I have a question about the proof of lemma 6.4.12 in the book Algebraic Operads (Loday-Vallette) which I do not seem to be able to fully complete on my own. Hopefully, somebody here can point out what I am not seeing.
Let me sketch the situation. Given a cooperad $(C,\Delta,\epsilon)$ and a operad $(P,\gamma,\eta)$, we consider its free right P-module $C \circ P$. Let $\alpha: C \longrightarrow P$ be a twisting morphism (of degree $-1$). The map $d^r_\alpha$ is then defined as follows $$ C \circ P \overset{\Delta_{(1)} \circ P}{\longrightarrow} (C \circ_{(1)} C) \circ P \overset{(C \circ_{(1)} \alpha) \circ P}{\longrightarrow} (C \circ_{(1)} P) \circ P \cong C \circ (P ; P\circ P) \overset{C \circ (P;\gamma)}{\longrightarrow} C \circ (P ; P) \cong C \circ P$$ In concrete terms, $$d^r_\alpha( c ; p_1,\ldots,p_n) = \sum_i \pm (c_{(1)}; p_1,\ldots, \alpha c_{(2)} \circ (p_i,\ldots,p_j), \ldots,p_n) $$ where $\Delta_{(1)}(c) = c_{(1)} \otimes (i, c_{(2)})$ denotes the sweedler notation.
The proof of the lemma then states the following $$ [ d^r_\alpha , d^r_\alpha ] = d^r_{[\alpha,\alpha]}$$ which reduces to $$ (d^r_\alpha)^2 = d^r_{\alpha * \alpha}$$ The book says the computation is similar to the case of twisted complex for associative (co)algebras. However, I think extra terms appear on the left-hand side, namely $$ (c_{(11)}; p_1, \ldots, \alpha c_{(12)} \circ (p_i,\ldots,p_j), \ldots ,\alpha c_{(2)} \circ (p_k,\ldots,p_l), \ldots ,p_n)$$ and $$(c_{(11)}; p_1, \ldots, \alpha c_{(2)} \circ (p_i,\ldots,p_j), \ldots ,\alpha c_{(12)} \circ (p_k,\ldots,p_l), \ldots ,p_n) $$ Even using coassociativity, I still do not see how these terms cancel as they are not of the right shape? Perhaps I am overlooking something simple?
Thanks in advance for taking your time to read and help!
$\renewcommand{\inf}{\circ_{(1)}}$ In this case it is probably better to stick to diagrams and axioms and avoid element wise proofs, since the notation quickly becomes unmanageable. Let me consider $\alpha : C\to P$ and let me write $$-\frown\alpha : C\circ P\longrightarrow C\circ P$$ for the unique (infinitesimal) derivation that extends $$C\to C \circ_{(1)} C \to C\circ_{(1)} P \subseteq C\circ P.$$
Then what you want to prove is that for any element $x$ in $C\circ P$ we have that $$(x\frown \alpha)\frown \beta) = x \frown (\alpha\smile \beta)$$ where $\smile$ is the star product in $\hom(C,P)$.
The computation is a bit cumbersome anyways, but it involves some neat "cable working" as in the proof for algebras. Essentially, you need to move some "loose cables" around to exchange $\Delta$ and $\alpha$, $\Delta$ and $\mu$, and $\mu$ and $\beta$. Concretely, you need to use that the following three diagrams commute, where I use juxtaposition to denote usual composition, but keep infinitesimal ones, and use $x$ and $y$ instead of $C$ and $P$ for readability:
First, the following diagram commutes:
$$\require{AMScd} \begin{CD} xy @>{\Delta_{(1)}\circ 1}>> x(y,xy) @>{1(1,\alpha\circ 1)}>>x(y,yy) @>{1(1,\mu)}>> x(y,y) = xy\\ & &@V{\Delta_{(1)}(1,1)}VV @V{\Delta_{(1)}(1,1)}VV @VV{\Delta_{(1)}(1,1)}V \\ & &(x\inf x)(y,xy) @>{1(1,\alpha\circ 1)}>>(x\inf x)(y,yy) @>{1(1,\mu)}>> (x\inf x)y = x(y,xy)\\ & &\color{white}{X} @V{(1\inf \beta)(1,1)}VV @VV{1(1,\beta\circ 1)}V \\ & &\color{white}{X} & & (x\inf y)(y,yy) @>{1(1,\mu )}>> (x\inf y)y = x(y,yy)@>{1(1,\mu )}>>xy \end{CD}$$
To see it, look at the three squares and observe that they all commute for trivial reasons: the arrows do not interact (this is what I think of as moving loose cables).
These takes care of all the loose cables, and brings two $\Delta$s first, then $\alpha$ and $\beta$ in the middle, and then two $\mu$s at the end, if you follow the lower boundary of the diagram.
To finish, you need to use (co)associativity to show you get $-\frown (\alpha\smile \beta)$, and note that the composition
$$(x\inf x)(y,xy) \stackrel{1(1,\alpha\circ 1)}\longrightarrow (x\inf x)(y,yy) \stackrel{(1\inf \beta)(1,1)}\longrightarrow (x\inf y)(y,yy)$$
acts like $\alpha\inf \beta$. For the endpoints, you get on the one hand
$$(x\inf y)(y,yy) \stackrel{1(1,\mu)}\longrightarrow (x\inf y)(y,y) = x(y,yy) \stackrel{1(1,\mu)}\longrightarrow x(y,y) = xy$$
and
$$xy \stackrel{\Delta_{(1)} \circ 1 }\longrightarrow x(y,xy) \stackrel{\Delta_{(1)} \circ 1 }\longrightarrow (x\inf x)(y,xy).$$
As you observe, there is a minor issue, which is that $(x\inf x)\inf x$ is larger than $x\inf (x\inf x)$. The way this is fixed here is the parallel axiom (which of course is new for operads, and did not exist for associative algebras): the extra terms in $(x\inf x)\inf x$ all appear as parallel compositions.
Once you apply $\alpha$ and $\beta$, you will create elements with homological degrees, and then the parallel axiom for $\mu$, which in the dg setting takes signs, will make these elements cancel in pairs (the element where you applied $\alpha$ first and then $\beta$ will cancel with the one where you applied $\beta$ first and then $\alpha$).
What this means is that a part of one of the diagrams is zero (a direct sum) and then you can get what you wanted. I have not checked all the little details in this last part, but hopefully the diagrams above help you write it all down.