Is $(0,1)$ homotopic to $(0,1) \times (0,1)$?

Question

Is there a homotopy between a map with $(0,1) \in R^2$ as image and one with $(0,1) \times (0,1)\in R^2$ as image? Both have domain $(0,1)$. Or the same question for closed intervals.

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$

Why doesnt spacefilling curves imply there is a homeomorphism between $(0, 1) \subset \Reals^{2}$ and $(0, 1) \times (0, 1) \subset \Reals^{2}$?

If there were a bijective (continuous) space-filling curve from $X = [0, 1]$ to $Y = [0, 1] \times [0, 1]$, then $X$ and $Y$ would be homeomorphic. (With the open interval and open rectangle, you have to work harder, because a continuous bijection from $(0, 1)$ is not generally a homeomorphism.)

Since an interval (open or closed) is not homeomorphic to a square (open or closed) for other reasons, there exists no bijective continuous map $I \to I \times I$ for $I = [0, 1]$ or $I = (0, 1)$.