Let $S\subseteq\Bbb R$. Assume a function $f: S\to S$ has these properties:
With a $90^\circ$ counter-clockwise rotation of its graph around the origin, we get the same graph $\Gamma_f$ again. Prove that the function must be bijective.
I've seen the parametric formula:$$x'=x\cos\theta-y\sin\theta$$$$y'=x\sin\theta+y\cos\theta$$In my case:$$\theta=\frac{\pi}{2}$$
My first attempt was to find the properties of a function that satisfies this condition:$f(x)=y\in\Gamma_f \wedge f(-y)=x\in\Gamma_f,$ but it didn't help me. I'm not sure if I understand what "the same" means in this task.
$\endgroup$ 21 Answer
$\begingroup$Alert: just a long comment with the purpose of giving a visual insight
According to the post itself and what @conditionalMethod has already stated, let $r_O:\Bbb R^2\to\Bbb R^2$ be our rotation for $90^\circ$ around the origin $O(0,0)$, then: $\color{brown}{r_O:(x,f(x))\mapsto (-f(x),x))}$.
Here is a concrete example of such a function:
$$f(x)=\begin{cases}x+1,&x\in(-2,-1)\\-x+1,&x\in(-1,0)\\-x-1,&x\in(0,1)\\x+1,&x\in(1,2)\end{cases}$$so, e.g.$$S=(-2,2)\setminus\{-1,0,1\}$$I also high-lighted a squre with edges of length $a=\sqrt{2}$ which I found interesting. $\Gamma_f$ is black in the picture:
Another example is the function $$\begin{cases}x=\cos\left(\alpha+k\frac\pi2\right)\\y=\sin\left(\alpha+k\frac\pi2\right)\end{cases},\quad\alpha\in\left(\arctan a,\frac\pi2\right),k\in\{1,2,3,4\},$$
whose graph consists of the four red arcs on the unit circle:
