Let $f:B \longrightarrow \mathbb{R}^n$ a continuous and injective function from closed ball in $\mathbb{R^n}$ to $\mathbb{R}^{n}$. I'd like to know $f$ has to maps the boundary $\partial B$ in the boundary $\partial f(B)$?
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It seems the following. The opposite inclusion should be non-trivial. I consider $B$ as a subset of a space $\Bbb R^n $. By the Open Domain Principle, each continuous injective map $f:Uā\Bbb R^n$ defined on an open subset $U\subset \Bbb R^ n$ is an open topological embedding. So $f(\operatorname{int} B)\subset \operatorname{int} f(B)$. Then $$f(\partial B)=f(B\setminus \operatorname{int} B)=f(B)\setminus f(\operatorname{int} B)\supset f(B)\setminus\operatorname{int} f(B)=\partial f(B).$$ ā Alex Ravsky Dec 17 '14 at 17:42
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Note that $f(B)^{\circ}$, the interior of $f(B)$, is open so $f^{-1}(f(B)^{\circ})$ is an open subset of $B$. Furthermore, $f^{-1}(f(B)^{\circ}) \subseteq f^{-1}(f(B))$ and $f^{-1}(f(B)) = B$ as $f$ is injective, so $f^{-1}(f(B)^{\circ}) \subseteq B^{\circ}$.
Now let $x \in \partial B =\overline{B}\setminus B^{\circ}$, then $x \notin f^{-1}(f(B)^{\circ})$ so $f(x) \notin f(B)^{\circ}$. However $f(x) \in f(B)$ and hence $f(x) \in \overline{f(B)}$, so $f(x) \in \overline{f(B)}\setminus f(B)^{\circ} = \partial f(B)$.
Michael Albanese
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