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This interesting answer raised a question in my mind:

Suppose that $M$ and $N$ are smooth manifolds and $f:$ $M \rightarrow N$ is a diffeomorphism. Can we conclude that $C^{\infty} (M)$ and $C^{\infty} (N)$ are isomorphic? In other words, is there a one to one correspondence between smooth functions on the two smooth manifolds? ($C^{\infty} (M)$ denotes the set of smooth functions from $M$ to $\mathbf{R}$).

Thanks

Ali Koohpaee
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    Yes. This follows from the fact that $M \mapsto C^\infty(M)$ is a contravariant functor. – Mark Saving Jun 03 '22 at 20:01
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    The function\begin{align}C^\infty(N)&\to C^\infty(M)\\phi&\mapsto \phi\circ f \end{align}is invertible (the function\begin{align}C^\infty(M)&\to C^\infty(N)\\psi&\mapsto \psi\circ f^{-1} \end{align}is its inverse). Does this answer your question? – Filippo Jun 03 '22 at 20:29
  • @MarkSaving: Thanks. So this follows from the fact that there is a contravariant functor from Top (category of topological spaces) to Set. – Ali Koohpaee Jun 05 '22 at 18:37
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    @Filippo: Yes, thanks very much for your answer. That was exactly what I was looking for. – Ali Koohpaee Jun 05 '22 at 18:38

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