Lee Mosher's comment to this question suggests that all charts on a manifold are compatible (or I think his answer suggests that), but I can't understand it. Why are any two charts on a manifold necessarily compatible?
EDIT:
This one, same as the question I linked:
Given two charts, ($U_{1},φ_{1}$), ($U_{2},φ_{2}$), on a n-dimensional topological manifold M, such that: $U_{1} \cap U_{2}\neq \emptyset$, we get transition maps:
$φ_{1}\circ φ_{2}^{-1} : φ_{2}(U_{1}\cap U_{2}) \rightarrow φ_{1}(U_{1}\cap U_{2})$, and
$φ_{2}\circ φ_{1}^{-1} : φ_{1}(U_{1}\cap U_{2}) \rightarrow φ_{2}(U_{1}\cap U_{2})$
Two charts, as above, are called compatible if the transition maps, as above, are homeomorphisms. If $U_{1} \cap U_{2} = \emptyset$, then they are compatible.
compatibility of any two charts $(U,\phi),(V,\psi)$ of an atlas on a $n$-dim. manifold $M$ means that the composite $\phi\circ\psi^{-1}:\psi(U\cap V) \rightarrow \phi(U\cap V)$ is a homeomorphism. If you have a smooth manifold, you claim the transition maps to be diffeomorphisms.
The thing is that you have a priory no differentiable structure or even topological structure on $M$, since $M$ is simply some set with an atlas.
If the transition functions of an atlas are smooth as functions defined on subsets of $\mathbb{R}^n \rightarrow \mathbb{R}^n$, then you call $M$ a smooth manifold.
Edit: the comment on the post you mentioned simply refers to the fact that the transition map is the composite of two homeomorphisms (diffeomorphisms), hence a homeomorphism (diffeomorphism) itself.