§ Equivalent Definitions of One-forms
Definition I. Let \(M\) be a manifold and consider the cotangent bundle \((T^*M, \pi,M)\).
A one-form on \(M\) is a section \(\omega\in\Gamma(T^*M):=\Omega^1(M)\).
Definition II. Let \(M\) be a manifold and consider the cotangent bundle \((T^*M, \pi,M)\).
A one-form on \(M\) is a map \(\omega\colon \Gamma(TM)\to \mathcal{C}^\infty(M)\).
Proposition. Both definitions of one-forms are equivalent.
Proof. Suppose that a one-form is a section \(\omega\in \Gamma(T^*M)\). Then, \(\omega\colon M\to T^*M\),
so that \(\omega(p):=(p, \omega_p)\), where \(\omega_{p}\in T^*_pM\). We can define a map \(\Omega \colon \Gamma(TM)\to\mathcal{C}^\infty(M)\)
such that for a section \(X\in\Gamma(TM)\), given by \(X(p)=(p,X_p)\) where \(X_p\in T_pM\), we have:
$$\begin{equation}
\Omega(X):= \omega_p(X_p) \quad \forall p\in M
\end{equation}$$
After which we simply rename \(\Omega\equiv\omega\) and we have shown one implication.
It remains to show, that if \(\omega\colon \Gamma(TM)\to \mathcal{C}^\infty(M)\), then we can define a smooth section
\(\Omega\in\Gamma(T^*M)\). We do so as follows: for each \(p\in M\), we define \(\Omega(p)=(p,\Omega_p)\), where
\(\Omega_p\in T^*_pM\) is defined via \(\Omega_p(X_p):=\omega(X)(p)\), where the vector field \(X\) is constructed
such that \(X(p)=(p,X_p)\). \(\blacksquare\)