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Let $M$ be a smooth $n$-dimensional manifold. Denote by $\Lambda^{k}(T^*M)$ the $k$-th exterior power of the cotangent bundle of $M$, and let $\Omega^{k}(M)=\Gamma\big( \Lambda^{k}(T^*M)\big)$ be the space of differential $k$-forms on $M$.

Given $\omega \in \Omega^k(M)$, we can view it as a map $\omega: M \to \Lambda^{k}(T^*M)$, and then take its standard derivative (i.e derivative of a map between manifolds): $$ T\omega:TM \to T \big( \Lambda^{k}(T^*M) \big)$$ We can view $T\omega$ as a section of the vector bundle $T^*M \otimes w^*T\big( \Lambda^{k}(T^*M) \big)$ (the Hom bundle over $M$).

Question: Is there a map $S:\Gamma\bigg(T^*M \otimes w^*T\big( \Lambda^{k}(T^*M)\big) \bigg) \to \Omega^{k+1}(M)=\Gamma\big( \Lambda^{k+1}(T^*M)\big)$ such that $d\omega = S ( T\omega)$?

If so, does it come from a bundle-homomorphism $T^*M \otimes w^*T\big( \Lambda^{k}(T^*M) \big) \to \Lambda^{k+1}(T^*M)$?

In other words, I am asking whether or not all the information of the exterior derivative is already encoded in the "standard" derivative.

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    I don't understand how you want to view $\omega$ as a function $M\to\Omega^k(M)$. You can view $\omega$ as a smooth map $M\to T^*M$ and then $d\omega$ is determined by $T\omega:TM\to TT^*M$. An efficient formulation of properties of this type is that $d\omega(x)$ only depends on the 1-jet of $\omega$ in $x$. – 2017-01-11
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    You are right, it was a typo: (fixed now) - I meant to view $\omega$ as a map $\omega: M \to \Lambda^{k}(T^*M)$. I am not sure how are you viewing it as a map $M \to T^*M$ though... – 2017-01-11
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    You are completely right, it should be $\omega:M\to\Lambda^kT^*M$ and $T\omega:TM\to T(\Lambda^kT^*M)$, but $d\omega$ is determined by $T\omega$. This is basically just the statement that $d$ is a first order operator. – 2017-01-12

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