The following question naturally originates from this question and this one.
While the usual $C^{0}$ Gelfand duality involves a topology on the function algebras considered (it relates compact Hausdorff topological spaces to unital $C^{*}$-algebras, which in particular are Banach algebras), why the "smooth Gelfand duality" seems, according to what I understood from the above questions, to see only the "pure" algebraic structure of certain algebras over $\mathbb{R}$ ?
Edit: I've just read the introduction of this. The topology actually enters the picture, but not in the form of a structure of topological algebra on the function spaces that locally model those $C^{\infty}$-differentiable spaces; it enters the picture when defining a "differentiable algebra" as the quotient of the algebra of smooth functions on $\mathbb{R}^n$ by a Fréchet-closed ideal.
But a question still stands: would it be possible to define compact Hausdorff topological spaces in the analogous way? Perhaps the answer is "no because of a lack of a universal local model $C^{0}(...)$", but "yes in the case of topological manifolds". Does it make sense?