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?
 A: If $(x,y)$ is on the line passing through $(0,t)$ and $(1,t^2)$ then 
$$y=t+x(t^2-t)=xt^2+(1-x)t,$$
so by the quadratic formula 
$$t=\frac{x-1}{2x}\pm\frac{1}{2x}\sqrt{(x-1)^2+4xy}.$$
Because $x$ and $y$ are nonnegative, the negative sign is impossible so there is at most one such $t$.
A: The line connecting $(0,t)$ and $(1, t^2)$ consists of all points $(x,y)$ satisfying $y = (t^2 - t)x + t$.  Consider the function
$$
f(t,x) = (t^2 - t) x + t
$$
We have $f(0,x) = 0$ for all $x$ and $f(1,x) = 1$ for all $x$.  Finally, we have 
$$
\left.\frac{\partial f}{\partial t}\right|_x = (2t - 1)x + 1. 
$$
But we also have 
$$
(2t - 1) \geq - 1 \geq -\frac{1}{x} \quad \Rightarrow \quad (2t - 1)x \geq -1
$$ for all values of $x\in [0,1]$ and $t\in [0,1]$.  This implies that 
$$
\left.\frac{\partial f}{\partial t}\right|_x \geq 0,
$$
i.e., we therefore have that $f(t,x)$ is monotonic in $t$.  Since we also have that $f(0,x) = 0$ and $f(1,x) = 1$ for all $x$, we conclude that for all $(x,y) \in [0,1]^2$, there exists a unique $t$ such that $f(t,x) = y$.
A: 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 $(x,y)$.
