Manifold has uncountable many smooth stuctures if it has one

Question

This is the Problem 1-6 of John Lee's Introduction to smooth manifold:

Let $M$ be a nonempty topological manifold of dimension $n\geq1$. If $M$ has a smooth structure, show that it has uncountably many distinct ones. [Hint: first show that for any $s>0$, $F_s(x)=|x|^{s-1}x$ defines a homeomorphism from $\mathbb{B}^n$ to itself, which is a diffeomorphism if and only if $s=1$.]

What I tried:

It can be proved there is a atlas $\mathcal{A}$ (not maximal) which is compact with the original smooth sturcture of $M$ and has the following property: $\forall(U,\psi)\in\mathcal{A}$, $\psi(U)=\mathbb{B}^n$. I tried to define $\psi'=F_s\circ\psi$ and hope $\{(U, \psi')\}$ to form a new atlas for $M$. But $$\varphi'\circ(\psi')^{-1}=F_s\circ\varphi\circ\psi^{-1}\circ F_s^{-1}$$ may not be diffeomorphism. Any help, thanks.

Answer 1

Given a smooth structure, we would like to find a coordinate ball $(U_0,\phi_0)$ such that the center $p$ of $U_0$ is covered by only this chart. It's not hard to see that we need only find a point $p$ covered by one chart. Then we can replace this chart with $(U_0,F_s \circ \phi_0)$ and get a new smooth structure, which is not smoothly compatible with the original one.

By Thm 1.15 and Lemma 1.10, we can find a countable, locally finite open refinement of the smooth structure consisting of precompact coordinate balls. This refinement is also a smooth structure, let's work with it. Then choose an arbitrary point $q$ on the manifold, it has a neighborhood intersects finitely many smooth charts denoted them as $U_1$ ~ $U_k$.

1)If $k=1$, then $q$ is only covered by $U_1$. We can replace $(U_1,\phi_1)$ with $(U_1,F_s\circ\frac{\phi_1 - \phi_1(q)}{r+|\phi_1(q)|})$, where r is the radius of $\phi(U_1)$.

2) If $k>1$, then repeat the following procedure starting from $i=1$: If $U_i$ is covered by the rest charts, then remove it from the refinement and get a new smooth structure otherwise stop the procedure. Eventually, there is going to be a point $q'$ covered by only one precompact coordinate ball. If we stop before $i=k$, then $q' \neq q$ otherwise $q'=q$. Apply 1).

By 1) and 2), we find a smooth structure distinct from the original one. Since there are uncountably many $F_s$, we have proved that given any smooth structure of a topological manifold, there exists uncountably many distinct smooth structures on the manifold.

Answer 2

I found the other answer here a little difficult to interpret, so I'm posting one in case it helps.

Given a smooth structure $\mathcal{A}$ on $M$ we shall construct a new smooth structure $\mathcal{B}$ on $M$ as follows. Fix a smooth coordinate ball $B$ (in the original atlas $\mathcal{A}$) whose image under its smooth coordinate map $\varphi$ is $\mathbb{B}^{n}$ (without loss of generality). Let $x = \varphi^{-1}(0)$. Since the smooth coordinate balls form a basis for $M$, for each $y\neq x$ we may pick a smooth coordinate ball $B_{y}$ containing $y$ such that $B_{y}\subseteq M\setminus\{x\}$. Let $\varphi_{y}$ denote the corresponding smooth coordinate maps. Define $\mathcal{B}$ to be the smooth structure (the unique maximal smooth atlas) determined by the smooth atlas $$\{(B_{y}, \varphi_{y}): y\neq x\}\cup\{(B, F_{s}\circ\varphi)\}$$ where $s\neq 1$. Now use the idea in Anthony Carapetis's comment.