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In a nutshell, I wish to understand this post by Ted Shiffrin, the relevant bit of which reads

For example, suppose you have a surface $S$ in $\mathbb R^3$ that you can locally write as $f=0$, but you don't know how to do so globally. You can cover $S$ with open sets $U_i\subset\mathbb R^3$ on which you have smooth functions $f_i\colon U_i\to\mathbb R$ with $S\cap U_i = \{x\in U_i: f_i(x)=0\}$. Consider $\Phi = \{\phi_i\}$, where $\phi_i$ is supported in $U_i$. Then $f=\sum \phi_if_i$ will define a smooth function with $f=0$ on $S$. If you want $f$ to be zero only on $S$, you can take an additional open sets $U_0 = \mathbb R^3 - S$, set $f_0 = 1$, and throw $\phi_0f_0$ into your sum.

Formally, what does the bold statement mean?

Sam
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    I.e., locally the surface is the zero set of some function. The unit sphere is $f^{-1}(0)$ for $f(x,y,z)=x^2+y^2+z^2-1$. – pancini Jan 04 '24 at 04:03
  • "Locally" means "in a neighbourhood of any of its points." That is, for every $p \in S$ there is a function $f$ defined on a neighbourhood $U$ of $p$, such that $S \cap U = {x \in U: f(x) = 0}$. – Robert Israel Jan 04 '24 at 04:08
  • Perhaps off-topic in your question, but I would have written $f=\sum_i \phi_if_i^2$ instead. (I think this fixes most of the issue referenced in a comment) – Brian Moehring Jan 04 '24 at 04:24

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