When is the image of $\mathbb{R}^n$ a smooth submanifold?

Question

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be an injective $C^{k}$-function, for some positive integer $k$. Under what conditions is the complement of the open set (I know its open by Brower's domain invariance theorem) a finite union of smooth submanifold of $\mathbb{R}^n$?

Answer 1

Here is a partial answer. Recall that every simply connected open subset of $\mathbb{C}$ is biholomorphically equivalent to the unit disk $\Delta = \{z\in \mathbb{C}~|~ |z| <1\}$.

Let $\gamma : \mathbb{S}^1 \to \mathbb{C}$ be a continuous simple curve that is nowhere differentiable (for example, the Von Koch curve), and let $\Omega$ be its interior (by Jordan's theorem). Then $\Omega$ is a simply connected open subset of $\mathbb{C}$ different from $\mathbb{C}$ : there exists a biholomorphism $g : \Delta \to \Omega$. Moreover, $\Delta$ is diffeomorphic to $\mathbb{R}^2$, say $h : \mathbb{R}^2 \to \Delta$ is such a diffeomorphism. Then \begin{align} f = g\circ h : \mathbb{R}^2 \to \mathbb{R}^2 \end{align} is an injective smooth function with image $\Omega$. But $\mathbb{R^2}\setminus \Omega$ is not a smooth manifold as otherwise, its boundary would be smooth, that is not the case.

Here has been shown that even if $f$ is $\mathcal{C}^{\infty}$, there is no reason for smoothness to extend to the boundary.

Notice that we could have choosen $\Omega$ to be the interior of a triangle, and the boundary would not be smooth either because of corners. My example shows that it can happen everywhere.

Edit It is asked in comment if, whenever $f$ has a dense image, $\mathbb{R}^{n} \setminus \mathrm{Im}(f)$ would be a $n-1$ dimensional smooth manifold. The answer is no. Here is a counter example.

Let $\gamma : \mathbb{R}_+ \to \mathbb{C}$ be an injective curve such that $\gamma$ is nowhere differentiable, and goes to infinity (imagine the trajectory of a brownian motion). Then $ \Omega =\mathbb{C}\setminus\gamma(\mathbb{R}_+)$ is a simply connected open subset of $\mathbb{C}$ different from $\mathbb{C}$. The above construction show there is a smooth function $f : \mathbb{R}^2 \to \mathbb{R}^2$ with image $\Omega$, which is dense in $\mathbb{R}^2$ but with complement that is nowhere smooth.