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.

Solution 1:

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.

Solution 2:

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.