Playing with functions and their inverses, I noticed you cannot really obtain the full inverse of any surjective function.
The full inverse of an application $f\left(x\right) = y$ would be the function $g\left(y\right) = x$. Which is just impossible with any surjective applications. For instance, $f\left(x\right) = x^2$ has a partial inverse (in red) which is the square root function $g\left(y\right) = sqrt\left(x\right)$. But we lost all the negative inverse images.
Same for the cardioid defined by $r\left(\theta\right) = 3-3\cos\left(\theta\right)$. We agree that $r\left(\theta\right)$ can take any values of $\theta + k2\pi$ where $k \in \mathbb{Z}$. So the radius will have the same positive value for $r\left(\frac{3}{4}\pi\right)$ and $r\left(\frac{5}{4}\pi\right)$ since the radius $r$ in polar coordinates cannot be negative.
So if we define (in red) the partial inverse $t(r) = \arccos\left(-\frac{r}{3}+1\right)$ and we plot it as we did for the square function, we notice that we get half of our cardioid, as angular images appear all on quadrant II. Again, we lost the angular inverse images on quadrant III.
However, we can restore the values of any surjective application by just combining the inverse function and its opposite. So the injective inverse of $f\left(x\right)$ would be $\{g\left(y\right);-g\left(y\right)\}$, and the injective inverse of $r\left(\theta\right)$ would be $\{t\left(y\right);-t\left(y\right)\}$.
I need advices on my empirical approach, which may be incorrect.