I am trying to understanding a proof of a proposition about the tangent space to an open manifold.
Let $M$ be a smooth manifold, $\iota:U\rightarrow M$ the inclusion map for some open $U\subset M$. For every $p\in U$ the differential $d\iota_p:T_pU\rightarrow T_pM$ is an isomorphism.
When proving the surjectivity of $d\iota_p$ the following is said
Suppose $w\in T_pM$ is arbitrary. Define an operator $v:\mathcal{C}^\infty(U)\rightarrow\mathbb{R}$ by setting $vf=w\tilde{f}$, where $\tilde{f}$ us any smooth function on all of $M$ with $f_{|\overline{B}}=\tilde{f}_{|\overline{B}}$, where $B$ is a neighbourhood of $p$, such that $\overline{B}\subset U$. Because $vf=v\tilde{f}$ according to the last proposition, $vf$ is independent of the choice of $\tilde{f}$, so $v$ is well defined.
I do not understand the last bit. The proposition talks about one and the same derivation $v$ and states that it is locally defined - not about two different derivations $v,w$.
Can somenody elaborate?
The author define an operator $v : C^{\infty}(U) \to \mathbb{R}$ by setting $vf:=w\tilde{f}$, where $\tilde{f}$ is $\textbf{any}$ smooth function on $M$ that agree with $f$ on some closure of a nbhd $\bar{B} \subseteq U$ and $w \in T_pM$.
Now, there are many extension of a function $f : U \to \mathbb{R}$ allowed (by definition), say we have $\tilde{f}_1,\tilde{f}_2 \in C^{\infty}(M)$.
If $w\tilde{f}_1 \neq w\tilde{f}_2$, then we have different value for $vf$ and hence the map $v$ is not well-defined. But, by Prop.3.8, we saw that any tangent vector $V \in T_pM$ is act locally. That is $Vf=Vg$ for any function $f,g \in C^{\infty}(M)$ that agree on some nbd of $p$. In our case $w\tilde{f}_1 = w\tilde{f}_2$ since by hypothesis $\tilde{f}_1|_{\bar{B}} =f|_{\bar{B}}= \tilde{f}_2|_{\bar{B}}$. Therefore any choices of $\tilde{f}$ in our definition $vf:=w\tilde{f}$ will do, so $v$ is well-defined.