Show that two $C^{∞}$ vector fields $X$ and $Y$ on a manifold $M$ are equal if and only if for every $C^{∞}$ function $f$ on $M$,we have $Xf =Yf$.
I have sone one direction of the proof.
let $p ∈ M$. To show that $Xp = Yp$, it suffices to show that $Xp[h] = Yp[h]$ for any germ $[h]$ of $C^{∞}$ functions in $C^{∞}_{p} (M)$. Suppose $h: U → \Bbb R$ is a C^{∞} function that represents the germ [h]. We can extend it to a C^{∞} function $h : M → \Bbb R$ by multiplying it by a C∞ bump function supported in U that is identically 1 in a neighborhood of p. By hypothesis, $X h ̃ = Y h ̃$ . Hence, $X_{p} h ̃ = ( X h ̃)_{p} = ( Y h ̃ )_{ p} = Y _{p }h ̃$ . Because $h ̃ = h $ in a neighborhood of p, we have Xph = Xph ̃ and Yph = Yph ̃. It follows from (14.1.1) that $X_{p}h = Y_{p}h$. Thus, $X_{p} = Y_{p}$. Since p is an arbitrary point of $M$, the two vector fields X and Y are equal.
I have done the second direction of the proof. Hopefully, it is true. Please can you show my mistakes? Thank you
Let $(x_1, \dots, x_n)$ be a coordinate system centred on $p$ and define $f^p_i(x) = x_i$ locally around $p$ and multiply it by a suitable bump function to extend it to a $C^{\infty}$ function over $M$. Then $Y_i = (Y f^p_i)_p = (X f^p_i)_p = X_i$ and so $X_p = Y_p$ for all $p$. Therefore $X = Y$.