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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$_\infty$)?

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  • $\begingroup$ I'm very far from an expert of $A_\infty$-algebras, but I assume that this is in order for a map to be an equivalence iff the induced map in $Ho(A_\infty-alg)$ is an isomorphism $\endgroup$ – Denis Nardin Apr 12 '16 at 15:00
  • $\begingroup$ Well, the thing is that when computing the cohomology of a homotopy algebra, e.g. Hochschild cohomology of an $A_\infty$-algebra, we built the Hochschild complex with differentials in terms of the whole $A_\infty$ structure, using higher arity maps. Intuitively, I would think that to construct a weak equivalence in the category A$_\infty$, I would need higher arity maps to detect again what happens in cohomology. (but seems like it's not) --I can understand that if a map $f$ induces an iso in Ho(A$_\infty$), $f_1$ should be a quasi-iso; but I can't see why it should hold the other way around. $\endgroup$ – Marcel Rubió Apr 12 '16 at 15:33
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    $\begingroup$ The answer to this more specific question is, basically, "because the spectral sequences starting from the cohomology along the differential induced by $m_1$ converge to the cohomology of the total differential, and a quasi-isomorphism on the $E_1$ pages implies a quasi-isomorphism on the limit pages". See my more general answer to Q1 below for a further elaboration. $\endgroup$ – Leonid Positselski Apr 12 '16 at 20:21
  • $\begingroup$ That makes sense, thanks! I am in fact more interested in your answer below; but it is nice to see these methods appearing here. $\endgroup$ – Marcel Rubió Apr 12 '16 at 23:36
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Q1: The conventional theory of homotopy algebras is built on the premise that the lower-degree operations dominate over the higher-degree ones, in some sense. A discussion of this can be found in the introduction to my preprint "Weakly curved $\mathrm A_\infty$-algebras over a topological local ring", http://arxiv.org/abs/1202.2697. This does not answer your question fully, but explains the underlying ideology to some extent.

Q2: The theory of curved DG-algebras/curved DG-coalgebras and the co/contraderived categories of CDG-modules/comodules/contramodules over them is built on the premise that the (co)multiplication dominates over the differential (and the differential dominates over the curvature). So the higher-degree operations dominate over the lower-degree ones in these "theories of the second kind". On CDG-coalgebras or DG-coalgebras, there is even a model structure with such weak equivalences (though a precise definition is more complicated and maybe does not accord to what you describe in the question). See my memoir "Two kinds of derived categories, ...", http://arxiv.org/abs/0905.2621.

Q3: It is indeed true that the homotopy category of DG Lie algebras is equivalent to the homotopy category of $\mathrm L_\infty$-algebras, though perhaps for reasons more complicated than described in the question. Similarly, the homotopy category of associative DG-algebras is equivalent to the homotopy category of $\mathrm A_\infty$-algebras.

Basically, with any $\mathrm A_\infty$-algebra one can naturally associate a much bigger DG-algebra quasi-isomorphic to it; while for any DG-algebra one can, if one wishes, construct a (generally speaking) much smaller $\mathrm A_\infty$-algebra quasi-isomorphic to it (in addition to the obvious option of viewing a DG-algebra as an $\mathrm A_\infty$-algebra with vanishing higher operations).

But of course, one cannot just forget the higher operations of an $\mathrm A_\infty$-algebra and obtain a quasi-isomorphic DG-algebra, firstly because the multiplication in an $\mathrm A_\infty$-algebra need not be associative, and secondly, even if it is, the identity map cannot be extended to an $\mathrm A_\infty$-morphism (in most cases).

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  • $\begingroup$ thank you for your answers! I appreciate you sharing your papers. I did not know about curved $A_\infty$-algebras, but I will try to read more about them --I imagine that deformation theory in the setting of curved $A_\infty$-algebras would be difficult (because of gauge/homotopy equivalence?). $\endgroup$ – Marcel Rubió Apr 12 '16 at 23:22
  • $\begingroup$ In particular, in the first paper you linked, at 0.24 of the introduction, you talk a bit about deformation theory. From what I read, is it right to think that there are statements like: 'given a weak equivalence between two curved algebras A and B (weak equivalence in the sense that you described for Q2), the respective deformation functors are isomorphic' ? $\endgroup$ – Marcel Rubió Apr 12 '16 at 23:28
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My favorite explanatory analogy for the first question, along the line of Leonid's answer, is that a power series in one variable with no constant term, $a_1x +a_2x^2 +\cdots$ has an inverse under composition if and only if $a_1$ has an inverse in the ground ring.

