Defining the universal twisting morphism of the bar construction

Question

I'm currently studying bar-cobar adjunction in the simplest case of algebras and coalgebras. I'm stuck in understanding of universal twisting morphism. $\pi: BA \to A$, more explicitly - $\pi: BA = T_c(s\overline{A}) \to s\overline{A} \to A$. So,I guess $\pi(a_1 \otimes ...\otimes a_n) = 0;\pi(a) = a$ is it true? And second question is that why is $\pi$ is a twisting morphism, There are two conditions for it $\pi$ is of a degree -1 and $\pi$ is a solution to a Maurer-Cartan equation. But I dont get why is $deg(\pi)=-1$ my literature is Algebraic Operads by J.-L. Loday and B.Valette. page 37.

Answer 1

The confusion arises because homogeneous elements of $T(s\bar A)$ are not of the form $a_1\otimes \cdots \otimes a_n$, but of the form $sa_1\otimes \cdots \otimes sa_n$. That's why it's the tensor coalgebra on $s\bar A$, not on $\bar A$. Thus, if $a$ is a degree $n$ element of $A$, then $sa$ is a degree $n+1$ element of $sA$. So, your definition of $\pi$ is correct, except for the suspension symbols.

Regarding the Maurer--Cartan equation, just recall the definition of the product and differential in the Hom chain complex $\operatorname{Hom}(BA,A)$, and you are done. If $(M,d)$ is a chain complex, and $(N,\delta)$ is another one, then the differential $D$ in the chain complex $\operatorname{Hom}(M,N)$ is given on homogeneous elements by $D(f) = \delta \circ f - (-1)^{|f|} f \circ d$. If $M$ is furthermore a coalgebra with diagonal $\Delta$, and $N$ an algebra with product $\mu$, then the product of two homogeneous maps $f,g$ in the Hom complex is defined as $f*g:= \mu \circ (f\otimes g)\circ \Delta$. With this information, you should be able to compute the Maurer--Cartan equation on $\pi$.