Let $M$ be a smooth manifold, and let $U\subset M$ be an open set. Let $C^\infty_M$ be the sheaf of real-valued smooth functions on $M$ and denote $C^\infty(M)=\Gamma(M,C^\infty_M)$. Define the multiplicative subset $$ S=\{s\in C^\infty(M)\mid s(x)\neq 0,\,\forall x\in U\}. $$ I am wondering whether $C^\infty(U)=S^{-1}C^\infty(M)$. To decide the truth value of this assertion, one must prove or disprove the hypotheses on the restriction homomorphism $C^\infty(M)\to C^\infty(U)$ of the following result of Atiyah, Macdonald, Introduction to Commutative Algebra:

Corollary 3.2. If $g: A \rightarrow B$ is a ring homomorphism such that:

  1. $s \in S \Rightarrow g(s)$ is a unit in $B$;
  2. $g(a)=0 \Rightarrow a s=0$ for some $s \in S$;
  3. Every element of $B$ is of the form $g(a) g(s)^{-1}$;

then there is a unique isomorphism $h: S^{-1} A \rightarrow B$ such that $g=h \circ f$ (where $f:a\in A\mapsto a/1\in S^{-1}A$).

Point 1 is clear. To prove point 2, take a smooth function $s\in C^\infty(M)$ whose set of zeroes is exactly $M\setminus U$ (such $s$ exists by Theorem 2.29 of J. M. Lee, Introduction to Smooth Manifolds, 2nd ed.). If $f\in C^\infty(M)$ is such that $f|_U=0$, then $sf=0$. However, I'm unable to show point 3, nor come up with a counterexample. If one now considers $f\in C^\infty(U)$ and takes a partition of unity $\{\rho_U,\rho_M\}$ subordinated to the open cover $\{U,M\}$, then one cannot write $f=\frac{f\rho_U}{\rho_U}$, since $\rho_U\not\in S$ in general.



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