Function from $[0,1]^2$ to $[0,1]$ Consider $(x,y)\in [0,1]^2$. Is it true, that there exists only one $t\in [0,1]$ such that $(x,y)$ belongs to the line passing through $(0,t)\in [0,1]^2$ and $(1,t^2)\in [0,1]^2$? Could you help me to prove it? Is there a way to write in an "explicit" way this function $f$ from $[0,1]^2$ to $[0,1]$?
What about $f^{-1}$? It seems to me that given a $t\in [0,1]$ there is an infinite number of pairs $(x,y)\in [0,1]^2$ such that $(x,y)$ belongs to the line passing through $(0,t)\in [0,1]^2$ and $(1,t^2)\in [0,1]^2$. Is this correct?

The line between $(0, t)$ and $(1, t^2)$ has the equation
$$y = t + (t^2-t)x$$
Given $(x,y)$ we can solve for $t$ and get
$$t = \frac{x-1}{2x} \pm \sqrt{\left(\frac{x-1}{2x}\right)^2 + \frac{y}{x}}
= \frac{(x-1) \pm \sqrt{(x-1)^2+4xy}}{2x}$$
Since $x,y \geq 0$ we have $(x-1)^2 + 4xy \geq (x-1)^2$ so taking the negative sign would make $t \leq 0$.
Thus, given $(x,y)$ there's only one $t$ such that the line between $(0,t)$ and $(0, t^2)$ passes $(x,y)$.
