Is it true that if $P \subseteq \mathbb{R}^2$ is homeomorphic to $[0,1]^2$, then $\partial P$ is a Jordan curve (i.e. homeomorphic to a circle)?

Question

Given a homeomorphism $h \colon P \to [0,1]^2$, it follows from the domain invariance theorem that $h$ maps the interior of $P$ onto an open subset of $\mathbb{R}^2$. At the same time, $h^{-1}$ maps the interior of $[0,1]^2$ onto an open subset of $\mathbb{R}^2$. Hence, $h$ maps the interior of $P$ onto $(0,1)^2$. Consequently, $h$ maps the boundary of $P$ onto the boundary of $[0,1]^2$, which is a Jordan curve.

Am I missing something? I just want to make sure the statement holds.

Answer 1

It is a bit more elementary: if $f:X\to Y$ is a homeomorphism, then so is its restriction $f\mathbin\upharpoonright A:A\to f[A]$ to every subspace of $X$. The restriction is a bijection, it is continous, and the restriction of its inverse to $f[A]$ is continuous too.

So the inverse of your homeomorphism $f:P\to[0,1]^2$ restricts to a homeomorphism on the boundary of $[0,1]^2$.