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Let $M$ be a smooth manifold and let $A=C^\infty(M)$. In this question, it is observed that the map $\Omega^1_k(A)\to \Omega^1(M)$ from the Kähler differentials of $A$ to the 1-forms of $M$ is not an isomorphism.

However, in the comments the following remark and question are raised:

Remark: $\Omega^1(M)$ can be identified with the double dual of $\Omega^1_k(A)$. This amounts to: a) vector fields on $M$ are derivations of $C^\infty(M)$, b) differential forms on $M$ are the $C^\infty(M)$-dual of vector fields, and c) the universal property of Kähler differentials.

Now, there is a canonical surjective map $\Omega^1_k(A)\to \Omega^1(M)$ described in Georges Elencwajg's answer. There is also the canonical map $\Omega^1_k(A) \to \Omega^1_k(A)^{**}$.

Question: do these maps commute with the isomorphism $\Omega^1(M)\cong \Omega^1_k(A)^{**}$?

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Bruno Stonek
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    I think it is easy to see that both these maps correspond to the same derivation $A\to T(M)^*$. The first one sends $f$ to $X\mapsto Xf$ by definition of exterior differential, while the second does the same thing by the identification of $T(M)$ with the dual of $\Omega^1_k(A)$. – Denis Nardin May 02 '16 at 17:19

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