Differential $1$ forms as linear maps Let $M$ be a smooth manifold and $\Gamma(M)$ denotes the collection of all vector fields on $M$ and $C^{\infty}(M)$ denotes the set of all smooth functions on $M$.
Let $\omega$ be a $1$ form on $M$ i.e., $\omega:M\rightarrow T^*M$.
Let $X$ be a vector field on $M$ then $\omega(X):M\rightarrow \mathbb{R}$ defined as $p\mapsto \omega(p)(X_p)$ is a smooth map. So, any differential one form gives a map $\Gamma(M)\rightarrow C^{\infty}(M)$. More over, this map is $C^{\infty}(M)$ linear.
I am trying to see if other direction is also true. That is, given a $C^{\infty}(M)$ linear map $F:\Gamma(M)\rightarrow C^{\infty}(M)$ do we get a $1$ form on $M$. I think it is true, I am just now able to write details.
We need to define $1$ form $\omega:M\rightarrow T^*M$. Let $p\in M$ then, we need to define $\omega(p):T_pM\rightarrow \mathbb{R}$. 
Let $v\in T_pM$ then, there exists a vector field $X$ on $M$ (I forgot how do you extend a tangent vector to a vector field. Any quick remark would be welcome but that is not the main question). Then, define $$\omega(p)(v)=F(X)(p).$$
I am not able to show that this is well defined, suppose there is another vector field $Y$ whose value at $p$ is $v$ then, I do not see why $F(X)(p)=F(Y)(p)$. 
Any suggestion is welcome. This is definitely not difficult but just that I am not able to see.
 A: This is a standard argument. It suffices to prove that if $X(p) = 0$, then $F(X)(p)=0$. 
First, if $X$ is actually $0$ on a neighborhood $U$ of $p$, choose a bump function $\phi$ with $\phi=1$ on a smaller neighborhood of $p$ and $\phi$ supported in $U$. Then $\phi X$ is identically $0$, so you can check that $F(\phi X) = 0$. In particular, $F(\phi X)(p) = \phi(p) F(X)(p) = F(X)(p) = 0$.
Now suppose just that $X(p)=0$ and choose local coordinates on $M$ on some open set $U$ containing $p$. Using the same bump function $\phi$, consider $\tilde X = X - \phi^2 X$. Then $\tilde X$ is $0$ on a neighborhood of $p$, and so (by our first paragraph) $F(\tilde X)(p)=0$. But $F(\tilde X)(p) = F(X)(p)-F(\phi^2 X)(p)$, so $F(X)(p) = F(\phi^2 X)(p)$. Write $X = \sum a^i \frac{\partial}{\partial x^i}$ on $U$ and note that $\phi^2 X = \sum (\phi a^i) \left(\phi\frac{\partial}{\partial x^i}\right)$ is globally defined. Then $F(\phi^2 X)(p) = \sum \phi(p)a^i(p) F(\phi\frac{\partial}{\partial x^i}) = 0$ since all $a^i(p) = 0$. Thus $F(X)(p)=0$, as we wished to show.
