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Let $M$ be a smooth ( real) manifold, if $ p\in M$ and $f\in C^{\infty}(U)$ ($U$ is an open subset of $M$), the symbol $[ f]_p$ indicates the smooth germ of $f$ at $p$ . Consider the following set $$S=\{[f]_p\; :\; f\in C^{\infty}(M)\}$$

If $C^{\infty}_p(M)$ is the algebra of all smooth germs at $p$, clearly $S\subseteq C^{\infty}_p(M)$, but is it true or not that $S=C^{\infty}_p(M)$ ? Does exist a smooth function defined over an open subset of $M$ that doesn't coincide around $p$ with an element of $C^{\infty}(M)$?

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up vote 4 down vote accepted

If $U$ is an open neighborhood of $p$ and $f\in C^\infty(U)$, then there is a function $g\in C^\infty(M)$ such that $f = g$ in some neighborhood of $p$. To prove this, one constructs a smooth function $\chi\colon M\to [0,1]$ such that $\chi$ is supported in a compact subset of $U$ and $\chi\equiv 1$ in a small neighborhood of $p$ (this $\chi$ is called a "bump" function). You can then set $g = \chi f$ on $U$ and $g\equiv 0$ outside of $U$. This function $g$ is smooth, and agrees with $f$ on a neighborhood of $p$.

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