Does the existence of exotic smooth structure in $\mathbb{R}^4$ imply the existence of an atlas which has a $C^0$ mapping to the Cartesian atlas, but not a $C^k$ mapping (for some finite $k$)? Does the nonexistence of exotic smooth structure in $\mathbb{R}^n$, $n\neq 4$ imply that all atlases therein have smooth mappings to the Cartesian atlas? 



exotic smooth structure clarification
Regarding your 1st question, perhaps you meant to ask something else? Any atlas can be composed with a non-smooth homeomorphism to produce an atlas that isn't smooth in the standard sense. For example, $\mathbb R \to \mathbb R$ defined by $t \longmapsto t^{1/3}$ is an atlas on $\mathbb R$ but it's not $C^1$. This answers your 2nd question in the negative. 


Alternatively, some exotic smooth $\mathbb R^4$'s are diffeomorphic to open subsets of the standard $\mathbb R^4$, so even for exotic smooth $\mathbb R^4$'s you could potentially have only a one-map atlas, which is smooth in the standard sense. 


You might like to read this article: <http://en.wikipedia.org/wiki/Exotic_R4>


