Making a homeomorphism a $\mathscr{C}^k$-map

Question

I'm solving the following exercise and I just need a little push for the one step I'm failing to do:

Let $M$ be a $\mathscr{C}^k$-manifold, $N$ be a topological manifold, and $\alpha: M\to N$ be a homeomorphism. Then there is a unique $\mathscr{C}^k$-structure on $N$ which makes $\alpha$ a $\mathscr{C}^k$-map.

So far, what I have done: if $\mathfrak{A}_M$ is a differential structure on $M$, we let $$\mathfrak{A}=\{(\alpha[U],\varphi\circ \alpha^{-1})\mid (U,\varphi)\in \mathfrak{A}_M\}. $$Then $\mathfrak{A}$ is a $\mathscr{C}^k$-atlas for $N$, inducing a differential structure which makes $\alpha$ a $\mathscr{C}^k$-map.

Ok, now assume that we have another differential structure $\mathfrak{A}'$ which makes $\alpha$ a $\mathscr{C}^k$- map. It is enough to check that $\mathfrak{A}'$ is $\mathscr{C}^k$-compatible with $\mathfrak{A}$. Given $(V,\psi)\in \mathfrak{A}'$ and $(\alpha[U],\varphi\circ \alpha^{-1})\in \mathfrak{A}$, we'd have to see both that $$\psi\circ (\varphi\circ \alpha^{-1})^{-1}:\varphi[\alpha^{-1}[V]\cap U]\to \psi[V\cap \alpha[U]] $$ and $$(\varphi\circ \alpha^{-1})\circ \psi^{-1}:\psi [V\cap\alpha [U]]\to \varphi [\alpha^{-1}[V]\cap U] $$are $\mathscr{C}^k$ (as maps between open sets of Euclidean spaces). The first one is just the local representation of $\alpha$, which is $\mathscr{C}^k$ according to $\mathfrak{A}'$, hence is $\mathscr{C}^k$. But I don't know how to argue for the second one. Help?

Answer 1

Ok, we figured it out sometime ago but only now I could manage to come back here and write an answer. It goes like this: the result is not true as stated, we must have $\mathscr{C}^k$-diffeomorphism instead of $\mathscr{C}^k$-map. This is because the atlas $\mathfrak{A}$ defined in the original post actually makes $\alpha$ a $\mathscr{C}^k$-diffeomorphism (smoothness of the inverse is also trivial, the problem is that I missed this). This being said, if we assume that $\mathfrak{A}'$ also makes $\alpha$ a $\mathscr{C}^k$-diffeomorphism, then the smoothness of the second transition map follows from the exact same argument used in the smoothness of the first transition map (in other words, merely assuming that $\mathfrak{A}'$ makes $\alpha$ a $\mathscr{C}^k$-map is not strong enough to prove that $\mathfrak{A}$ and $\mathfrak{A}'$ are $\mathscr{C}^k$-compatible).

Answer 2

Suppose that we have, in any category, invertible functions $$ A \overset{f}{\longrightarrow} B \overset{g}{\longrightarrow} C. $$ Then we have the formula $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

In your situation, this shows that the first map is the inverse of the second. Hence, you are able to apply the following standard theorem from calculus:

Diffeomorphism from Inverse function theorem