Krdytkyu

I want to understand why Kähler manifolds are formal.

This was first proved by Deligne, Griffiths, Morgan, Sullivan

Let $mathcal M$ be a minimal differential
algebra and $H^*(mathcal M)$ the cohomology
of $mathcal M$, viewed as a
differential algebra with $d =0$.

Definition.

1. 
   $mathcal M$ is formal if there is a map of differential algebras $psi: mathcal M to H^*(mathcal M)$ inducing the identity
   on cohomology.


2. The homotopy type of a differential
   algebra $mathcal A$ is a formal consequence of its cohomology if its minimal model is formal.


3. The real (or complex) homotopy
   type of a manifold $M$ is a formal consequence of the
   cohomology $M$ if the de Rham homotopy type of the real (or
   complex) forms
   $mathcal E$ is a formal consequence of its cohomology.

In section 6, the following (main) theorem is proved:

Let $M$ be a compact complex manifold for which the
$dd^c$-lemma holds (e.g. $M$ Kähler, or $M$ a Moisezon space).
Then the real homotopy type of $M$ is a formal consequence of
the cohomology ring $H^*(M; mathbb R)$

Let ${mathcal E^*_M,d}$ be the de-Rham complex on $M$, ${mathcal E^c_M,d}$ the subcomplex of $d^c$-closed forms and ${H_{d^c},d}$ the quotient complex $mathcal E_M^c/d^c mathcal E_M$.

Using the $dd^c$-lemma ($partial barpartial$-lemma), it is an easy calculation, that the natural maps
$${mathcal E^*_M,d} stackrel ileftarrow {mathcal E^c_M,d} stackrel pto {H_{d^c},d} $$
are quasi-isomorphisms and that the differential on $H_{d^c}(M)$ vanishes.

In the proof, the theorem follows immediately from the above. I don't see how and I am a little confused that there were no reasons given why the theorem follows. Only "This proves the claim and consequently [part (1)] of the theorem".

The maps seem to be maps of differential algebras, so this is fine.

But as far as I see, the theorem only follows if we can replace ${H_{d^c},d=0}$ with ${H_M,d=0}$.

Update/ Solution:

I repeat: The theorem only follows if we can replace ${H_{d^c},d=0}$ with ${H_M,d=0}$.

But the above argument shows, that the cohomologies are isomorphic. More precisely, the isomorphism is induced by $i$ and $p$.

A dg-algebra with vanishing differential is its own cohomology. The morphisms $i_*,rho_*$ are isomorphisms between the cohomology of $(mathscr{E}^*_M,d)$ -- which is $H^*(M)$ -- and the cohomology of $(H_{d^c}, d = 0)$ -- which is $H_{d^c}$, because the differential vanishes.

More generally if a dg-algebra $A$ is quasi-isomorphic to any dg-algebra $B$ with vanishing differential, then $B$ is isomorphic to the cohomology of $A$, pretty much by definition; hence $A$ is formal.