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This is almost the same problem as in this question. However, the OP there was looking for a solution where we could assume any number of things, while I want to stick with just the given assumption (continuity). All the answers there addressed the OP's intent, making certain assumptions that might not be necessary. So none of them apply here.

Here's the question:

A continuous, injective curve $C$ goes from $(-1,-1)$ to $(1,1)$ in the Euclidean plane. Rotating $C$ in either direction by $90$ degrees with respect to the origin, we obtain a new curve $C'$ from $(-1,1)$ to $(1,-1)$. Prove that there exists a point $X$ that lies on both $C$ and $C'$.

In the linked question, the accepted answer by Arthur assumes every straight line through the origin intersects $C$ at exactly one point (of course, except the line $y=x$), and that $C$ goes strictly counterclockwise. Assuming these, he provides a proof. The other two answers are incomplete, each missing a case that they did not discuss.

Here's an approach that seems promising. Look at $C$ as a set of points, which we represent in polar coordinates $(r,\theta)$. There exists a point $P_0=(r_0,\theta_0)$ so that $r(P)\geq r_0$ for all $P\in C$ (by continuity), and similarly we have a upper bound with $P_1=(r_1,\theta_1)$. If we assume that there exists precisely two points in $C$ of distance $r$ to the origin for every $r$ so that $r_0<r<r_1$, then the difference of their $\theta$s is uniquely determined by $r$. Let's call this number $\theta(r)$. Since $\theta(\sqrt2)=\pi>\pi/2$ and $\theta(r_0)=0<\pi/2$, and $\theta$ is continuous, by IVT there is some $r$ so that $\theta(r)=\pi/2$. So we have a point on $C$ which is exactly $90$ degrees away from another point on $C$ of the same distance to the origin. Hence, rotating this point by $90$ degrees, the result still stays on the curve. We are done.

This approach still assumes more than just continuity. In particular, it is not in general true that there are exactly $2$ points of each $r$ on $C$. If we don't assume that, this proof breaks down since $r$ does not determine $\theta$. Any thoughts appreciated!

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  • $\begingroup$ It seems to me that a rotation by any angle will always give an intersection. Is there some counter-example for angles different from $90^\circ$? $\endgroup$ – Jens Feb 11 at 21:26
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    $\begingroup$ It seems to me that it might be simpler to approach the question topologically. Consider the space $S = I^2 \setminus C$. This consists of two connected components. $C'$ is a path (that is, the image of a continuous mapping $I\to S$) that starts at $\langle1,-1\rangle$ and ends at $\langle-1, 1\rangle$. $C'$ is therefore path-connected, and it therefore lies entirely inside one of the two components of $S$. But its endpoints lie in separate components, which is a contradiction. $\endgroup$ – MJD Feb 11 at 21:28
  • $\begingroup$ @Jens yes, if you rotate unit half-circle by 180 degrees, the only common points will be $(1, 1)$ and $(-1, -1)$, so depending on how intersection is defined, it might be false. $\endgroup$ – enedil Feb 11 at 21:30
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    $\begingroup$ BTW, that claim about injective curves within curve images? I was right -- it is tricky. See Lee Mosher's answer here: math.stackexchange.com/questions/857066/… $\endgroup$ – John Hughes Feb 12 at 2:53
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    $\begingroup$ @Jens I've posted a counterexample here. I still think the proposition is true for angles not greater than $90^\circ$ in either direction - but I must admit that my judgement on this question has repeatedly been proved to be terrible! $\endgroup$ – Calum Gilhooley Feb 23 at 1:41
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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.

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  • $\begingroup$ How do you know such a $t_{\mathrm{min}}$ and $t_{\mathrm{max}}$ are unique? For example, what if our curve is one half of the unit circle? $\endgroup$ – YiFan Feb 11 at 22:40
  • $\begingroup$ I believe he merely says that there are numbers $t_{min/max}$ at which it attains these values, not that they are unique. $\endgroup$ – John Hughes Feb 11 at 23:39
  • $\begingroup$ @YiFan I did't say they were unique and they need not be $\endgroup$ – Hagen von Eitzen Feb 12 at 0:15
  • $\begingroup$ A diagram would be very helpful. $\endgroup$ – Jens Feb 12 at 21:49
  • $\begingroup$ @Botond Thank you, but I was thinking of a diagram with annotations relevant to the argument, i.e. a diagram showing points, paths, distances, axes, etc which are mentioned in the argument. At present, the argument here remains opaque to me. $\endgroup$ – Jens Feb 12 at 22:58
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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, $[\gamma] \cap [i\gamma] \ne \emptyset$.

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Not a solution, but the broad strokes of one

I suspect that by compactifying to $S^2$, and then adding an arc through the north pole and down to two points on the curve, one can show the complement has multiple components via Alexander duality, which would then do the job. I also suspect that this is not the proof that OP wants, so I'm not working out the details.

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