Why are quasi-isomorphisms of homotopy algebras only defined for arity 1?

Question

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$_\infty$)?

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

$\text{}$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.

Answer 5

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).