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Here is a down to earth explanation, by way of analogy.

A group homomorphism $\varphi:G\to H$ is an isomorphism of groups iff it is a bijection of sets. Why do we not need to say anything about the group operations coinciding? The reason is simply because the fact that $\varphi$ is a group homomorphism already implies this.

There is an essentially endless list of similar situations, and quasi-isomorphisms of $A_\infty$-algebras is one of them. Of course, the fact that quasi-isomorphisms of $A_\infty$-algebras (over a field!) are invertible up to homotopy is by no means automatic, but it is straightforward to check.

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I will try to answer Q1.

One simple reason why $A_\infty$-algebras appear is that usual DG-algebras are not homotopy stable, in the sense that if $A$ is a DG-algebra and $V$ is a complex which is a deformation retract of $A$ (in the usual topological sense, using the notion of homotopy for complexes), then $V$ does not inherit a DG-algebra structure, but well, an $A_\infty$-algebra structure. In fact, it suffices, that there is a map of complex $V\to A$ that is right homotopy invertible, see this paper. One usually likes to do this when $V$ is quasi-isomorphic to $A$ via some map $V\to A$ (for example, $V=H(A)$), and one can produce a retraction for example when the ground ring is a field.

Now one problem which $A_\infty$-algebras fix is that of localising the category of DG-algebras at quasi-isomorphisms. If $A$ is an $A_\infty$-algebra, one can consider its bar construction $BA$, which is a honest DG-coalgebra, and take $\mathsf{DASH}$ the category with objects the $A_\infty$-algebras and hom-sets the maps DG-coalgebra maps between bar constructions. Every DG-algebra is an $A_\infty$-algebra, and one can think of the category $\mathsf{DA}$ of DG-algebras sitting inside $\mathsf{DASH}$ as a (non-full, and that's the jist of it all) subcategory via the obvious inclusion. There is a usual notion of homotopy between maps of DG-coalgebras which yields the homotopy category $\mathsf{DASH}/\sim$, and it turns out the map $\mathsf{DA} \to \mathsf{DASH}$ gives an equivalence of categories $\mathsf{DA}[\text{Qis}^{-1}] \to \mathsf{DASH}/\sim$. This is explained, and I suppose first proven in this paper by Munkholm. Thus, $A_\infty$-algebras "fix" the problem of quasi-isomorphism of complexes by replacing perhaps a big complex by a small one, usually its homology, at the cost of giving you an ugly differential.

To get back to Q1, note that in the category $\mathsf{DC}$ of conilpotent DG-coalgebras, we also have a notion of quasi-isomorphism. Thus, say that a map $f:A\to A'$ of $A_\infty$-algebras is a weak equivalence if its image in $\mathsf{DASH}$, i.e. $Bf : BA\to BA'$, is a quasi-isomorphism. It turns out that, because $BA$ and $BA'$ are quasi-free (i.e. free as coalgebras, with some funny differential), weak equivalences are homotopy equivalences (there is a model category on $\mathsf{DC}$ where all objects are cofibrant and the fibrant objects are the quasi-free coalgebras, as proven here), and in fact one can show that $Bf$ is a quasi-isomorphism if and only it is an homotopy equivalence, if and only if $f_1$ is a quasi-isomorphism. This means that, so far as quasi-isomorphism (equivalently, homotopy) of $A_\infty$-algebras goes, the only relevant map is the one of length $1$. You can find a proof of this here. As mentioned in the comments, the implication "$f_1$ a quasi-isomorphism then $Bf$ a quasi-isomorphism" is a simple spectral sequence argument, but the reverse implication is a bit more involved, and is proven in loc. cit. with comparison theorems (Zeeman/Moore) of spectral sequences and constructions (Cartan).

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  • $\begingroup$ (Usually the new differential has nice combinatorial interpretations, for example, but from a computational point of view, it is still a bit ugly to work with!) $\endgroup$ – Pedro Tamaroff Jan 7 '18 at 5:05

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