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See the article Kähler C^∞-differentials of smooth functions are differential 1-forms for the necessary background for this article, including the notions of C^∞-ring, C^∞-derivation, and Kähler C^∞-differential.
In algebraic geometry, (algebraic) differential forms on the Zariski spectrum of a [commutative ring (or a commutative -algebra ) can be defined as the free commutative differential graded algebra on .
This definition does not quite work for smooth manifolds: as already explained in the article Kähler C^∞-differentials of smooth functions are differential 1-forms, the notion of a Kähler differential must be refined in order to extract smooth differential 1-forms from the C^∞-ring of smooth functions on a smooth manifold .
Thus, in order to get the algebra of smooth differential forms, the notion of a commutative differential graded algebra must likewise be adjusted.
\begin{definition} A commutative differential graded C^∞-ring is a real commutative differential graded algebra whose degree 0 component is equipped with a structure of a C^∞-ring in such a way that the degree 0 differential is a C^∞-derivation. \end{definition}
With this definition, we can recover smooth differential forms in a manner similar to algebraic geometry, deducing the following consequence of the Dubuc–Kock theorem for Kähler C^∞-differentials.
\begin{theorem} The free commutative differential graded C^∞-ring on the C^∞-ring of smooth functions on a smooth manifold is canonically isomorphic to the differential graded algebra of smooth differential forms on . \end{theorem}
The Poincaré lemma becomes a trivial consequence of the above theorem.
\begin{proposition} For every , the canonical map
is a quasi-isomorphism of differential graded algebras. \end{proposition}
\begin{proof} (Copied from the MathOverflow answer.) The de Rham complex of a finite-dimensional smooth manifold is the free C^∞-dg-ring on the C^∞-ring . If is the underlying smooth manifold of a finite-dimensional real vector space , then is the free C^∞-ring on the vector space (the real dual of ). Thus, the de Rham complex of a finite-dimensional real vector space is the free C^∞-dg-ring on the vector space . This free C^∞-dg-ring is the free C^∞-dg-ring on the free cochain complex on the vector space . The latter cochain complex is simply with the identity differential. It is cochain homotopy equivalent to the zero cochain complex, and the free functor from cochain complexes to C^∞-dg-rings preserves cochain homotopy equivalences. Thus, the de Rham complex of the smooth manifold is cochain homotopy equivalent to the free C^∞-dg-ring on the zero cochain complex, i.e., in degree 0. \end{proof}
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