A continuous curve intersects its 90 degrees rotated copy?

Some notation: For a curve (map) $\phi\colon[0,1]\to \Bbb C$, define the curve (set) $[\phi]:=\phi([a,b])$. Also, if $\phi(0)=0$, define $\phi^\pm$ as the concatenation of the given curve with its reversed negative: $$\begin{align}\phi^\pm\colon[a-b,b-a]&\to\Bbb C \\ t&\mapsto\begin{cases}-\phi(a-t)&t\le 0\\\phi(t+a)&t\ge0\end{cases}\end{align}$$

Now let $\gamma\colon[0,1]\to\Bbb C$ be our curve with $\gamma(1)=-\gamma(0)$. We can concatenate $\gamma$ with $-\gamma$ to a closed curve $\tilde\gamma$.

Fig 1. From $\gamma$ to $\tilde\gamma$.

Assume we can find a point $z_0\in [\tilde \gamma]\cap i[\tilde\gamma]$. Then $z_0$ is in one of $\pm[\gamma]$ as well as in one of $\pm i[\gamma]$. Therefore, one of the points $z_0, iz_0,-z_0,-iz_0$ is $\in [\gamma] \cap i[\gamma]$, as desired. Thus our goal is to find such an intersection $z_0$ of $[\tilde \gamma]\cap i[\tilde\gamma]$.

As the map $r\colon [0,1]\to \Bbb R$, $ t\mapsto |\gamma(t)|$ is continuous with compact domain, there exist $t_\min, t_\max\in[0,1]$ where it attains its minimum $r_\min$ and its maximum $r_\max$, respectively. If $i\gamma(t_\min)\in[\tilde\gamma]$, we can let $z_0=i\gamma(t_\min)$ and are done. Hence we assume from now on that $i\gamma(t_\min)\notin [\tilde\gamma]$. In particular, $r_\min >0$. Similarly, we may assume that $i\gamma(r_\max)\notin [\tilde\gamma]$.

Fix $r>r_\max$ and let $D$ be the open disk around $0$ of radius $r$. We say that a path $\eta\colon[a,b]\to\Bbb C$ escapes from $z$, if $\eta(a)=z$, $|\eta(b)|=r$, and $[\eta]\cap[\tilde\gamma]=\emptyset$.

Assume there exists a path $\eta\colon[0,1]\to\Bbb C$ that escapes from $0$. Because the complement of $[\tilde\gamma]$ is open, we can adjust $\eta$ locally to our liking without changing the end points or the disjointness to $[\tilde \gamma]$; therefore, we may assume that $\eta$ is a polyline (of finitely many segments). Now let $\zeta\colon[0,\ell]\to \Bbb C$ be a shortest path parametrized by length (so of length $\ell$) among all those with $\zeta(0)=0$, $|\zeta(\ell)|=r$, and $[\zeta]\subseteq [\eta^\pm]$. This is possible because $[\eta^\pm]$ is a simple graph with straight edges.

Fig 2. From a) general curve to b) polyline $\eta$, to c) shortest $\zeta$, to d) simple $\zeta^\pm$.

Then $\zeta^\pm$ is a point-symmetric path with end points on $\partial D$ and otherwise living in $D$. Also, $\zeta^\pm$ is simple: Any self-intersection comes from $t_1\ne t_2$ (wlog. $t_1<t_2$) with $\zeta(t_1)=\pm\zeta(t_2)$. But then the concatenation of $\zeta|_{[0,t_1]}$ and $\pm\zeta|_{[t_2,\ell]}$ would be shorter than $\zeta$, contradiction. The end points of $\zeta^\pm$ split $\partial D$ into two semicircle arcs. Together with either of these arcs, $\zeta^\pm$ forms a simple closed curve - a nice and friendly Jordan curve. Moreover, the interior regions of these two Jordan curves are disjoint, point symmetric to each other, and their union is $D\setminus [\zeta^\pm]$. By the point symmetry, $\gamma(0)$ and $\gamma(1)=-\gamma(0)$ are not in the same Jordan curve interior. It follows that $[\gamma]$ intersects $[\zeta^\pm]$, which is absurd.

We conclude that no path escapes from $0$.

