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If $G$ and $H$ are groups, then the map $BG\vee BH\to B(G\ast H)$ is a weak equivalence by van Kampen's theorem. However the classifying space $BM$ of a monoid can have arbitrary homotopy type so higher homotopy groups are involved.

If $M$ and $N$ are monoids, is the map $BM\vee BN\to B(M\ast N)$ still a weak equivalence? If there are counterexamples, is there still some criterion that insures such a weak equivalence?

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I don't think Van Kampen's theorem suffices to prove that the first map is a weak equivalence. VK's theorem shows that it is an isomorphism on $\pi_1$, but an argument is needed to prove that the wedge of aspherical spaces is aspherical. –  Fernando Muro May 9 '12 at 14:38
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One can prove that BG∨BH is aspherical using Ganea's homotopy fiber sequence ΣG∧H → BG∨BH → BG×BH. The fiber and base are K(π,1)-spaces (the fiber is a wedge of circles), so therefore is the total space. –  John Klein May 9 '12 at 18:01

3 Answers 3

up vote 11 down vote accepted

Here is a high-tech point of view.

The inclusion functor $\mathrm{Groups}\to \mathrm{Monoids}$, has a left adjoint $F\colon \mathrm{Monoids}\to\mathrm{Groups}$, which is the group completion functor. You know that $$\pi_1(BM) = FM.$$

The claim is that if we instead consider the total left derived functor $LF$ of $F$ ("derived group completion"), then we get a sharper result, namely $$BM \approx B(LF(M))$$ where $\approx$ is weak equivalence. This should apply for any simplicial monoid $M$ (or topological monoid, if you prefer).

Being a left derived functor, $LF$ must commute with homotopy colimits. If you also know that:

  1. The free product of any two discrete monoids is weakly equivalent to their homotopy coproduct as simplicial monoids, and

  2. the homotopy theory of simplicial groups is equivalent to the homotopy theory of pointed connected spaces,

then the result follows.

I think the paper "Simplicial localizations of categories" by Dwyer and Kan (http://www.ams.org/mathscinet-getitem?mr=579087) has a nice treatment of these kinds of ideas. (They actually discuss the derived functor of the construction $(C,W)\mapsto C[W^{-1}]$, which associates a category of fractions to a category with a distinguished subcategory. In the case where $W=C$, this is groupoid completion; they show that the classifying space of the derived groupoid completion of $C$ is weakly equivalent to the nerve of $C$.)

Added: It appears that the statement you want is proven (in simplicial language) as Proposition 3.8 of the Dwyer-Kan paper I linked to. In that paper, they work with categories which all have the same object set $O$; when $O$ is a singleton, the lemma exactly says that $N(D*E)\approx N(D)\vee N(E)$, where $N$ is the nerve.

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The answer is yes. This is a special case of Theorem 4.1 of http://www.jstor.org/stable/10.2307/2374307 which gives a sufficient condition for a pushout diagram of monoids to be transformed into a pushout diagram of classifying spaces.

For those without access, the result is that if the $M_i$ are a family of monoids with a common submonoid $W$ such that $\mathbb ZM_i$ is flat over $\mathbb ZW$ for all $i$, then the pushout of the $BM_i$ over $BW$ is a classifying space for the pushout out of the $M_i$ over $W$. Clearly, if $W$ is trivial (the free product case), the condition holds.

Notice the flatness condition is free for groups (pun intended) since the group algebra is free over the group algebra of a subgroup. Also the proof is very close to the one I tried in my previous (now deleted) answer.

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Charles has given a very good answer to the question.

The following is not meant to be an answer, but just a heuristic argument which I cannot make into a proof.

There should be an operation, "free product," denoted $\sharp$, in the category of associative topological monoids. If $X,Y$ are based spaces, then $$\Omega (X \vee Y)$$ (Moore loops), should decompose (at least up to homotopy) in this category as $$(\Omega X) \sharp (\Omega Y) .$$ The reason I find this to be plausible is that a loop in $X \vee Y$ is clearly a word of loops of $X$ and $Y$, and a word is supposed to represent an element of the free product.

Supposing this to be the case, we could take $X = BM$ and $Y = BN$, then we would have $$ \Omega (BM \vee BN) \simeq (\Omega BM) \sharp (\Omega BN) $$ It should also be the case that the inclusion $$M \ast N \to (\Omega BM) \sharp (\Omega BN)$$ is group completion, since $ (\Omega BM)$ and $ (\Omega BN)$ are group-like and the operation $\sharp$ should preserve grouplike monoids (furtheremore, we also should have $M \ast N \simeq M\sharp N$). If this is true, then $(\Omega BM) \sharp (\Omega BN) \simeq \Omega B (M\ast N)$.

If the above works, then the homomorphism $$ \Omega (BM \vee BN) \to (\Omega BM) \sharp (\Omega BN) $$ is an equivalence. Now apply the classifying space to get the desired equivalence $BM \vee BN \simeq B(\Omega BM) \sharp (\Omega BN) \simeq B(M \ast N)$.

Question: Can this heuristic sketch be made into a proof?

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I'd be tempted to look at the RPT approach to modeling loop spaces (developed by my advisor, S. Husseini, in the early 1960s). –  Jeff Strom May 9 '12 at 19:32
    
@Jeff: I did a mathscinet search of Husseini's work. No dice. –  John Klein May 9 '12 at 20:26

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