In Topology from the Differentiable Viewpoint an (embedded) smooth manifold is defined as follows.
A subset $M \subseteq \mathbb{R}^k$ is called a smooth manifold of dimension $m$ if each $x \in M$ has a neighbourhood $W \cap M$ that is diffeomorphic to an open subset $U$ of the euclidean space $\mathbb{R}^m$
In the language of charts and atlases we can convert this definition to the following
A subset $M \subseteq \mathbb{R}^k$ is called a smooth manifold of dimension $m$ if there is a collection of charts called a smooth atlas $\mathcal{A} = \{ (W_i, \psi) \ |\ W_i \text{ is open in $M$ and $\psi : W \to \mathbb{R}^m$ is a diffeomorphism\}}$, and $\bigcup_{i}W_i = M$
Then in Introduction to Smooth Manifolds by John Lee (abstract) smooth manifolds are defined in the following way.
A smooth manifold is a pair $(M, \mathcal{A})$ where $M$ is a topological manifold and $\mathcal{A}$ is a smooth structure on $M$. A smooth structure is a maximal smooth atlas, and a smooth atlas is a collection of charts whose domains cover $M$ and where any two charts $(U, \phi), (V, \psi)$ in $\mathcal{A}$ are smoothly compatible with each other. Those charts are smoothly compatible if either $U \cap V = \emptyset$ or the transition map is a diffeomorphism.
Now my question is the following, in the second definition, the functions $\phi$ and $\psi$, are homeomorphisms, so they are continuous, bijective and have continuous inverse, but are they diffeomorphisms?. Are they necessarily smooth and is their inverse necessarily smooth?
The definition doesn't explicitly state that they need be diffeomorphisms.
If $(U, \phi)$ and $(V, \psi)$ are two charts in $\mathcal{A}$ such that $U \cap V \neq \emptyset$, the transition map from $\phi$ to $\psi$ is defined to be the composition $\psi \circ \phi^{-1} : \phi[U \cap V] \to \psi[U \cap V]$
So that reduces my question to the following, is $\psi \circ \phi^{-1}$ a diffeomorphism if and only if both $\psi$ and $\phi^{-1}$ are diffeomorphisms?
Best Answer
There's a reason that definition does not require that the map $\phi$ in a chart $(U,\phi)$ be a diffeomorphism: that would require knowing already that $M$ is a smooth manifold, but since that is what is being defined, the definition would become circular.
However, once a smooth manifold $(M,\mathcal{A})$ is defined, then one can move forward and define smooth functions on open subsets of $M$. Namely, for each open set $W \subset M$, a function $\xi : W \to \mathbb{R}^k$ is smooth if and only if for each chart $(U,\phi)$ in the atlas $\mathcal{A}$ the map $\xi \circ \phi^{-1} : \phi(W \cap U) \to \mathbb{R}^k$ is smooth. And then, by applying the definition of a smooth atlas, it is now an easy lemma to prove that if $(U,\phi)$ is a chart in the atlas $\mathcal{A}$ then $\phi : U \to \mathbb{R}^m$ is indeed smooth.