It is a kind of folklore but I would like to see the proof of the following fact: given two smooth manifolds $M$ and $N$ if we assume that the algebras $C^{\infty}_0(M)$ and $C^{\infty}_0(N)$ are isomorphic (as algebras) then $M$ and $N$ are diffeomorphic.


1 Answer 1


Theorem 2.8 on p. 30 of Models for smooth infinitesimal analysis by Moerdijk and Reyes. The theorem works with $C^\infty$ rings, which are a bit more general than algebras of functions. The advantage is that the proof is short.

  • $\begingroup$ I have already accepted your answer, however I realized that there is one subtle thing. As you mentioned the notion of $C^{\infty}$ ring is more general than just smooth functions on manifold however the notion of morphism in this category appears to be more restrictive. So how one can prove that any homomorphism of algebras is automatically a $C^{\infty}$ morphism (as defined in Moerdijk's book)? $\endgroup$
    – Totentanz
    Feb 10, 2017 at 17:30
  • 1
    $\begingroup$ I don't know a direct proof. My answer slyly suggested that looking at the algebra of functions on a smooth manifold as an algebra is a bit silly and that $C^\infty$ rings provide a better framework. But since I cheated, here is an answer to the question you asked: Theorem 2.3 on p. 34 of $C^\infty$-Differentiable Spaces by Juan A. Navarro Gonzalez and Juan B. Sancho de Salas (Springer LNM 1824). $\endgroup$ Feb 10, 2017 at 20:54

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