Is there a continuous bijection between an interval $[0,1]$ and a square: $[0,1] \times [0,1]$? Is there a continuous bijection from $[0,1]$ onto $[0,1] \times [0,1]$?
That is with $I=[0,1]$ and $S=[0,1] \times [0,1]$, is there a continuous bijection
$$
f: I \to S?
$$
I know there is a continuous bijection $g:C \to I$ from the Cantor set $C$ to $[0,1]$.
The square $S$ is compact so there is a continuous function 
$$
h: C \to S.
$$
But this leads nowhere.
Is there a way to construct such an $f$?
I ask because I have a continuous functional $F:S \to \mathbb R$.
For numerical reason, I would like to convert it into the functional
$$
G: I \to \mathbb R, \\
G = F \circ f ,
$$
so that $G$ is continuous.

No, such a bijection from the unit interval $I$ to the unit square $S$ cannot exist. Since $I$ is compact and $S$ is Hausdorff, a continuous bijection would be a homeomorphism (see here). But in $I$ there are only two non-cut-points, whereas in $S$ each point is a non-cut-point.
