For the Kodaira-Spencer complex $\Omega^{0,*}(T^{1,0})[[t]]$ on a compact complex manifold, with a Hermitian metric. It is well known that finding formal series solution $\Xi = \Xi_1 t^1 + \Xi_2 t^2 + \dots$ for the Maurer-Cartan (MC) equation with prescribed harmonic first order term $\Xi_1$ can be achieved by solving the equation: $$ \Xi = \Xi_1 -\frac{1}{2} \bar{\partial}^*G[\Xi, \Xi]. $$ The Green's operator is used to find 'good' representative for solving $\bar{\partial} \mu = \nu$ in the iteration process.
In a general DGLA $(L^*,d)$, together with an homotopy $H : L^* \rightarrow L^*[-1]$ retracting $L^*$ to its cohomology, like the operator $\bar{\partial}^* G$ in the case of Kodaira-Spencer DGLA. Suppose a MC element $\Xi$ is known to gauge equivalent to $0$, can one use the homotopy to fix a 'good' gauge equivalent $a = a_1 t + a_2 t^2 + \dots \in L^*[[t]]$ solving $$ e^{a} * 0 = -\frac{e^{ad_a} - id}{ad_a}(da) = \Xi\;\;?? $$ It seems to me that a naive thing to do would be solving $$ a = -H \big( (\frac{ad_a}{e^{ad_a} - id})^{-1} \Xi \big), $$ while I cannot show immediately the solution of second solution solve first one, as I cannot show the $$ d\big( (\frac{ad_a}{e^{ad_a} - id})^{-1} \Xi \big)=0, $$ properly possibly due to lack of knowledge in differential graded Lie algebras (dgLa). It should follows from the fact that $\breve{\Xi}$ satisfying the MC equation while I cannot have a systematic way of showing that.
