Let $M$ be a smooth manifold and let $A=C^\infty(M).$ 

We consider module of Kahler differentials $\Omega_k(A)$ and module of 1-forms $\Omega^1(M).$ Denote Kahler differential by $d_k$ and classical differential on manifold $M$ by $d$.

From universal property of $\Omega_k(A)$ we know that $\Omega_k(A)^*\cong Der(A).$ More precisely there is an isomorphism of $A-$modules $\beta:Der(A)\to\Omega_k(A)^*$ such that

$$[\beta(X)](d_kf)=X(f)$$
for all $X\in Der(A)$ and $f\in A.$

On the other hand we know that $\frak{X}$$(M)\cong Der(A)$ and $\Omega^1(M)=\frak{X}$$(M)^*.$ Hence
$$\Omega_k(A)^{**}\cong Der(A)^*=\Omega^1(M).$$

Using $\beta$  we naturally get a map $\gamma:\Omega_k(A)\to\Omega^1(M)$ defined by composition:
$$\Omega_k(A)\stackrel{**}{\to}\Omega_k(A)^{**}\stackrel{\beta^*}{\to}\Omega^1(M).$$
If we compute $\gamma$ we see that
$$[\gamma(d_kf)](X)=[\beta^*(d_kf)^{**}](X)=(d_kf)^{**}(\beta(X))=[\beta(X)](d_kf)=X(f)=(df)(X)$$
for every $X\in Der(A)$ and $f\in A.$

> **Main problem.** How to prove that $\gamma$ is surjective?



Georges Elencwajg in his answer [to this question][1] wrote that

> There is an obvious surjective map $\Omega_k(A)\to\Omega^1(M)$ because the relations displayed above are valid in the classical interpretation of the calculus (Leibniz rule).

I do not understand this argument. In my opinion surjectivity of $\gamma$ holds for completely different reason. I would say that $\gamma$ is surjective because any 1-form $\alpha$ can be written as a finite sum
$$\alpha=\sum_{i=1}^p g_idf_i$$
for some $f_1,\dots,f_p,g_1,\dots,g_p\in A.$ From this we get that
$$\gamma\left(\sum_{i=1}^p g_id_kf_i\right)=\sum_{i=1}^p g_idf_i=\alpha.$$
Presenting $\alpha$ as a such finite sum I would explain using Serre-Swan theorem.

> **Actual question.** Is there a simpler argumentation of surjectivity of $\gamma$ or do I really need to rely on specific form of every 1-form $\alpha?$



  [1]: https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials