# Counterexample: $\phi$ is a diffeomorphism iff $(U, \phi)$ belongs to the maximal smooth atlas of $M$.

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.

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.
• 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! Commented Jan 5, 2023 at 18:25