A function whose inverse relation is not a function

Question

I need to come up with a function $f:X \to X$ where $X=\{1,2,3,4,5\}$ whose inverse relation is not the function $X \to X$. I’ve tried with for example $f(x)=|x|$, but this clearly doesn’t work.

Answer 1

Here's an easy example$$f(x)=1+|3-x|$$

Answer 2

You can simply take an ambiguous expression, such as the ln, and then invert that into a unique one, which is then your function.

e.g.:

$$ \begin{align*} k &\in \mathbb{Z}\\ f&: \left\{1, ~2, ~3, ~4, ~5 \right\} \to \left\{1, ~2, ~3, ~4, ~5 \right\}\\ f(x) &= e^{2 \cdot k \cdot x \cdot \pi \cdot \mathrm{i}} = e^{2 \cdot x \cdot \pi \cdot \mathrm{i}} = \cos(2 \cdot k \cdot x \cdot \pi) + \sin(2 \cdot k \cdot x \cdot \pi) \cdot \mathrm{i} = 1\\ f^{-1}(x) &= \frac{\ln(x)}{2 \cdot \pi \cdot \mathrm{i}} = \frac{2 \cdot k \cdot x \cdot \pi \cdot \mathrm{i}}{2 \cdot \pi \cdot \mathrm{i}} = k \cdot x \\ \\ f(x) &= 1\\ f^{-x}(f(x)) &= k \cdot x \end{align*} $$

You can also just look up an inverse of a function that isn't a function such as arctan2, where the inverse of the inverse of the function is the function itself.

e.g.

$$ \begin{align*} k &\in \mathbb{Z}\\ f&: \left\{1, ~2, ~3, ~4, ~5 \right\} \to \left\{1, ~2, ~3, ~4, ~5 \right\}\\ f(x) &= \tan(\pi \cdot x + \frac{\pi}{4}) = 1\\ f^{-1}(x) &= \frac{\arctan2(x, x)}{\pi} - \frac{1}{4} + 1\\ \\ f(x) &= 1\\ f^{-x}(f(x)) &= 1 + k\\ \end{align*} $$