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The twisting universal morphism $\tau: BA\rightarrow A$ induces a differential $\partial_{\tau}$ on $BA\otimes_{\tau}A$, we have: $$\partial_{\tau}(x\otimes y)=\partial x\otimes y+(-1)^{\lvert x\rvert}x\otimes\partial y+\displaystyle\sum{(-1)^{\lvert x_{(1)}\rvert}}x_{(1)}\otimes \tau(x_{(2)})y$$ According to some references as Algebraic Operads from Loday-Vallette, the proof of proposition 2.2.13 says that $BA\otimes_{\tau}A$ (as a complex) is identified with a Hochschild complex where the differential of $A$ dissapears and only the product of $A$ remains. I mean, let's consider ($BA,\partial$) with $\partial=d_{1}+d_{2}$. I need to 'kill' $d_{1}$, which is induced by the differential of $A$, but I do not find it evident. Could anybody please give me a hint?.

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An element of $BA\otimes A$ looks like $u = [a_1 \vert \cdots\vert a_n]a$. Let us assume for a minute that $A$ has trivial differential. Then the map $\partial_\tau$ acts as follows on such a basis element: $\tau$ is nonzero only on elements of the form $[a]$, so the only term in the coproduct of $[a_1 \vert \cdots\vert a_n]$ is $[a_1 \vert \cdots\vert a_{n-1}]\otimes [a_n]$, and then what you obtain for $\partial_\tau(u)$ is $[a_1\vert\cdots\vert a_{n-1}]a_na$, so here what you get is the complex which is a reoslution of the trivial module.

To get the 'cyclic' Hochschild complex you not only need to use $-\frown\tau = \partial_\tau$ (check the definition of the cap products, say in Proute's thesis) but also $\tau\frown -$ so what you are doing is twisting the differential $d_{BA}\otimes 1$ of $BA\otimes A$ by the 'commutator' $[\tau,-]$, giving the terms where $a_n$ and $a_1$ pop out of the bars and act on $a$.

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  • $\begingroup$ Thank you @Pedro, I was some confused about which Hochschild complex among the ones that may be found in literature (apparently different from each other, mainly by the differential, see for example section 2.2 here and compare with Algebraic Operads page 37). I think this Hochschild complex $(BA\otimes_{\tau} A)$ should correspond to the first example. $\endgroup$ – Victor TC Feb 28 at 6:01

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