Assume there is a path that escapes from $i\gamma(t_\min)$. Then together with the line segment from $0$ to $i\gamma(t_\min)$, we'd obtain a path that escapes from $0$, which we know does not exist. On the other hand, a line segment radially outward from $i\gamma(t_\max)$ escapes (here we use that $i\gamma(t_\max)\notin[\tilde\gamma]$). We conclude that any path from $i\gamma(t_\min)$ to $i\gamma(t_\max)$ intersects $\tilde \gamma$. In particular, this holds for the path between these points along $i\gamma$, thereby giving us an intersection point $z_0$, as desired.

Here is a quick and easy proof - thanks in good part to a helpful hint given by Moishe Cohen, in a comment on my recent question cited below. (I repeat here the quotation included in that question, to make this answer self-contained, although unfortunately this makes the proof appear longer than it is!)

Let $[a, b]$ be a compact interval in $\mathbb{R}$. For any continuous function $\gamma \colon [a, b] \to \mathbb{C}$, denote the compact, connected set of points $\gamma([a, b])$ by $[\gamma]$, and define: \begin{align*} i\gamma \colon [a, b] \to \mathbb{C}, \ & t \mapsto i(\gamma(t)), \\ -\gamma \colon [a, b] \to \mathbb{C}, \ & t \mapsto -(\gamma(t)), \\ -i\gamma \colon [a, b] \to \mathbb{C},\ & t \mapsto -i(\gamma(t)). \end{align*} If $0 \in [\gamma]$, then also $0 \in i[\gamma] = [i\gamma]$, so the curves $\gamma, i\gamma$ intersect. We assume from now on that $0 \notin [\gamma]$.

By Theorem 7.2.1 of A. F. Beardon, Complex Analysis (Wiley, Chichester 1979), there exists a branch of$\operatorname{Arg}\gamma$ on $[a, b]$, i.e. a continuous function $\theta \colon [a, b] \to \mathbb{R}$ such that: $$ \gamma(t) = r(t)e^{i\theta(t)} \quad (a \leqslant t \leqslant b). $$ The functions $\theta + \frac{\pi}{2}$, $\theta + \pi$, $\theta - \frac{\pi}{2}$ are branches of $\operatorname{Arg}(i\gamma)$, $\operatorname{Arg}(-\gamma)$, $\operatorname{Arg}(-i\gamma)$, respectively, on $[a, b]$.

Quoting from the same book:

Definition 7.2.1 Let $\gamma \colon [a, b] \to \mathbb{C}$ be any curve and suppose that $w \notin [\gamma]$. We define the index $n(\gamma, w)$ of $\gamma$ about $w$ by $$ n(\gamma, w) = \frac{\theta(b) - \theta(a)}{2\pi}, $$ where $\theta$ is any branch of $\operatorname{Arg}(\gamma - w)$ on $[a, b]$. If $\gamma$ is closed then $n(\gamma, w)$ is an integer.

The index $n(\gamma, w)$ is sometimes called the winding number of $\gamma$ about $w$, for it represents the number of times that a point $z$ moves around $w$ as it moves from $\gamma(a)$ to $\gamma(b)$ along $\gamma$. [...]

The index can be used to clarify the difficult question of what is meant by the 'inside' and 'outside' of a closed curve $\gamma$. We shall say

(a) that $z$ is inside $\gamma$ if $z \notin [\gamma]$ and $n(\gamma, z) \ne 0$,

(b) that $z$ is on $\gamma$ if $z \in [\gamma]$, and

(c) that $z$ is outside $\gamma$ if $z \notin [\gamma]$ and $n(\gamma, z) = 0$.

[...] Observe that [...] the outside of $\gamma$, say $O(\gamma)$, is the union of those components of $\mathbb{C} \setminus [\gamma]$ on which the index is zero. Thus $O(\gamma)$ is an open set. Further [...] $O(\gamma)$ contains the complement of some closed disc. If we denote the inside of $\gamma$ by $I(\gamma)$, then $$ \mathbb{C} \setminus O(\gamma) = [\gamma] \cup I(\gamma), $$ and so the set of points which lie inside or on $\gamma$ is a compact set.

Quoting now from D. J. H. Garling, A Course in Mathematical Analysis, vol. III (Cambridge 2014) - the slight clash of notation should cause no confusion -

If $\gamma \colon [a, b] \to X$ and $\delta \colon [c, d] \to X$ are [curves], and $\gamma(b) = \delta(c)$, the juxtaposition $\gamma \vee \delta$ is the [curve] from $[a, b + (d - c)]$ into $X$ defined by $(\gamma \vee \delta)(x) = \gamma(x)$ for $x \in [a, b]$ and $(\gamma \vee \delta)(x) = \delta(x + (c - b))$ for $x \in [b, b + (d - c)]$. [...]

