TITLE: Why are quasi-isomorphisms of homotopy algebras only defined for arity 1?
QUESTION [10 upvotes]: When reading about homotopy algebras (e.g. $L_\infty$-algebras, $A_\infty$-algebras), an $\infty$-morphism $f$ is called an $\infty$-quasi-isomorphism if $f_1$ is a quasi-isomorphism.
Recall/Example ($A_\infty$-algebras):
An $A_\infty$-morphism between two $A_\infty$-algebras $(A,\mathfrak{m})$ and $(A', \mathfrak{m}')$ (here $\mathfrak{m}$ and $\mathfrak{m}'$ are the structure maps) is a collection $\{f_k\}_{k\geq1}:(A,\mathfrak{m}) \rightarrow (A',\mathfrak{m}') $ of degree zero (degree preserving) multilinear maps
\begin{equation*}
f_k: A^{\otimes k}\rightarrow A', \hspace{1cm}k\geq 1
\end{equation*}
that satisfy the following relation for $n\geq1$:
\begin{equation*}
\sum_{k+l=n+1}\sum^k_{i=0} (-1)^{a_1+\dots+a_n}f_k(a_1, \dots, a_i, m_l(a_{i+1}, \dots, a_{i+l}), a_{i+l+1}, \dots, a_n).
\end{equation*}
\begin{equation*}
=\sum_{\substack{1\leq k_1\leq \dots \leq k_j \\ k_1+\cdots+k_j= n}} m'_j(f_{k_1}(a_1, \dots, a_{k_1}), f_{k_2}(a_{k_1+1}, \dots, a_{k_1+k_2}), \dots, f_{k_j}(a_{k_{j-1}+1}, \dots, a_n))
\end{equation*}
Furthermore, we call such morphisms $A_\infty$-quasi-isomorphisms if $f_1$ induces isomorphism in cohomology.
Q1: Why do we normally omit higher arity maps when talking about quasi-isomorphisms? 
Q2: Would it be possible to have a weak equivalence that only appears in higher arity maps?
Q3: In case we only care about $f_1$, wouldn't that imply an equivalence at the level of homotopy categories between, for example, Ho(DGLA) and Ho(L$_\infty$), as all the higher arity maps between $A$ and $B$ in L$_\infty$ with the same $f_1$ give isomorphism in Ho(L$_\infty$)?

REPLY [3 votes]: $\text{}$Hi Marcel, I don't believe you're confused about this anymore, but I stumbled upon this question and wanted to add a remark that I think is missing from the other answers.
I claim that before understanding what a quasi-isomorphism of $A_\infty$-algebras is, one should understand what an isomorphism of $A_\infty$-algebras is.
But there is absolutely no ambiguity in what it means for a morphism in a category to be an isomorphism. An isomorphism always means a morphism with a two-sided inverse. And one can easily prove the following result:
Proposition: If $f \colon A \to B$ is a morphism of $A_\infty$-algebras, given by components $f_n \colon A^{\otimes n} \to B$, then $f$ is an isomorphism in the category of $A_\infty$-algebras if and only if $f_1$ is an isomorphism in the category of chain complexes.
Given this result it is very natural that quasi-isomorphisms are defined the way they are.