local property of tangent vector Given a smooth manifold $M$ and a point $p \in U$, $U \subset M$ open, my book says there is a linear isomorphism between vector spaces $T_pU$ and $T_pM$ by:
$v \rightarrow \tilde{v}$ where for any $f \in C^\infty(M)$, $\tilde{v}(f) = v(f|_U)$. Now I can see the linear part, but I cannot see the bijection.
Edit: Background lemma that might help:
Let $v \in T_pM$. If $f,g \in C^\infty(M)$ are equal on a neighborhood of $p$, then $v(f) =  v(g)$. If $h \in C^\infty(M)$ is constant on a neighborhood of $p$, then $v(h) = 0$
 A: You can use the following lemma to prove the surjectivity of $ \bar {v} $:
Let $M$ be a differentiable manifold, $U \subset M$ open and $g: U \longrightarrow \mathbb{R}$, with $g \in C^{\infty}(U)$. If $p \in U$ then there exists a function $h \in C^{\infty} (M)$ such that $h = g$ in a neighborhood of $p$ contained in $U$.
And use that if $ f, g \in C^{\infty} (M) $ are equal on a neighborhood of $ p $, then $ v (f) = v (g) $ to show the injectivity of $ \bar {v} $.
A: The "background lemma" answers your question.
A tangent vector in $T_pM$ is a derivation $v : C^\infty(M) \to \mathbb R$ at the point $p$. The lemma says that that we may regard $v$ as a function $C^\infty(M)/\sim_p \phantom{.} \to \mathbb R$, where $f \sim_p g$ if $f$ and $g$ are equal on a neighborhood of $p$.
You have already defined a linear map $\phi: T_pU \to T_pM, \phi(v)(f) = v(f|_U)$. Let us define $\psi : T_pM \to T_pU$ as follows:
Let $v \in T_pM$ and $f \in C^\infty(U)$. Choose an open neigborhood $V$ of $p$ in $M$ such that $\bar V \subset U$. You know that $f \mid_{\bar V}$ has a smooth extension $\tilde f : M \to \mathbb R$. There are many choices of $V$ and many choices of extensions $\tilde f$ of $f \mid_{\bar V}$, but clearly all these $\tilde f$ are equivalent with respect to $\sim_p$. Thus $v(\tilde f)$  is well-defined. Now define  $\psi(v)(f) = v(\tilde f)$. It is clear that $\psi \circ \phi =id$ and $\phi \circ \psi =id$ which proves that $\phi$ is an isomorphism.
