Abstract Smooth Manifolds vs Embedded Smooth Manifolds 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?
 A: 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).
A: The point is that since your manifold is abstractly defined and not embedded, it's not completely obvious what you mean by diffeomorphic (before you define the smooth structure!).
In the case of an embedded submanifold, you can just ask that $\phi$ and $\psi$ extend to diffeomorphisms defined on open subsets of the ambient space, but in the second definition you cannot do that. So what you do is you define a map $f: M \to \mathbb{R}^n$ to be smooth if on every chart domain $(U,\phi)$, the map $f \circ \phi^{-1}$ is smooth.
To be able to do that consistently, you need the condition that the transition maps are smooth, and a posteriori, yes, the charts are smooth for the smooth structure they induce.
A: 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.
