Let $X$ be a real manifold, and $\mathcal{O}_X$ be the sheaf of smooth functions. We define a sheaf of $1$-forms as \begin{align*} \Omega_X\colon=\operatorname{\mathcal{Hom}}(\mathcal{O}_X(TM),\mathcal{O}_X), \end{align*} where the tangent bundle of $X$ is denoted by $TX$, and $\mathcal{O}_X(E)$ is the sheaf of sections considered as an $\mathcal{O}_X$-module for an arbitrary vector bundle $E$.
And so we define a sheaf of $p$-forms as \begin{align*} \Omega_X^p\colon=\bigwedge^p\Omega_X, \end{align*} which is the sheaf associated to the following preshef: \begin{align*} U\mapsto\bigwedge^p_{\Gamma(U,\mathcal{O}_X)}\Gamma(U,\Omega_X). \end{align*}
Question: Is $\Omega_X^p$ the sheaf associated to the preshef $U\mapsto\operatorname{Hom}_{\Gamma(U,\mathcal{O}_X)}\left(\bigwedge^p_{\Gamma(U,\mathcal{O}_X)}\Gamma(U,TX),\Gamma(U,\mathcal{O}_X)\right)$,or are these two (pre)sheaves equal?
If this statement is not true, the definition of an exterior derivative is seemingly not consistent. \begin{equation*} d\omega(V_0,\ldots,V_p)=\sum_i(-1)^iV_i(\omega(V_0,\ldots,\widehat{V_i},\ldots,V_p))\\ +\sum_{i<j}(-1)^{i+j}\omega([V_i,V_j],V_0,\ldots,\widehat{V_i},\ldots,\widehat{V_j},\ldots,V_p) \end{equation*}
Now using these facts:
- There exists an equivalence between the category of vector bundles on $X$ and the category of locally free sheaves of finite rank
- Since $X$ is a smooth, morphisms $E\rightarrow F$ between any vector bundles can be identified with morphisms of $\Gamma(X,\mathcal{O}_X)$-modules, $\Gamma(X,E)\rightarrow\Gamma(X,F)$
we can deduce these isomorphisms: \begin{align*} \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(E),\mathcal{O}_X(F))\cong\operatorname{Hom}_{}(E,F)\cong\operatorname{Hom}_{\Gamma(X,\mathcal{O}_X)}(\Gamma(X,E),\Gamma(X,F)) \end{align*} This results for any chart $U\subset X$, we get \begin{align*} \Gamma(U,\Omega_X^p)=\bigwedge^p_{\Gamma(U,\mathcal{O}_X)}\Gamma(U,\Omega_X), \end{align*} and \begin{align*} \Gamma(U,\Omega_X^p)=\operatorname{Hom}_{\Gamma(U,\mathcal{O}_X)}\left(\bigwedge^p_{\Gamma(U,\mathcal{O}_X)}\Gamma(U,TX),\Gamma(U,\mathcal{O}_X)\right). \end{align*} However, I don't know whether these are true for arbitrary open subset $U\subset X$.
Edit 2019-10-23
By the isomorphisms \begin{align*} \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X(E),\mathcal{O}_X(F))\cong\operatorname{Hom}_{}(E,F)\cong\operatorname{Hom}_{\Gamma(X,\mathcal{O}_X)}(\Gamma(X,E),\Gamma(X,F)), \end{align*} we can deduce that the three functors $\mathcal{O}_X$, ${}^*$, $\bigwedge^p$ defined on the category of vector bundles commute with each other (${}^*$ stands for a dualization). This implies that \begin{align*} \Omega_X^p=\bigwedge^p\mathcal{O}_X\left(TX\right)^*=\mathcal{O}_X\left(\bigwedge^pTX\right)^*, \end{align*} and that \begin{align*} \Gamma(U,\Omega_X^p)=\operatorname{Hom}_{\Gamma(U,\mathcal{O}_X)}\left(\Gamma\left(U,\bigwedge^pTX\right),\Gamma(U,\mathcal{O}_X)\right) \end{align*} holds for arbitrary open subset $U\subset X$. So the only thing to show is the following equation, but I cannot show this.
Question: Does the equality holds? \begin{align*} \bigwedge^p_{\Gamma(U,\mathcal{O}_X)}\Gamma(U,TX)=\Gamma\left(U,\bigwedge^pTX\right) \end{align*}
Could anyone help me?