Let $X$ be a set. One can define a non-symmetric colored version of the non-unital $A_\infty$ operad as follows. The set of colors is the set of ordered pairs in $X$. Let $(x_1,y_1),\dots,(x_n,y_n)$ be an ordered list of $n\geq 2$ colors. The space of $n$-ary operations with domain $(x_1,y_1),\dots,(x_n,y_n)$ and codomain $(x,y)$ is empty unless $x = x_1$, $y = y_n$, and $y_i = x_{i+1}$ for $i=1,\dots,n-1$. When the colors do satisfy these relations, the space of colors is the finite cell complex whose cells are labeled by planar rooted trees, each vertex having at least two inputs ("rooted" means each vertex has a unique output, namely the half-edge connecting the vertex towards the root), and whose regions are labeled, in left-to-right order along the top (the "leaf" end of the planar tree), by the colors $x=x_1,y_1=x_2,\dots,y_n=y$. The dimension of a cell is $\sum_v (\deg v - 2)$, where the sum ranges over the vertices in the tree, and $\deg v$ is the number of inputs to the vertex $v$. The boundary of a cell consists of all cells that can be formed by blowing up a vertex into two vertices. The combinatorics of this "boundary" operation are precisely described by some Stasheff polytopes, and so is a sphere. Composition in this colored operad is by grafting of trees: ignoring the boundary operation, this colored operad is free on the trees with one vertex. So I believe that this colored operad is cofibrant and has the homotopy type of the colored operad whose algebras Set are categories.

A representation of this colored operad is nothing but an $A_\infty$ category with objects indexed by $X$.

Let $\mathcal S$ be a cartesian-closed presentable $(\infty,1)$-category. I'm pretty sure that every $A_\infty$ category in $\mathcal S$ is also an $\mathcal S$-enriched $(\infty,1)$-category.

Is the opposite true? I.e. is the inclusion of $A_\infty$ categories among all $\mathcal S$-enriched categories a homotopy equivalence? My impression from Bergner, Models for $(∞,n)$-categories and the cobordism hypothesis, Mathematical Foundations of Quantum Field Theory and Perturbative String Theory, 2011 was that at least at the time this was expected but unknown.

But my impression could be off, or there could be new results in the intervening years, or I could have misunderstood what an "$A_\infty$ category" is.

Assuming that the answer is (at least after necessary corrections have been made to my definition) "it's not yet known": where does the difficulty lie?

  • $\begingroup$ Shouldn't this only be true "rationally" (cf, arxiv.org/pdf/1308.2587.pdf for dg-categories)? $\endgroup$ Dec 19 '13 at 21:48
  • $\begingroup$ @AaronBergman: I don't know. But I would expect that the part that's only true "rationally" is a strictification from "A_\infty" to "dga". $\endgroup$ Dec 19 '13 at 23:49
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    $\begingroup$ I thought that any $A_infty$ category was quasi-equivalent to a dg-category. Unless my memory's completely failed me, it's true for algebras via the bar-cobar construction. $\endgroup$ Dec 20 '13 at 2:25
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    $\begingroup$ There are too many uses of $A_\infty$. Theo's question is nonlinear, not even stable. Yes, $A_\infty$ should always be equivalent to associative. If you specialize Theo's question to $S$ = chain complexes, you get the situation where $A_\infty$ chain categories are equivalent to DG categories, which is the same as $A_\infty$-algebras are equivalent to DGAs. The scenario in which things are not equivalent is when you don't balance enrichments, because enriching over $\mathbb Z$-modules is extra structure, unless they are rational. $\endgroup$ Dec 20 '13 at 3:57
  • $\begingroup$ If enrichments are strict, then any enriched A_∞-category can be rectifiied to an enriched category, see Proposition 10.2.3 in arxiv.org/abs/1410.5675. $\endgroup$ Mar 19 '15 at 21:30

It seems to me that some progress has been made in the recent paper of Giovani Faonte, but I don't think that the result you are looking for has been proved.

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    $\begingroup$ That paper is about a different notion of $A_\infty$-category, the linear one. $\endgroup$ Dec 26 '13 at 17:14

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