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I am asked to prove the following:

Let $M$ be a smooth manifold of dimension $n$, and $\phi: U \to \phi(U)$ a homeomorphism from an open subset $U$ of $M$ to an open subset of $\mathbb{R}^n$. Show: $\phi$ is a diffeomorphism iff $(U, \phi)$ belongs to the maximal smooth atlas of the differentiable manifold $ M $.

However, I don't think this statement is true at all, and I think I have a counterexample:

Consider $\psi: \mathbb{R} \to \mathbb{R}: x \mapsto x^3$ and $\textrm{Id}_\mathbb{R}: \mathbb{R} \to \mathbb{R}: x \mapsto x$. Then it is quite clear that both $(\mathbb{R}, \psi)$ and $(\mathbb{R}, > \textrm{Id}_\mathbb{R})$ are charts, and $\textrm{Id}_\mathbb{R}$ is a diffeomorphism. However, $\psi$ is not a diffeomorphism, as its inverse is not differentiable at $0$.

Now we can define $\mathcal{A}$ to be the maximal smooth atlas containing $(\mathbb{R}, \psi)$, making $(\mathbb{R}, \mathcal{A})$ into a smooth manifold. Then, according to the statement, $(\mathbb{R}, \textrm{Id}_\mathbb{R}) \in \mathcal{A}$. But the two charts are not smoothly compatible, as

$$\psi \circ (\textrm{Id}_\mathbb{R})^{-1} = \psi$$

is not a diffeomorphism, contradicting $(\mathbb{R},\textrm{Id}_\mathbb{R}) \in \mathcal{A}$.

Conversely, $(\mathbb{R}, \psi)$ is trivially in $\mathcal{A}$, but $\psi$ is not a diffeomorphism, disproving both implications of the statement.

Did I make a mistake somewhere? My first thought was that the statement is only about proper subsets $U$, but the above argument can easily be changed to work for $U = (-1, 1)$ as well. The statement is true however if all charts in $\mathcal{A}$ are diffeomorphisms themselves, but this is not mentioned in the question.

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Consider $\Bbb R$ endowed with the maximal atlas containing $(\Bbb R,\psi)$. $\DeclareMathOperator{\Id}{Id}$ Then $(\Bbb R, \Id)$ is not a chart of this atlas, precisely because $\Id\circ \psi^{-1}\colon \Bbb R \to \Bbb R$ is not smooth in the usual sense. It follows that your claimed counterexample is not one. Moreover, the statement you are asked to prove is indeed true.

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  • $\begingroup$ I see! We are considering $\phi$ here a diffeomorphism between the manifolds $M$ and $\phi{U}$, not as a diffeomorphism on $\mathbb{R}$, invalidating my counterexample. Thanks! $\endgroup$
    – BrakkoFTW
    Commented Jan 5, 2023 at 18:25

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