# Does the differential of an augmented dga algebra fix the augmentation ideal?

I am reading about the bar/cobar construction in the book Algebraic Operads. The differential on the bar construction of a augmented dga algebra $A$ is a sum of two differentials $d_1+d_2$ where $d_1$ is somehow induced from the differential $d_A$ on $A$. I know we get a differential on $A^{\otimes n}$ by taking $\sum_i 1^{\otimes i-1}\otimes d_A\otimes 1^{\otimes n-i}$. But this is only a differential on $\bar{A}^{\otimes n}$ if $d_A(\bar{A})\subset \bar{A}$ (where $\bar{A}$ is the kernel of the augmentation map). My question is if this is true in general and if not how do we get a differential on $\bar{A}^{\otimes n}$?

Much grateful for any answer=)

• $d_A(\bar{A}) \subseteq \bar{A}$ provided $d_A$ commutes with the augmentation map.
– JHF
Sep 3 '18 at 20:52
• OK that makes sense, but do you know then how to get a differential on $\bar{A}^{\otimes n}$ from $d_A$? Or do we have to assume that the augmentation map of a dga algebra $A$ commutes with the differential $d_A$ in order to construct $BA$? Sep 4 '18 at 6:17

An augmentation of a nonnegatively graded chain complex $A$ is defined as a chain map to the complex that is $k$ in degree zero and $0$ otherwise (see section 1.5.5 in Loday's Algebraic Operads). Thus, it commutes with the boundary map $d_A$ by definition. As mentioned in the comments, this means the augmentation ideal is preserved by $d_A$. So there's no problem in defining the differential in the bar construction as you have.