Is the chart function of a smooth manifold a differomorphism, not just a homeomorphism It's clear that a smooth chart on a manifold is a diffeomorphism.  To me, the fact that smoothness of a manifold implies the smoothness of the transition function between the representation of two charts (whose domains overlap in M) should also imply that each plain old chart function is also a diffeomorphism.  But is this correct? 
 A: Let's suppose we have an $n$-dimensional manifold $M$ equipped with a smooth atlas $\mathcal{A}$. Then the transition function for any pair of charts in $\mathcal{A}$ is indeed smooth, as a map between open subsets of $\mathbb{R}^n$. Now, you could consider a larger atlas (i.e. a collection of charts in the topological sense) $\mathcal{A}'$ containing $\mathcal{A}$, but there is no guarantee that the transition functions for pairs of charts drawn from $\mathcal{A}'\setminus\mathcal{A}$ are smooth. The charts in $\mathcal{A}'$ are homeomorphisms, but they need not be diffeomorphisms.
Here's a somewhat silly example. Let $M=\mathbb{R}$ with the usual smooth structure. Then the chart $(U,\phi_U)$ defined by $U=\mathbb{R}$ and $\phi_U(x) = x^3$ is a topological chart (i.e. a homeomorphism) but it is not a diffeomorphism, as its inverse $x\mapsto x^{\frac 13}$ is not smooth.
A: It is true that each chart map is a local diffeomorphism, but perhaps not for the reason you think. 
When one defines manifolds, one starts with a topological space $X$. For a topological space it makes sense to talk about homeomorphisms and local homeomorphisms. Thus it makes sense to say that $X$ is covered by $U_i$ with each $U_i$ homeomorphic to $\mathbb{R}^n$ via chart map $\phi_i$. The notion of smoothness, however, does not make sense at this stage - $X$ (and also $U_i$'s) does not have any structure that would let one talk about smoothness and hence about diffeomorphisms. If the chart maps are such that transition functions are smooth (as maps between subsets of $\mathbb{R^n}$), then we say that the atlas of these charts is smooth and only then does $X$ obtain a smooth structure from the $\phi_i$'s. Now we can talk about smoothness of maps to or from $X$ - but the very definition is to compose with chart map (for maps to $X$) or to precompose with inverse of chart map (for maps from $X$).
In particular, smoothness of $\phi_i:U_i \to \mathbb{R}^n$ is by definition the same as the smoothness of the composition $\phi_i \cdot\phi_i^{-1}: \mathbb{R}^n \to \mathbb{R}^n$. Of course, that map is identity, so the maps $\phi_i$ are smooth. Similarly, by definition, smoothness of $\phi_i^{-1}$ is the same as smoothness of $\phi_i \cdot\phi_i^{-1}$. So indeed, after one makes $X$ into a smooth manifold by supplying a smoothly compatible collection of charts, the notion of local diffeomorphism makes sense for $X$, and the chart maps are diffeomorphisms. 
