$
Thus, we cannot possibly have the equality $g^2(L_0)=L_0$. A contradiction. qed

A version of this proof works in greater generality and one obtains:

Proposition. Suppose that $S$ is a connected orientable surface and $c$ is a simple loop in $S$ representing an element $h\in \pi_1(S)\setminus \{1\}$. Then there is no $n\ge 2$ and $g\in \pi_1(S)$ such that $h=g^n$.

Note that this proposition is false for non-orientable surfaces, the Moebius band is the simplest example: You take $g$ represented by the "middle loop" of the Moebius band. Then $g^2$ is represented by a simple loop. But, essentially, this is the only counter-example. All other examples come from squares of central loops in Moebius bands in non-orientable surfaces.

Solution 2:

We’ll assume WLOG that $\gamma$ lies in the interior of the annulus, since we can always contract along $[0,1]$ without changing self intersections. Now, identify $S^1\times (0,1)$ with $B^2-\{0\}$. Then WLOG, $\gamma$ has winding number $2$ about the origin $0$. Let $$r=\min_{s\in S^1}|\gamma(s)|$$ and let $c$ be the circular curve of radius $r$ about the origin with winding number $-1$. Let $s_0$ be such that $\gamma(s_0)=r$. Lastly, let $g=\gamma+c$, where addition is done with $\gamma (s_0)$ as the basepoint. Then $g$ has winding number $1$ about $0$. Let’s assume $\gamma$ is simple. Then $\gamma$ is simple and $g$ is “almost simple” in that it still encloses an interior region, and is homotopic to a jordan curve that coincides with $g$ in all but an $\varepsilon$-neighborhood of $\gamma(s_0)$ for any $\varepsilon>0$. Then $0$ is not in the region bounded by $g$, so $g$ has winding number $0$ about $0$. This is a contradiction, so we conclude that $\gamma$ isn’t simple. Clearly, this analysis came be done where $\gamma$ has winding number $\geq 2$ about $0$.