If $(M^{2n}, \omega^2)$ is a symplectic manifold, then for each tangent vector $v \in T_xM$ for some $x \in M$ we may associate a $1$-form $\omega_v^1 \in T_x^* M$ by $\omega_v^1(u) = \omega^2(u,v)$.
The map carrying a vector to it's associated $1$-form is an isomorphism. My issue is with proving this.
Showing that it is injective is simple enough by looking at the kernal. It is surjectivity that I am stuck on. I'm trying to show that given an arbitrary $1$-form on $T_x M$, there is a vector, such that the image of this vector under the map constructed above, is the form.
If you can show that it is injective, then it is surjective since both spaces have the same dimension.