Let $M$ be a symplectic manifold with a non-degenerate closed $2$-form $\omega.$ Then for any $f \in C^{\infty} (M)$ and for any vector field $u \in \mathfrak {X} (M)$ there exists a vector field $V_{f} \in \mathfrak {X} (M)$ such that $df (u) = \omega (u, V_{f}).$
I know that any differential $2$-form on $M$ can be thought of as a $C^{\infty} (M)$-bilinear, skew-symmetric map from $\mathfrak {X} (M) \times \mathfrak {X} (M) \longrightarrow C^{\infty} (M).$ So the $1$-form $\omega (u, \cdot) : \mathfrak {X} (M) \longrightarrow C^{\infty} (M)$ is a $C^{\infty} (M)$-linear map for any vector field $u \in \mathfrak {X} (M).$ Now how do I show that for any $f \in C^{\infty} (M)$ there exists a vector field $V_{f} \in \mathfrak {X} (M)$ such that $df (u) = \omega (u, V_{f})$ or in other words the smooth function $df (u)$ is actually in the image of the $1$-form $\omega (u, \cdot)\ $? The above exercise would help us concluding that any symplectic manifold is a Poisson manifold.
Any help in this regard would be greatly appreciated. Thanks for your time.
Source $:$ Lecture Notes on CQG by Pavel Etingof.