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I saw the following result: " If $A\subseteq \mathbb{R}^{n+p}\to \mathbb{R}^n$ of class $C^1$ in the open set $A$ (with $n,p$ being positive integers), then $f$ is not injective."

I wonder if the statement above remains true if we weaken the hypothesis "$f$ of class $C^1$" to the hypothesis "$f$ continuous".

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If such a map existed, consider the restriction $f_{|C} : \mathbb{R}^{n} \to \mathbb{R}$ with $C$ being the closed ball of center $0$ and radius $1$. Called $X = f(C)$ we have that $f_{|C}$ is a bijection between this two sets.

Furthermore, cloed subset $K \subseteq C$ are compact, so $f_{|C}(K)$ is compact in $X$. Since $X$ is a subspace of a $T_2$ space, it is $T_2$ and $f_{|C}$ is a closed map (A compact in a Hasdorf space is closed). But $C$ can not be omeomorphic to $X$ for $n > 1$ since removing one point on $X$ disconnects it while $C$ minus one point is still connected.

Edit. A similar results hold for $f : \mathbb{R}^{n+p} \to \mathbb{R}^{n}$ for positive $p$, but it makes use of higher homotopy groups

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  • $\begingroup$ Could you please illustrate the ideia that solves the problem for any $n,p>0$? $\endgroup$ Sep 2 '20 at 0:19
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Let $A$ be an open subset of $\mathbb R^n$. It contains a compact ball $C$. If there exists a continuous injective map $\phi: C\to \mathbb{R}$, then it should be a homeomorphism onto its image since $\phi(C)$ is Hausdorff. That is impossible because $C$ does not have cut points while $\phi(C)$ does. See this relevant Wikipedia page

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