An exercise from Loday and Vallette about Koszul morphism I tried to solve the following exercise from Loday and Vallette's Algebraic Operad. The first three parts are straightforward, however I have no idea how to solve the last part. I can't find any statements in that chapter which can be applied directly. Any help or hints would be really appreciated. 

 A: Recall: $\alpha$ being Koszul means that $C\otimes_\alpha A$ is acyclic, menaning that the augmentation map $C\otimes_\alpha A\overset{\epsilon\otimes\epsilon}{\longrightarrow}\mathbb{K}$ is a quasi-isomorphism. 
$\underline{\Rightarrow}$
Observe that $\epsilon\otimes\epsilon$ is the image of $\xi$ through the functor $\mathbb{K}\otimes_A-$, which is exact on free $A$-modules. 
Therefore, if $\xi$ is a quasi-isomorphism, then so is $\epsilon\otimes\epsilon$. 
$\underline{\Leftarrow}$
Using the weight grading on (the left most copy of) $A$, you can put a filtration on $A\otimes_\alpha C\otimes_\alpha A$, so that $\xi$ is filtration preserving. Then consider the associated spectral sequence. On the level of the $E_0$ page (i.e. associated graded) you get the following morphism: 
$$
A\otimes_\epsilon C\otimes_\alpha A\overset{id\otimes\epsilon\otimes \epsilon}{\longrightarrow}A
$$ 
Therefore, if $\epsilon\otimes\epsilon$ is a quasi-isomorphism, then we get a quasi-iomorphism at the level of $E_0$. Hence the spectral sequence (of the cone of $\xi$, lets say) degenerates and is $0$ at $E_1$. Thus the cone of $\xi$ is acyclic, which means that $\xi$ is a quasi-isomorphism. 
