This question arose out of [this stack exchange post](https://math.stackexchange.com/questions/3287675/taming-mathbbr-diffeomorphisms). I am wirting a thesis about the $s$-cobordism theorem and Siebenmann's work about end obstructions. Combined they give a quick proof of the uniqueness of the smooth structure for $\mathbb{R}^n$, $n \geq 6$. Roughly Siebenmann's theorem says that for $n \geq 6$ a contractible $n$-manifold $M$ that is simply connected at infinity embeds (smoothly) as the interior of a compact manifold. Since this compact manifold is contractible, by the $s$-cobordism theorem, it is diffeomorphic to the standard $n$-disk $D^n$ (see Minor's *Lectures on the $h$-cobordism theorem* for example). It follows that $M = \text{int } D^n$ is diffeomorphic to $\mathbb{R}^n$.


The problem is that the case $n = 5$ is not covered. I am aware of Stallings beautifully written *On the Piecewise-Linear Structure of Euclidean Space* but I am searching for a way to deal with the $n = 5$ case via Siebenmann's end theorem and the proper $s$-cobordism theorem (see link to the mse question). This leads me to the following question, which is interesting in itself



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> Given a smooth properly embedded codimension 1 submanifold $S \subset \mathbb{R}^{n+1}$, is there a self diffeomorphism $\mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ which carries $S$ into a one dimensional bounded region $\mathbb{R}^n \times (-1, 1)$ ?
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Now if $M$ is a manifold that is homeomorphic to $\mathbb{R}^5$, the product $M \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^6$, and hence also diffeomorphic. Granted the existence of the diffeomorphism in my question, we could find a diffeomorphism $f : M \times \mathbb{R} \rightarrow \mathbb{R}^6$ that maps $M \times 0$ into $\mathbb{R}^5 \times (-1, 1)$. This would produce a proper $h$-cobordism between $M$ and $\mathbb{R}^5$ by taking the region between $f(M \times 0)$ and $\mathbb{R}^5 \times 1$ in $\mathbb{R}^5 \times \mathbb{R}$. Since $M$ is simply connected, the proper $s$-cobordism theorem applies and shows that $M$ and $\mathbb{R}^5$ are actually diffeomorphic.



Wildness of codimension 1 submanifolds of euclidean space
For $n\geq 5$, such a diffeomorphism exists iff the complement of $S$ has an unbounded connected component. Indeed, the existence of such a diffeomorphism implies there is an unbounded component of the complement of $S$. Conversely, suppose that the complement of $S$ has an unbounded component. Choose a proper arc $\gamma:[0,\infty)\to\mathbb R^n$ in the complement of $S$ (i.e. going to infinity). Since $n\geq 5$, the sweepout of a generic proper homotopy from $\gamma$ to a standard coordinate ray is properly embedded, hence is an isotopy (this might still work for $n=4$, but it's definitely a bit more delicate, and I haven't thought about it carefully). Using this isotopy on $S$, we reduce to the case where $S$ is disjoint from a standard coordinate ray, which is easy.


A sufficient (but not necessary) condition for the complement of $S$ to have an unbounded component is that $S$ have finitely many components (this implies the complement of $S$ also has finitely many components, hence one must be unbounded).


