Smooth Manifolds: Charts or Parametrizations? Do I have the following correct?  
If you define manifolds as subspaces of $\mathbb{R}^N$ as is done in Guillemin and Pollack, then you don't need charts and atlases. All you need are smooth parametrizations to describe a topological manifold as smooth. However, if you are dealing with abstract manifolds where the topological space is not a subset of $\mathbb{R}^N$, then you need charts with smooth "overlaps" to show a that manifold is smooth.
Here's a link that discusses why the two different characterizations of smooth are ultimately the same, Guillmin & Pollack's Definition of a Manifold. My question is more of a practical nature.
 A: Definition: A manifold $M$ of dimension $k$ in the sense of Guillemin and Pollack is a subset $X$ of $\mathbb{R}^N$ such that for all $x\in X$ there is an open neighborhood $V\subset\mathbb{R}^N$ of $x$ so that $V\cap X$ is diffeomorphic to an open subset $U$ of $\mathbb{R}^k$. Such a diffeomorphism $\phi:U\to V$ is called a parametrization.
The trick here is that a diffeomorphism $f$ from $X\subset\mathbb{R}^N$ to $Y\subset\mathbb{R}^M$ means a smooth bijection with smooth inverse. But smooth means that, locally, $f:X\to Y$ always extends to a map $F:W\to Y\subset\mathbb{R}^M$, $W\subset\mathbb{R}^N$ which is smooth in the calculus 3 sense (all partials of all orders exist).
If you have a collection of parametrizations $\phi_\lambda$ that cover $X$, you have an atlas $\lbrace \phi_\lambda^{-1}\rbrace$. It is worth trying to show this so that you understand. The non-trivial condition that smooth maps have these local extensions is what allows us to not have to "check the overlaps". It trades one thing to check for another.
