Do infinite products commute with functor of smooth sections? Similarly to my previous question about direct limits, I have now basically the same question about inverse limits. It seems in fact, that I only need the result for products.
Question: Is there a natural smooth structure on $\prod \mathbb{R}$ such that $\mathcal{C}^\infty(U,\prod \mathbb{R}) = \prod\mathcal{C}^\infty(U,\mathbb{R})$?
 A: This works in the Convenient Calculus of Kriegl and Michor ("A Convenient Setting for Global Analysis").  There, the linear category concerned is the space of bornological locally convex topological vector spaces and this is the linear subcategory of the category of Frolicher spaces.  So $\prod \mathbb{R}$ being the bornological product is the categorical product as well and hence has the properties that you require.
A: As Martin Brandenburg said, cannot exist a (locally euclidean) smooth variety structure, but in the more large category  $(C^\infty-Ring)^{op}$ I think that the product exist (see "Models for Smooth Infinitesimal Analysis" Ieke Moerdijk, Gonzalo E. Reyes, see T. 2.8 pag. 30).
It is a funtor  $P: C^\infty\to Set$ that map $\mathbb{R}^n$ on the set of funtion like 
$f\circ\pi_J:\prod_i\mathbb{R}_i \to \prod_J\mathbb{R}_j\to \mathbb{R}^n $ where $J\subset I$ is a finite subset, $\pi_J$ is the natural projection, and $f$ a smooth map (and by composition on morphisms). We have to show that $P$ is the sum of $I$ copies of $\mathbb{R}$ (where $\mathbb{R}$ is see as a $C^\infty$ spaces naturally). Give a manifold M, a morphism of  $C^\infty$-spaces $\mathbb{R}\to M$ is uniquely represented by a smooth map $M\to \mathbb{R}$. Then give morphisms $g'_i: \mathbb{R}\to M$ i.e.  smooth maps $g_i: M\to \mathbb{R}$ i.e. a map $g=\prod_i g_i: M\to \prod_I \mathbb{R}$ follow the morphism $g': P\to M$ such that   on argument $g'_n: P(\mathbb{R}^n)\to M(\mathbb{R}^n)$ map $f\circ \pi_J $ (where $J=${$j_1,...,j_n$}) on $(f\circ g_{j_1}, ..., f\circ g_{j_n}): M\to \mathbb{R}^n$.
(I hope this work...)
