Is there an injective continuous function from an open set of $\mathbb{R}^{n+p}$ to $\mathbb{R}^n$?

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".

• The body of your question and the title are asking different things Jul 30 '20 at 23:03
• Yes, this is Brouwer's "Invariance of Domain" theorem: en.wikipedia.org/wiki/Invariance_of_domain Jul 31 '20 at 1:35

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
• Could you please illustrate the ideia that solves the problem for any $n,p>0$? Sep 2 '20 at 0:19
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