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?
Regarding your 1st question, perhaps you meant to ask something else? Any atlas can be composed with a nonsmooth 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 onemap atlas, which is smooth in the standard sense.
You might like to read this article: http://en.wikipedia.org/wiki/Exotic_R4

$\begingroup$ I'm confused by your $t^{1/3}$ example. This atlas seems nondiffeomorphic to the trivial one. But, I thought the lack of exotic differentiable structure in $\mathbb{R}$ implied that all atlases on $\mathbb{R}$ had to be diffeomorphic to each other. I thought this followed from the fact that a differentiable structure is an equivalence class, under diffeomorphism, of atlases, and only one differentiable structure is allowed for $\mathbb{R}$. What am I missing? $\endgroup$ – Outis Jul 28 '10 at 17:51

3$\begingroup$ $\mathbb R$ with the standard atlas and the $t \longmapsto t^{1/3}$ atlas are diffeomophic. The nonintuitive thing is that the diffeomorphism isn't the identity map  it's the $t \longmapsto t^3$ map. $\endgroup$ – Ryan Budney Jul 28 '10 at 17:58

$\begingroup$ I guess I need clarification on what it means for two atlases to be diffeomorphic. You're saying that the atlas represented by $t \longmapsto t$ and the atlas represented by $t \longmapsto t^{1/3}$ are diffeomorphic to each other. The domains of each of these maps are trivially diffeomorphic, but the ranges are not. $f: t \mapsto t^{1/3}$ is not smooth at $t=0$, although its inverse is. $\endgroup$ – Outis Jul 29 '10 at 1:43
What do you mean by "smooth mapping" of atlases? I think you maybe using the terminology a bit different from how I do.
A chart or a coordinate chart on a topological manifold $M$ is a pair $(U,\psi)$ where $U\subset M$ is open and $\psi: U \to V\subset \mathbb{R}^n$ is a homeomorphism. Given two charts $(U_1,\psi_1)$ and $(U_2,\psi_2)$ with nonempty intersection $U_1\cap U_2$, we say that the charts are $C^k$ compatible if the transition function $\psi_2\circ\psi_1^{1} _{\psi_1(U_1\cap U_2)}$ is $k$times continuously differentiable. Then a $C^k$ atlas on $M$ is a collection of charts $\{(U_\alpha,\psi_\alpha)\}_{\alpha \in A}$ such that the union $\cup_{\alpha \in A} U_\alpha$ covers $M$ and all the charts are $C^k$compatible.
Two atlases are said to be $C^k$ compatible if all of their corresponding charts are pairwise compatible, and hence if $\mathcal{A},\mathcal{B}$ are two compatible atlases, their union also is an atlas. An atlas is said to be maximal if any other compatible atlas must be a subset. It always exists by Zorn's lemma.
A differential structure on the manifold $M$ is a choice of a maximal $C^ \infty$ atlas, we write it as $(M,\mathcal{A})$. Two differentiable manifolds $(M,\mathcal{A})$ and $(N,\mathcal{B})$ are said to be diffeomorphic if there exists a homeomorphism $\Psi: M\to N$ such that for any $(U_{\alpha},\psi_\alpha) \in \mathcal{A}$ and $(V_{\beta}, \phi_{\beta})\in\mathcal{B}$ we have that the function $$ \phi_\beta \circ \Psi \circ \psi_\alpha^{1} _{\psi_\alpha(U_\alpha \cap \Psi^{1}(V_\beta))} $$ is smooth.
Ryan already gave you an answer to your question. But let me elaborate a bit on your question two. Let $\mathbb{R}^2$ be your manifold. Define two charts on it $$ U_1 := \mathbb{R}\times (1,\infty), U_2 := \mathbb{R}\times (\infty,1) $$ and define $\psi_1(x) = x$ if $x\in \mathbb{R}\times (1,2)$ and $\psi_1(x_1,x_2) = (x_1 + x_2  2, x_2)$ if $x_2 \geq 2$. Similarly $\psi_2$. Clearly $(U_1,\psi_1), (U_2, \psi_2)$ cover $\mathbb{R}^2$ and the transition function on the strip $\mathbb{R}\times (1,1)$ is equal to the identity, and hence is smooth, so this gives a smooth atlas. But this smooth atlas is not compatible with the standard Cartesian atlas. What that does not mean that $\mathbb{R}^2$ with this atlas is exotic!
Let $(\mathbb{R}^k,\mathcal{E})$ denote the standard Euclidean space. An exotic smooth structure $(\mathbb{R}^k,\mathcal{A})$ requires that every homeomorphism from $\mathbb{R}^k$ to itself to not extend to a diffeomorphism from $(\mathbb{R}^k,\mathcal{E})\to (\mathbb{R}^k,\mathcal{A})$. What we have here is just that our stupid choice of homeomorphism is nonsmooth. It is simple to change our homeomorphism such that the mapping is smooth relative to the fixed atlases.
So the nonexistence of exotic smooth structure just say that for any two fixed smooth atlases, there exists some homeomorphism from $\mathbb{R}^k$ to itself such that it extends to a diffeomorphism. If this is what you meant, then the answer to your second question is "yes". But again, I don't understand what you mean by "smooth mapping of atlases".

$\begingroup$ Willie, thanks, but let me make sure I understand. If I'm interpreting your construction correctly you've defined the function $\psi_1: (x,y) \mapsto (x,y)$ if $y<2$ and $(x,y) \mapsto (x,y2)$ if $y\geq 2$. But this function is discontinuous at $y=2$. Doesn't this function then fail to satisfy the definition of a homeomorphism? $\endgroup$ – Outis Jul 28 '10 at 21:44

$\begingroup$ Oops, you are right, I forgot to type in the second component of the map. It is not supposed to be a discontinuity, but a "kink", so that $\psi_1$ is not a $C^1$ map in the given coordinates. $\endgroup$ – Willie Wong Jul 28 '10 at 23:17