 Solution

Define the pointwise representation of one-forms by vector-field functionals byIt is $\mathbb{R}$-linear, its value is a smooth function, and $X_p=0$ implies $\theta (\omega )(X)(p)=0$. Moreoverso it is a map of $C^{\infty }(M)$-modules. This is the properly typed scalar-multiplication property: the same symbol for a form and its associated functional is understood through $\theta$.

To construct its inverse, let $\alpha \in W$ and $v\in T_{p}M$. Choose a global smooth vector field $X$ with $X_p=v$; a local coordinate field multiplied by a smooth bump function supplies such an extension. SetThis is well-defined: if $X_p=Y_p$, then $X-Y$ vanishes at $p$, so $\alpha (X-Y)(p)=0$. Real linearity of $\alpha$ makes ${\omega }_p$ a covector. In fact the pointwise vanishing assumption also forcesbecause $(f-f(p))X$ vanishes at $p$. Thus $\alpha (fX)=f\alpha (X)$ for every smooth $f$, even though only real linearity was initially imposed.

It remains to prove smoothness, rather than assume continuity of an abstract linear map. Near a fixed point, choose global fields $E_1,\dots ,E_n$ equal to the coordinate basis on a smaller neighborhood, using a bump function equal to one there. On that neighborhood,All coefficients are smooth by the definition of $W$, so $\omega$ is a smooth differential one-form. The construction holds near every point, hence gives $\omega \in {\Omega }^1(M)$ with $\alpha (X)=\omega (X)$ globally.

Every tangent vector can be realized by a global field, so $\theta (\omega )=0$ forces $\omega =0$. The inverse just constructed gives surjectivity. The evaluation map is a natural $C^{\infty }(M)$-module isomorphism ${\Omega }^1(M)\cong W$. No auxiliary metric or connection was chosen.