Suppose that $\gamma \colon [a, b] \to X$ is a [curve], and that $w \notin [\gamma]$. [...] If $\gamma = \alpha \vee \beta$ is the juxtaposition of two [curves] then $$ n(\gamma, w) = n(\alpha, w) + n(\beta, w). $$

Using the hypothesis that $\gamma(b) = -\gamma(a)$, we form the juxtapositions \begin{gather*} \sigma = \gamma \vee (-\gamma) \colon [a, 2b - a] \to \mathbb{C}, \\ \tau = (i\gamma) \vee (-i\gamma) \colon [a, 2b - a] \to \mathbb{C}, \end{gather*} and observe that, by the same hypothesis, these are closed curves. Clearly $\tau = i\sigma$, i.e. $\tau(t) = i(\sigma(t))$ ($a \leqslant t \leqslant 2b-a$). Using the hypothesis $\gamma(b) = -\gamma(a)$ for a third time, we have: $$ n(\gamma, 0) = n(-\gamma, 0) = n(i\gamma, 0) = n(-i\gamma, 0) = m + \tfrac{1}{2}, \text{ for some } m \in \mathbb{Z}, $$ and consequently: $$ n(\sigma, 0) = n(\tau, 0) = 2m + 1. $$ All we need retain from this are the implications $n(\sigma, 0) \ne 0$, $n(\tau, 0) \ne 0$, i.e., \begin{equation} \label{3109299:eq:1}\tag{1} 0 \in I(\sigma) \cap I(\tau). \end{equation}

Define $r(t) = |\gamma(t)| = |i\gamma(t)|$ ($a \leqslant t \leqslant b$). Being a continuous function on a compact set, $r$ attains a maximum value, $r(t_0)$. Extending $r$ continuously to the interval $[a, 2b-a]$, by writing $r(t) = r(t + a - b)$ ($b \leqslant t \leqslant 2b -a$), we see that $r(t_0)$ is also the maximum value of $|\sigma(t)| = |\tau(t)|$ ($a \leqslant t \leqslant 2b-a$).

Suppose that $[\sigma], [\tau]$ are disjoint. A straight line segment connects the point $\sigma(t_0) = \gamma(t_0)$ to the point $2\gamma(t_0)$, and it lies entirely in the complement $\mathbb{C} \setminus [\tau]$. As the second point clearly lies in the unbounded component of $\mathbb{C} \setminus [\tau]$, so must the first; therefore, the whole of the connected set $[\sigma]$ lies in the unbounded component of $\mathbb{C} \setminus [\tau]$. Similarly, $[\tau]$ lies in the unbounded component of $\mathbb{C} \setminus [\sigma]$. All we need retain from this are the implications: \begin{equation} \label{3109299:eq:2}\tag{2} [\sigma] \subset O(\tau) \text{ and } [\tau] \subset O(\sigma). \end{equation}

That \eqref{3109299:eq:1} and \eqref{3109299:eq:2} stand in contradiction to one another (for any closed plane curves $\sigma, \tau$, and with an arbitrary point of the plane in place of $0$) was precisely the content of the conjecture in my question yesterday. I can now proudly report that my weedy little conjecture has grown up into a beefy big theorem! Thus, the supposition that $[\sigma], [\tau]$ are disjoint must be false.

For $S \subseteq \mathbb{C}$, let $-S, iS, -iS$ denote the sets $\{-s : s \in S\}$, $\{is : s \in S\}$, $\{-is : s \in S\}$, respectively. We have just established that the sets \begin{gather*} [\sigma] = [\gamma] \cup [-\gamma] = [\gamma] \cup -[\gamma], \\ [\tau] = [i\gamma] \cup [-i\gamma] = i[\gamma] \cup -i[\gamma] \end{gather*} intersect. Let $z$ be a point common to both. If $z \in [\gamma] \cap [i\gamma]$, we are done right away. If $z \in [-\gamma] \cap [-i\gamma]$, then $-z \in [\gamma] \cap [i\gamma]$. If $z \in [\gamma] \cap [-i\gamma]$, then $iz \in [\gamma] \cap [i\gamma]$. If $z \in [-\gamma] \cap [i\gamma]$, then $-iz \in [\gamma] \cap [i\gamma]$. In all cases, $[\gamma] \cap [i\gamma] \ne \emptyset$.