I have a question that seems almost too trivial to ask, yet it confuses me often.
Suppose now that we have a dg-algebra $(A,m,d)$ with multiplication $m: A \otimes A \longrightarrow A$. If we look at its suspension $(sA,d)$, which signs appear in the suspended multiplication $$ m^{sA} : sA \otimes sA \longrightarrow sA$$ I know $m^{sA}$ is not a dg map as it has degree $1$. However, in the interpretation of the suspension $sA$ as the tensor product $\mathbb{K}s \otimes A$ for $s$ a degree $1$ element, we have that $$m^{sA} : sA \otimes sA \cong \mathbb{K}s \otimes \mathbb{K}s \otimes A \otimes A \longrightarrow \mathbb{K} s \otimes A$$ which comes with a sign because of the shuffle $m^{sA}(sa, sb) = (-1)^{|a|}s m(a,b)$ , but is this the only sign or does the isomorphism $\mathbb{K}s \otimes \mathbb{K}s \cong \mathbb{K}s$ also come with a sign? Is it $-1$?
It does not matter too much. One of the reasons that the shifted product on $sA$ is defined the way it is defined is that the unique extension of the map
$$\mu : sA\otimes sA\to sA$$
to a coderivation $d_2$ of $BA = T^c(s\overline{A})$ squares to zero. The reason it squares to zero is that $\mu$ is anti-associative, and you can do a computation to check this, which only depends on correctly using the braiding on $s$ and $a$, plus the fact $\mu$ is of degree $-1$. The sign will change things when $A$ itself has a differential, which then induces $$\delta : sA\to sA \qquad \delta(sA) = -sd_A(a)$$ and a unique coderivation $d_1$ on $BA$.
The sign will then determine if you want $d_{BA} = d_1-d_2$ or $d_{BA} = d_1+d_2$, i.e. if $d_1$ and $d_2$ commute or anti-commute. Putting no sign makes them anti-commute, putting a sign makes them commute.