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The Invariance of Domain theorem states that

Given a continuous injection $f : U \to \mathbb{R}^n$, where $U$ is a nonempty open subset of $\mathbb{R}^n$, $f$ is an open map.

These slides (see last slide) state that as a consequence,

Given a continuous injection $f : \mathbb{R}^n \to \mathbb{R}^m$, $n \leq m$.

However, is the following statement true?

Given a continuous injection $f : U \to \mathbb{R}^m$, where $U$ is a nonempty open subset of $\mathbb{R}^n$, $n \leq m$.

It looks like a combination of the above statements. It naturally holds if $U$ is homeomorphic to $\mathbb{R}^n$.

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Since $U \subset \mathbb R^{n}$ is open, it contains some open ball $B$. If we restrict $f$ to $B$, which is homeomorphic to $\mathbb R^{n}$, then we still have a continuous injection, and we've reduced to the case you already know.

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Yes. Take an open ball $B=B(x,r)$ in $U$. Then $B$ is homeomorphic to $\mathbb{R}^n$, as $[0,1)$ is homeomorphic to $[0,+\infty)$ with $0\longmapsto 0$ (use polar coordinates and apply the latter to the radius). Denote $g:\mathbb{R}^n\longrightarrow B$ such a homeomorphism. Then apply the second result to $h:=f\circ g$.

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Suppose $n>m$, and consider the composite map $U \to \Bbb{R}^m \hookrightarrow \Bbb{R}^n$. This is still continuous and injective, so by invariance of domain the image is open. But then we have a nonempty open subset of $\Bbb{R}^n$ contained in $\Bbb{R}^m$, which is impossible.

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  • $\begingroup$ @julien: there is no circularity. It's trivial to show that (for $m<n$) $\Bbb{R}^m$ (considered as a subset of $\Bbb{R}^n$) contains no nonempty open set. $\endgroup$ Mar 15 '13 at 13:29

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