Proving a topological manifold homeomorphic to a smooth manifold is smooth.

Question

Problem: Prove that $S^n$ the n-sphere is homeomorphic to $\partial I^{n+1}$ the unit cube. Show that the unit cube admits a smooth structure hence can be turned into a smooth manifold("even though it has corners"). Generalize this by proving that any topological manifold homeomorphic to a smooth manifold admits a smooth structure.

The unit cube is the equivalent of unit sphere for the "max metric" so it can be written as $\partial I^{n+1}=\{x\in \mathbb{R}^n:||x||_{\infty}\leqslant 1\}$ while the n-sphere can be written using the euclidean norm $S^n=\{x\in\mathbb{R}^n:||x||\leqslant1\}$.

For the first part I thought of the following map that is homeomorphic(the intuition came from the fact the two metrics generated by the norms are equivalent):

$f:S^{n}\to \partial I^{n+1}\\ \:\:\:\:\:\:x\longrightarrow x\frac{||x||_{\infty}}{||x||}$

f is bijective, continuous(once the norms are continuous and the product of continuous functions are continuous)

$f^{-1}=g:I^{n+1}\to S^n \\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y\longrightarrow y\frac{||y||}{||y||_{\infty}}$

By the same argument g is bijective and continuous hence f is a homeomorphism.

To prove that $\partial I^{n+1}$ is a smooth manifold I thought of defining a chart of the following kind $f\circ\phi$ where $\phi:U\subset S^n\to\mathbb{R}^n $ is an diffeomomrphism from a subset of $\partial I^{n+1}$. However for this chart $f\circ\phi$ to be smooth I would need $f$ to be smooth which is not necessarily true once it is only required for $f$ to be a homeomorphism.

Question:

How can a topological manifold homeomorphic to a smooth manifold gain a smooth structure? How would it be proved?

Thnaks in advance!

Answer 1

Perhaps it is worthwhile recalling the definition of a smooth structure on a manifold: it is an atlas with smooth overlap maps. To be precise, a smooth structure on an $n$-dimensional manifold $N$ means an atlas of coordinate charts of the form $$\mathcal A = \{(\phi_i,U_i) \mid i \in I\} $$ such that for each $i \in I$, $\phi_i : U_i \to \mathbb R^n$ is a homeomorphism defined on an open subset $U_i \subset N$, and each overlap map $$\phi_j \circ \phi_i^{-1} : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j) $$ is smooth. Notice that I did not say "$\phi_i$ is a smooth map" because that's not the requirement. The key requirement of a smooth atlas is not the smoothness of the maps defining the coordinate charts, but instead the requirement is smoothness of the overlap maps.

If $g : M \to N$ is any homeomorphism, then using $g$ together the smooth atlas $\mathcal A$ for $N$, we get the following smooth atlas for $M$, $$g^*\!\mathcal A = \{(\phi_i \circ g, g^{-1}(U_i) \mid i \in I\} $$ You can easily check that formally this does indeed satisfy the definition of a smooth atlas for $M$.

So yes, $M$ has a smooth structure. Furthermore, using this atlas, the homeomorphism $g$ becomes a diffeomorphism.