I'm trying to show that, for every $d> 2$, there exists a proper continuous self-map $f$ on $\mathbb{R}^d$ such that $\partial f(\mathbb{R}^d)$ is a sub-manifold of $\mathbb{R}^d$ of positive co-dimension. I feel that the construction should be obvious, but I haven't managed to demonstrate its existence...
Am I wrong in assuming that such a map exists?
What I have in mind: I'm trying to build a map from $\mathbb{R}^d$ to $\mathbb{R}^d - \left\{(x_1,\dots,x_d):\, x_1=\dots = x_{d-1}=0,\, x_d \geq 0 \right\}$ but that's all I've got so far.
Here is an example if you are talking about the topological boundary: define \begin{align} f : \mathbb{R}^n &\longrightarrow \mathbb{R}^n \\ \left(x^1,\ldots,x^n\right) & \longmapsto \left(|x^1|,x^2,\ldots,x^n\right) \end{align} This is a continuous map. One can easily show that it is proper. As a topological statement, $\partial f(\mathbb{R}^n)$ is the subspace $\left\{\left(0,x^2,\ldots,x^n\right) ~|~ \left(x^2,\ldots,x^n \right)\in \mathbb{R}^{n-1}\right\}$ and then is a submanifold of codimension $1$.
If you want $f$ to be injective, and smooth, you can define $f(x^1,x^2,\ldots,x^n) = \left(e^{x^1},x^2,\ldots,x^n\right)$, but $f$ would not be proper.
Notice that, if you require for $f$ to be injective, and for $\partial f(\mathbb{R}^n)$ as the boundary of the manifold $f(\mathbb{R}^n)$, then obviously, the boundary will be empty as $f$ will be an open map, and its image will be boundaryless as a manifold.