Bijective Continuous Map

Question

Does there exist a map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ such that $f$ is bijective and continuous but not open?

Intuition tells me that this would contradict the invariance of domain. However, I stumble in arguing so.

Answer 1

No such map exists. Indeed, we shall show that $n = m$. If $m < n$, then composing $f$ with the natural embedding $\mathbb{R}^m \to \mathbb{R}^n$ yields an injective continuous map $\mathbb{R}^n \to \mathbb{R}^n$ whose range is contained in $\mathbb{R}^m$ and therefore has no interior, contradicting invariance of domain.

If $n < m$, consider the closed balls $B_r = \bar{B}(O, r) \subset \mathbb{R}^n$. These balls are compact, whence $f(B_r)$ is compact and $f$ restricts to a bijective continuous map $B_r \to f(B_r)$. We observe that a bijective continuous map between compact metric spaces is necessarily a homeomorphism. Now, since $\cup_{N=1}^\infty B_N = \mathbb{R}^n$, as $f$ is surjective, $\cup_{N=1}^\infty f(B_N) = \mathbb{R}^m$, so by Baire category theorem, there exists $N$ such that $U \subset f(B_N)$ where $U \subset \mathbb{R}^m$ is nonempty and open. But as $f|_{B_N}: B_N \to f(B_N)$ is a homeomorphism, $f^{-1}|_U: U \to f^{-1}(U)$ is a homeomorphism as well. In particular, $f^{-1}|_U$ is an injective continuous map from $U$, an open subset of $\mathbb{R}^m$ to $\mathbb{R}^n$. So again by composing with the natural embedding $\mathbb{R}^n \to \mathbb{R}^m$, we see that this contradicts invariance of domain.

Thus, the only possibility is $n = m$, in which case invariance of domain immediately yields that $f$ is open.