I understand the mapping,
$H:[0,1] \rightarrow [0,1] \times [0,1]$,
through a Hilbert curve $H$ is not one-one and therefore $H$ in general may not be invertible. However, here is a nice illustration of Hilbert curve mapping in both directions and the author has in fact provided an animation in which we have at least one point in $[0,1]$ mapped if we click on a coordinate $(x,y)$ in the Hilbert curve animation.
From the authors code I am able to generate the $(x,y)$ coordinate for a given point in $[0,1]$ by converting into a quadit. However, given a $(x,y)$ how do I recover at least one point in $[0,1]$ that maps to that coordinate, just like how the author has done in the animation?
Thanks!
As you said, the Hilbert curve is not "one to one"
That means you CANNOT find the number in $[0,1]$ such that the image of this number is your point.
However, given $(x,y) \in [0,1]^2$, you can find $t \in [0,1]$ such that the distance between $(x,y)$ and $H(t)$ is as small as you want, but still non null.
For this approximation, you can look at the wiki page: https://en.wikipedia.org/wiki/Hilbert_curve#Applications_and_mapping_algorithms
You can use the function ˋxy2dˋ with arbitrary precision and then remap the range $[0, 2^{2n}-1]$ to $[0,1]$
EDIT
I was totally wrong !
It is true that a space filling curve cannot be one to one (see here)
But as said in the wiki page, the Hilbert curve is surjective.
That means that for a point in the square, you will find at least one antecedent, and possibly many.
If you create an algorithm, it will compute one such antecedent
As an exemple, $H(1/12)=H(11/12)$ if I'm not mistaken