# Is an A-infinity thing the same the same as strict thing viewed through a homotopy equivalence?

If I have a topological monoid ($X,\mu$), a space $Y$ and a homotopy equivalence $f,g$ between them, then $Y$ has the structure of an $A_\infty$ space defined 'pointwise' by $$y_1 * y_2 := g \left(\mu(f(y_1),f(y_2))\right)$$ Also, I have heard someone say the reverse as "Every $A_\infty$-space is (weakly?) homotopy equivalent to a topological monoid."

My question is this: In how far is this the general case? Can I define/think of $A_\infty$ structures as strict structures as viewed through the distorting glasses of a homotopy equivalence? When is an $A_\infty$ structure of this type - i.e. is there always an equivalent strict version? (I guess no in general. Can you tell me more?)

FYI: I started out with the question: Is the based loop space of a space $X$ always on the other side of a homotopy equivalence of a (strict) topological group?

thanks, a.

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I'm having trouble parsing your question. You can think of A_infty structures as up-to-coherent-homotopy monoid structures. But you may be asking something different. But I think you are asking whether, in any symmetric monoidal model category, an A_infty algebra can be strictified into a strict monoid structure. Is that right? –  Clark Barwick Jan 11 '11 at 21:37
I think your question is the following: "Is every $\Box_\infty$ space homotopy-equivalent to a strict $\Box$ space, for every value of $\Box$?" Is this right? –  Theo Johnson-Freyd Jan 11 '11 at 22:23
@Theo: What does the "square" symbol stand for? @Clark: I think that that is exactly what he is asking. And I don't know the answer, even in the case of spaces, except in the case when the monoid of components of the space is a group. –  Tom Goodwillie Jan 11 '11 at 22:56
@Tom: I guess the square is a place holder for arbitrary structures. For example, it could stand for "boolean algebra". I wonder if boolean-algebras-up-to-coherent-homotopy ever show up :) –  Mariano Suárez-Alvarez Jan 12 '11 at 3:08
@Clark: I think that is indeed what I am asking. I am thinking of A_\infty as up-to-coherent-homotopy monoid/group/algebra/..., but I am trying to get a better feel of what exactly that means. It would be nice to be able to think that this was a strict (up-to-identity) monoid/... smudged by a homotopy equivalence. My question thus has two parts: (1) if I have a strict structure and smudge it through a homotopy equivalence, do I get an A_\infty structure? (2) If I have an A_\infty structure, can I assume it arose in this way? –  aleph0 Jan 13 '11 at 17:30

i) X should have the homotopy type of a CW complex, and

ii) When we say topological group, we mean with respect to products taken in compactly generated spaces.

With respect to these assumptions, the quickest method to producing the topological group would be:

Step 1) Take the simplicial total singular complex of $X$. This will be a based simplicial set $SX$.

Step 2) Take the Kan loop group of $SX$. This is a simplicial group; denote it by $G(SX)$.

Step 3) Take geometric realization $|G(SX)|$ of $G(SX)$. This will be a topological group object in the category of compactly generated spaces. There is a natural chain of weak homotopy equivalences from the classifying space $BG(SX)$ to $X$. Consequently, $|G(SX)|$ models the loop space of $X$.

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I wonder if there is a variant of the Kan loop group that only allows "monotone" paths that recovers an associative model from the classifying space of an $A_\infty$ space? –  Ben Wieland Jan 12 '11 at 16:56
I don't know the answer to your question. But there is a construction due to Pawel Gajer which associates an actual topological group $\Omega'X$ and a map $\Omega X \to \Omega'X$ which is a homotopy equivalence when $X$ is a compact smooth manifold with basepoint. Here $\Omega X$ denotes the space of piecewise smooth Moore loops and $\Omega'X$ is gotten by forcing $\Omega X$ to be a group by formally inverting piecewise smooth loops and also rescaling smooth parts of a piecewise smooth Moore loop. –  John Klein Jan 12 '11 at 17:01

This is basically just an elaboration - in particular, it just gives a canonical procedure for what Ben Wieland describes using a two-sided bar construction, and it's standard in the field.

Let $A$ be an $A_\infty$-operad (a non-Σ operad) acting on a space $X$.

