ag.algebraic geometry - Why "smooth Gelfand duality" does not involve a topology on the algebras?

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?