Why do free DGAs in a morphism often give the same (co) homology as other (co) homologies?

Here is an example:

**Example:** Leave $ R $ be a ring, let $ A $ bean $ R $-algebra, leave $ M $ bean $ A $-module and leave $ d: A rightarrow M $ Be a derivation. There is a categorical construction $ Lambda (d) $, the free anticomutative DGA given an element of $ text {Der} _R (A, M) $, which is a cochain complex with multiplication and unity in the category of cochain complexes, defined by the configuration $ Lambda (d) ^ i = 0 $ for $ i <0 $, $ Lambda (d) ^ i = Lambda ^ i_A M $ in degree $ n $ for $ n geq 0 $, and where the differential is specified exclusively by setting

$$ d (x wedge y) = d (x) y + (-1) ^ { text {deg} (x) text {deg} (y)} x d (y) $$

By free, I mean that it is part of an attachment between a category whose objects are derivations and a category whose objects are DGA.

In particular (and very interesting *in particular*), Yes $ X $ it is a smooth collector $ C ^ { infty} (X) $ that's $ mathbb {R} $-algebra of $ C ^ { infty} $-functions to $ mathbb {R} $Y $ T ^ * (X) $ is your cotangent space, then we have a derivation $ d: C ^ { infty} (X) rightarrow T ^ * (X) $. Taking the left attachment I defined above, you get the free anticomutative DGA in this derivation, which many of us know and love as the De Rham complex $ Omega ^ * (X) $.

The interesting thing is that the cohomology of this resolution is the same as the cohomology of others:

$$ H_ {dRh} ^ n (X) cong H ^ n_ {sing} (X; mathbb {R}) $$

$$ H_ {dRh} ^ n (X) cong hat {H} ^ n (X, mathbb {R}) $$

Where the last cohomology group is supposed to be Čech cohomology.

Here is another example:

**Example:** Leave $ R $ be a ring and leave $ A $ bean $ R $-algebra. Leave $ M $ bean $ A $-module and leave $ d: M rightarrow A $ bean $ A $-Map of modules. We can form the free DGA (this time, a chain complex, not a chain complex, and this time resulting in homology, not cohomology), with $ M $ in degree $ 1 $ Y $ R $ in degree $ 0 $. We can form the Koszul complex, which is the free DGA given the map $ d: M rightarrow A $. Is $ 0 $ in negative degrees, and $ Lambda ^ i_A M $ in degree $ i $. The differential is uniquely specified by requiring that $ d (x wedge y) = d (x) y + (-1) ^ { text {deg} (x) text {deg} (y)} x d (y) $.

Yes $ M $ it's free with base $ x_1, …, x_n $ Y $ x_i $ map to a regular sequence $ a_i $ in $ R $ so the resolution of koszul is acyclic except in degree $ 0 $and therefore it is almost isomorphic to $ R / (x_1, …, x_n) $. Therefore, this resolution can be used to calculate $ text {Tor} $ Y $ text {Ext} $.

These can only be related distantly. In my opinion, it is mainly the first thing that is explained.

I am particularly interested in any answer that can explain the first example in all its generality, that is, to include some theorem about the cohomology of free DGA given a derivation.