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.
Best Answer
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.