Abstract Smooth Manifolds vs Embedded Smooth Manifolds

Solution 1:

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.

Solution 2:

It wouldn't make sense for the definition to require $\phi$ and $\psi$ to be diffeomorphisms, because you can't define what it means for them to be diffeomorphisms until you already have a smooth structure on $M$. This is in contrast with the situation for embedded manifolds, where you can define what it means for a map to be smooth using the usual notion of differentiation of maps between subsets of $\mathbb{R}^n$.

That said, any chart of a smooth manifold is a diffeomorphism. This is pretty much immediate from the definitions: for a map between manifolds to be smooth, that means its compositions with charts give smooth maps between open subsets of $\mathbb{R}^n$. But in the case that your map is itself a chart (or the inverse of a chart), these compositions are exactly the maps of the form $\psi \circ \phi^{-1}$ which the definition requires to be diffeomorphisms (in particular, smooth).