# Is the ring of smooth fcts over an open set $U$ the localization of the smooth functions on $M$ with respect to the nowhere vanishing fcts on $U$?

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.