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.

Hint: Consider what happens to the connected $[0,1]$ if the point $\frac12$ is removed. What happens to $[0,1]\times[0,1]$ when $f(\frac12)$ is removed?
Hint: A continuous bijection from a compact space to a Hausdorff space is bicontinous.
