Without using the Invariance of Domain result, I want to show that such an f cannot exist. Here is what I did:
Assume that there is an $f: \mathbb{R}^2 \to \mathbb{R}^n \ , n>2$ that is homeomorphic.
Consider $U = \left\lbrace x\in\mathbb{R}^2: r\leq \|x\| \leq R \text{ for some } 0<r<R \right\rbrace$ and $ f: \mathbb{R}^2 \setminus{U} \to \mathbb{R}^n \setminus f(U)$.
The inverse of f is continuous, but $\mathbb{R}^n\setminus f(U)$ is simply connected and the pre-image is not!
What is wrong with this argument? I understand that $f(U)$ doesn't necessarily have to be an annulus, but slap on some continuous $g: \mathbb{R}^n \to \mathbb{R}^n$ and it will be one.
Also, I was suggested that creating a quotient space can reduce the case back to $\mathbb{R}^2$ to $\mathbb{R}$. I thought about two equivalence relation:
$x \sim y \iff \|x\|= \|y\|$ w.r.t. Euclidean norm.
$(x_{1},x_{2}) = x$; $x \sim y \iff x_{1}= y_{1}$
Would either of these be alright? I am aware the first one creates a half line. I have had no practice with quotient space so I'd rather not dabble in it, to be honest.
Thanks.
The issue with your argument goes like this: the annulus $U$ is homotopic to a circle, and since $f$ is a homeomorphism its image $f(U)$ is also homotopic to a circle. This means that unless $f$ is the constant map, $\mathbb{R}^n\setminus f(U)$ amounts to $\mathbb{R}^n$ minus a wedge of circles, which is not simply connected. For example, think of a loop that links nontrivially with one of these circles; this loop cannot contract to a point.
Like William suggests, taking away a single point is the better strategy here.