The set of all covectors fields and the dual to the set of all vector fields, isomorphic? Given a smooth manifold $M$, the space of smooth sections of the tangent bundle is denoted by $\Gamma(TM)$ and the space of smooth sections of the cotangent bundle $\Gamma(T^* M)$. Both $\Gamma(TM)$ and $\Gamma(T^* M)$ are $C^{\infty}(M)$-modules. Then, there is also the dual to $\Gamma(TM)$, that we denote $\Gamma(TM)^*$, another $C^{\infty}(M)$-module. 
How can I prove that $\Gamma(TM)^*$ and $\Gamma(T^*M)$ are isomorphic as $C^{\infty}(M)$-modules?
 A: There's a natural map $\Phi : \Gamma(T^* M) \to \Gamma(TM)^*$ given by
$$ \Phi(\alpha)(X) = \alpha(X) $$
This is obviously $C^\infty(M)$-linear, and it's not hard to show that it is injective. The (slightly) more difficult part is to show surjectivity. Consider some $\xi \in \Gamma(TM)^*$. We want to define $\alpha \in \Gamma(T^* M)$ the following way: given a tangent vector $X_{x_0} \in T_{x_0} M$, we extend it to a (smooth) vector field $X \in \Gamma(TM)$, and then set $\alpha(X_{x_0}) = \xi(X)(x_0)$.
This would clearly do the trick, but it's not obvious that $\alpha$ is well-defined, because we made an arbitrary choice of how to extend $X_{x_0}$ into the vector field $X$. If you ponder it a bit you'll see that to show $\alpha$ is well-defined it suffices to show that, if $X_{x_0} = 0$, then for any extension $X$ we will have $\xi(X)(x_0) = 0$. This is where $\xi$ being $C^\infty(M)$-linear comes in. I'm gonna be a bit sketchy.
Using a bump function argument it's not hard to show that $\xi(X)(x_0)$ only depends on the values of $X$ in a neighborhood of $x_0$. This allows you to assume there's a local basis of tangent vectors $e_1, \ldots, e_n$ and so we can express $X = X^1 e_1 + \cdots + X^n e_n$ with each $X^i \in C^\infty(M)$. Note that $X^1(x_0) = \ldots = X^n(x_0) = 0$. Since
$$ \xi(X) = X^1 \xi(e_1) + \cdots + X^n \xi(e_n) $$
we clearly have $\xi(X)(x_0) = X^1(x_0) \xi(e_1)(x_0) + \cdots + X^n(x_0) \xi(e_n)(x_0) = 0$, which finishes the argument.
Observation: Don't forget to prove that $\alpha$ is actually smooth.
