User aleph0 - MathOverflow

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

2011-01-11T17:27:51Z

If I have a topological monoid ($X,\mu$), a space $Y$ and a homotopy equivalence $f,g$ between them, then $Y$ has the strucutre 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.

(∞,1) vs Category weakly enriched over spaces

2010-04-23T03:40:15Z

What is the difference between:

($\infty,1$) categories - in which have for two objects you have an ($\infty,0$) category of morphisms (i.e. a space of morphisms)

and

categories weakly enriched over spaces - by that I mean categories such that hom(x,y) is always a space and composition is defined only up to (coherent) homotopy

Answer by aleph0 for Homological Algebra for Commutative Monoids?

2009-12-12T01:38:51Z

I used to think about this problem in relation to a chain theory for bordism ( as mentioned by Josh Shadlen above).
The problems you have with monoids is first and foremost that the category is not balanced. That means that you can have an epimorphism that is also a monomorphism but NOT an isomorphism. eg. the inclusion of N --> Z.
Subsequently most constructions that you would like to make - notably short exact sequences and the snake lemma - fail at some level.
I made a few notes on this as part of my investigation into bordism theory / homework assignment here.
In this case, we have complexes of free abelian monoids whose homology takes it's values in abelian groups and yet, the long exact sequence does not come from a short exact sequence of monoid complexes.

My references include:

[Bau89] Friedrich W. Bauer. Generalised homology theories and chain complexes. Annali di Mathematica pura ed applicata, CLV:143–191, 1989.

[Bau95] Friedrich W. Bauer. Bordism theories and chain complexes. Journal of Pura and Applied Algebra, 102:251–272, 1995.

[BCF63] R.O. Burdick, P.E. Conner, and E.E. Floyd. Chain theories and their derived homology. Proceedings of the AMS, 19(5):1115–1118, Oct. 1963.

[Koc78] S. O. Kochman. A chain functor for bordism. Transactions of the American Mathematical Society, 239:167–196, 1978.

Comment by aleph0

2011-10-31T20:01:24Z

These are not good books to learn from. I find these two to be to long and drawn out. They lack focus. They are missing and overall arc or plot and feel more like an amalgam of thousands of snippets written by different people with little regard to what the others were writing. These books may contain everything, but they also contain everything. On the plus side, I enjoy the historical annotations and stories.

Comment by aleph0

2011-10-31T19:41:36Z

This is similar to this question at the Physics Exchange: theoreticalphysics.stackexchange.com/questions/…

Comment by aleph0

2011-01-13T17:30:33Z

@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?