First, take singular complexes: $Sing(A)$ then becomes an operad in simplicial sets acting on $Sing(X)$. Taking geometric realization, we get a map of operads $B = |Sing(A)| \to A$, making $X$ a $B$-algebra, and a map $Y = |Sing(X)| \to X$ which is a weak equivalence and a map of $B$-algebras. So without loss of generality we may assume that both our operad and our space are CW; if $X$ was a CW-complex in the first place then $Y$ is actually homotopy eqivalent to $X$.

There is a natural map from the operad $B \to Assoc$ to the associative operad. Each of these operads has an associated monad, taking spaces to the free algebras on that space. I'll abuse notation and use the same letter for the associated monad, so $$B(Z) = \coprod_{n \geq 0} B(n) \times Z^n.$$ For any CW-complex $Z$, the natural map $B(Z) \to Assoc(Z)$ is a homotopy equivalence.

We then hit this with a two-sided bar construction. We have a simplicial space $$Bar(B,B,Y) = \{B(Y) \leftleftarrows B(B(Y)) \cdots \}$$ whose structure maps come from the monad's structure maps, and a natural augmentation $|Bar(B,B,Y)| \to Y$ which is a homotopy equivalence of $B$-algebras for formal reasons (the simplicial space has an "extra degeneracy").

We have another simplicial space $$Bar(Assoc,B,Y) = \{Assoc(Y) \leftleftarrows Assoc(B(Y)) \cdots\}$$ and a natural map of simplicial spaces $Bar(B,B,Y) \to Bar(Assoc,B,Y)$ which is a levelwise homotopy equivalence. Taking geometric realization, we get a map $|Bar(B,B,Y)| \to |Bar(Assoc,B,Y)|$ which is a homotopy equivalence and a map of $B$-algebras, where the latter is actually an algebra over the associative operad.

Net result, we get a diagram of algebras over the $A_\infty$-operad $B$ as follows: $$|Bar(Assoc,B,Y)| \leftarrow |Bar(B,B,Y)| \to Y \to X$$ Here the left two arrows are homotopy equivalences, the rightmost arrow is a weak equivalence, and the leftmost object is actually strictly associative rather than just an $A_\infty$-algebra.

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Thank you for denoting the associative operad by "$Assoc$." –  Craig Westerland Jul 23 at 14:59
@CraigWesterland Goodness, yes. There are impressionable young people around and I don't want them confusing it with the Adams spectral sequence. –  Tyler Lawson Jul 24 at 2:28

Given a map of operads $A\to B$, there is a restriction functor that turns a $B$-algebra into an $A$-algebra. This should have a left adjoint, which should be thought of as a tensor-with-$B$ or induction functor that produces a $B$-algebra from an $A$-algebra by gluing on $B$ along $A$. If these are operads without symmetry and for each $n$, $A(n)\to B(n)$ is an equivalence, then the induction functor should not change the underlying homotopy type (if applied to cofibrant objects, or something). This should cover both that every $A_\infty$-space is equivalent to an associative monoid and that every $A_\infty$-algebra is equivalent to DGA. The latter is written down in many places and usually the construction given generalizes to the induction functor I claim.

The induction functor from $E_\infty$-spaces to commutative monoids in spaces does not preserve homotopy type. That is because of the symmetric group action on operads. The induction functor involves the quotient by the symmetric group. For the induction functor not to change the underlying homotopy type one should ask for something like $A(n)\to B(n)$ is an equivalence not just of underlying objects, but objects-with-group action, which means an equivalence of fixed points for all important subgroups of the symmetric group. I'm not sure which subgroups are important for operads; perhaps symmetric groups or maybe products of them.

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What do you mean by all the shoulds in the first paragraph? –  Mariano Suárez-Alvarez Jan 12 '11 at 5:23
The first example of a rigidification theorem like this is that a monoidal category is equivalent to a strict monoidal category. One adds new objects with names like $A\Box B\Box C$, with morphisms by making them isomorphic to $(A\Box B)\Box C$. Another way to say this is that one forms the free strict monoidal category, but then changes the morphisms to make it equivalent to the old category. –  Ben Wieland Jan 12 '11 at 17:04

This is a tangential comment. In the monoidal category whose objects are topological monoids, an $A_\infty$ monoid is the same as a space with a little $2$-cubes action whereas a strict monoid is a commutative monoid.

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