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.

As the other answers state, there is no bijection. However, since you mention numerics, an approximation might be of interest:
The Lissajous curve $\begin{pmatrix}\sin(at+\delta)\\\cos(bt)\end{pmatrix}$ for an irrational ratio $a/b$, e.g. $a=1, b=\sqrt2$, is not closed and therefore maps $\mathbb R$ to a dense subset of $[0,1]^2$. Now take one of the usual $\mathbb [0,1]\to\mathbb R$ mappings, e.g. $t = \tan(\pi(u-\frac12))$ or $\operatorname{artanh}(2u-1)$, to obtain a map from $[0,1]$ to a dense subset of $[0,1]^2$. Now I wonder if there is an analysical formula to obtain the $t$ best approximating a given $(x,y)$...
