A∞-morphism and quasi-isomorphism

Let $A$ and $B$ be uncurved cohomological A∞-algebras in the convention of Uncurved cohomological A∞-algebra in the unsuspended convention. An A∞-morphism $f:A\to B$ is a family of homogeneous maps

$f_n:A^{{\otimes }_{k}n}⟶B,\mathrm{\mid }{f}_n\mathrm{\mid }=1-n,n\ge 1,$

such that, for every $n\ge 1$, the following equality of maps $A^{{\otimes }_{k}n}\to B$ holds:

${\sum }_{\begin{array}{c}r,s,t\ge 0,s\ge 1\\ r+s+t=n\end{array}}(-1{)}^{r+st}{f}_{r+1+t}(1^{\otimes r}\otimes m_s^A\otimes 1^{\otimes t})={\sum }_{q=1}^n{\sum }_{\begin{array}{c}{i}_1+\cdots +i_q=n\\ i_j\ge 1\end{array}}(-1{)}^{\eta (i_1,\dots ,i_q)}{m}_q^B(f_{i_1}\otimes \cdots \otimes f_{i_q}),$

where $\eta (i_1,\dots ,i_q)={\sum }_{j=1}^{q-1}(q-j)(i_j-1)$ and all tensor maps use Koszul sign rule for graded tensor products and permutations. The identity at $n=1$ says that $f_1$ is a cochain map from $(A,m_1^A)$ to $(B,m_1^B)$.

An A∞-morphism is an A∞ quasi-isomorphism if $f_1$ induces an isomorphism on cohomology, where $H^j(A,m_1^A)=\ker (m_1^A:A^j\to A^{j+1})\mathrm{/}\mathrm{im}(m_1^A:A^{j-1}\to A^j)$ and likewise for $B$. This definition imposes no unit condition on $f$.