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I'm reading through some notes (Link to pdf, page 2) proving the isomorphism $\mathfrak X(M)\to\operatorname{Der}(C^\infty(M,\mathbb R))$ between vector fields and derivations.

In order to prove that all derivations $D$ can be locally written as $D=\sum_i a^i \partial_i$ for the smooth functions $a^i=D(x^i)$, the author observes that any smooth function on $\mathbb R^n$ may be written as $$f(z) = f(y) + \sum_i (x^i(z)-x^i(y))g_i(z), \quad\text{where } g_i(y)=\partial_i f(y).$$ Where does this relation come from?

glS
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    Your description of $x^i$ is more not quite correct. Perhaps more simply, any map $\varphi:M\to\mathbb{R}^n$ can be equivalently thought of as $n$ separate smooth maps $x^i:M\to\mathbb{R}$. In terms of projections, this would be $x^i=\pi_i\circ\varphi$, where $\pi_i$ is the projection onto the $i$-th component. – Kajelad Jun 17 '20 at 07:16
  • @Kajelad thanks. So in my case, $\varphi=f$, right? Or in other words, $x^i$ are what I was denoting later in my post with $f_i$? – glS Jul 07 '20 at 14:32
  • By $\varphi$ I was refering to the coordinate chart; the functions $x^i$ are just the component functions of the chart. In $(1)$ $f$ seems to refer to a map in $C^\infty(\mathbb{R^n},\mathbb{R})$, so I'm not sure what you mean by splitting $f$ into components. – Kajelad Jul 07 '20 at 18:17
  • This is basically Taylor expansion, but there's no remainder term so it's not literally correct, I think. You should replace the constants $\partial_i f (y)$ with suitable functions $h_i (z)$. – Zhen Lin Jul 08 '20 at 12:35
  • Hint: Consider the function $h(t):= f(tz + (1-t)y) - f(y)$ and use the approach in my answer here: https://math.stackexchange.com/questions/62742/little-b%C3%A9zout-theorem-for-smooth-functions – Jason DeVito - on hiatus Jul 08 '20 at 13:57

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