Let $c\in[0,1]^2$ and $f_c: [0,1]^2\to[0,1]$ be defined as follows:
Given $u\in[0,1]^2\neq c$, let $d$ be the positive scalar such that $c+d(u-c)$ lies on the border of $[0,1]^2$. Then define $f(u):=(1/d)^2$ and $f(c):=0$. So the function on a single line from $c$ to the border of $[0,1]^2$ is just $x^2$ scaled down so it runs from $0$ to $1$.
My question: does $f$ transform $[0,1]^2$-uniformly distributed random variables into $[0,1]$-uniformly distributed random variables? In other words, do we have $\lambda(f_c^{-1}([a,b]))=b-a?$
My intuition is yes, I imagine $[0,1]^2$ to be sliced in thin triangles starting in $c$. So on each triangle, the thickness linearly scales with distance from $c$, so I want the derivative of my density function along the triangle to equal a multiple of $1/x$ and also go from $0$ to $1$. The square term seems to satisfy this.
Is this thinking correct?
Yes, this is correct.
Define $l(\theta)$ to be the length of the line between $c$ and the border of $[0, 1]^2$, at an angle $\theta \in [0, 2\pi]$. If we make a transformation to scaled polar co-ordinates about $c$,
$$x = c + r l(\theta) \cos(\theta),\qquad y = c + r l(\theta) \sin(\theta),$$
then the random variables $R \in [0, 1]$ and $\Theta \in [0, 2\pi]$ have joint density $$f(r, \theta) = f(x, y)\ \bigg\lvert\frac{d(x,y)}{d(r,\theta)}\bigg\rvert = rl(\theta)^2,$$ so $R \sim \text{Beta(2, 1)}$ independent of $\Theta$, with density $f(r) = 2r$.
Then letting $s = r^2 = 1/d^2$, the density of $S \in [0,1]$ is $$f(s) = f(r)\ \bigg\lvert\frac{dr}{ds}\bigg\rvert = 1,$$ as required.