Let $A$ be a unital commutative algebra (say over complex numbers). Consider the multiplication map $m:A \otimes A \to A$ and put $\Omega^1_u(A)=\ker m$ to be the space of universal differential forms. We turn $A \otimes A$ into $A-A$ bimodule: left action is the natural one, the right is such that $d_u$ defined via $d_u(a)=1 \otimes a -a \otimes 1$ satisfies Leibniz rule. In more detail: $(x \otimes y) \cdot z=x \otimes yz-xy\otimes z$. We take $M:=(\Omega^1_u(A))^2$ be a subbimodule of $\Omega^1_u(A)$ generated by the terms of the form $d_u(a)d_u(b)$. If $A$ is commutative one then shows that the quotient $\Omega_{ab}^1(A):=\Omega^1_u(A)/M$ is symmetric $A$ bimodule therefore can be regarded as left $A$ module. Put $d_{ab}:=1 \otimes a -a \otimes 1 (\mod M)$. This is easily seen to be a derivation. The pair $(\Omega^1_{ab}(A),d_{ab})$ is universal in the following sense: given any pair consisting of $A$ module $E$ and a differential (derivation) $d:A \to E$ one can find unique $A$-module map $\varphi_d$ such that $d=\varphi_d \circ d_{ab}$. This establishes a bijection $Hom_A(\Omega^1_{ab}(A),A)=(\Omega^1_{ab}(A))' \cong Der(A,A)$ which can easily be seen to be an $A$ module map (here $Der$ stands for the space of all derivations). One can consider $\Omega^*_{ab}(A)$ defined as an exterior power (over $A$) with the unique extension of differential (still to be denoted by $d_{ab}$) via (graded) Leibniz rule and the requirment $d^2_{ab}=0$. Let us now specify $A=C^{\infty}(M)$ where $M$ is a smooth manifold. Then $Der(A,A)=\mathcal{X}(M)$ is a space of vector fields. I found the following remarkable theorem which states that in this case one can identify $\Omega^*_{ab}(A)$ with the complex of de Rham forms. Not everything in the proof is clear for me:
Question 1: As I explained we have $\mathcal{X}(M) \cong (\Omega_{ab}^1(A))'$ where $A=C^{\infty}(M)$. Is it obvious that this would imply $\Omega^1_{ab}(A) \cong \Omega^1_{dR}(M)$? The problem is that I don't know whether $\Omega^1_{ab}(A)$ satisfies $\Omega^1_{ab}(A)=(\Omega^1_{ab}(A))''$.
Suppose that the answer for the first question is affirmative: we get as far as I understood only the isomorphism of $A$-modules. So my second qestion is the following:
Question 2 Does the fact $\Omega^1_{ab}(A) \cong \Omega^1_{dR}(M)$ imply the isomorphism of $\Omega^*_{dR}(M)$ and $\Omega^*_{ab}(A)$ and if the anser is positive-this is an isomorphism of what structures? I would be very suprised if we have an isomorphism of differential graded algebras.
