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I start with a longer discussion which will result in a precise version of the question. A am puzzled about an issue with the Quillen plus construction. I have seen outstanding experts being confused about this point. There are the following different ways of calling a map $f:X \to Y$ a homology equivalence:

  1. $f_\*:H_\*(X;\mathbb{Z}) \to H_*(Y;\mathbb{Z})$ is an isomorphism ("weak homology equivalence").
  2. For each abelian system of local coefficients $A$ on $Y$ ($\pi_1 (Y)$ acts through an abelian group), the induced map $H_* (X;f^* A) \to H_* (Y;A)$ is an isomorphism ("strong homology equivalence").
  3. For each system of local coefficients $A$ on $Y$, the induced map $H_* (X;f^* A) \to H_* (Y;A)$ is an isomorphism ("acyclic map").

The third condition is equivalent to each of

3'. The homotopy fibres of $f$ are acyclic. 3'''. $f$ is can be identified with the Quillen plus construction.

EDIT: Before, I included the statement ''3''. $f$ is weak homology equivalence, $\pi_1 (f)$ is epi and $ker(\pi_1 (f))$ is perfect.'' This is false (does not imply the other two conditions); in my answer to Spaces with same homotopy and homology groups that are not homotopy equivalent? I gave an example of a weak homology equivalence that is even an isomorphism on $\pi_1$, but whose homotopy fibre is not acyclic. END EDIT

The implications $(3)\Rightarrow (2)\Rightarrow ( 1)$ hold. If all components of $Y$ are simply connected, then all these notions coincide; if $\pi_1 (Y)$ is abelian (each component), then $(2)\Rightarrow(3)$. In that case, $\pi_1 (X)$ is quasiperfect (i.e., its commutator subgroup is perfect). If $\pi_1 (Y)$ is nonabelian, then $(2)$ does not imply $(3)$ (take the inclusion of the basepoint into a noncontractible acyclic space). Even if $Y$ is an infinite loop space, a weak homology equivalence does not have to be strong: Take $X=BSL_2 (Z)$, $Y=Z/12$. The abelianization of $SL_2 (Z)$ is $Z/12$, and the map $SL_2 (Z) \to Z/12$ is a weak homology equivalence. The kernel, however, is a free group on two generators.

Now, many cases of such maps arise in the process of ''group completion''. Here are some examples

  1. $X=K_0 (R) \times BGL (R)$ for a ring and $Y=\Omega B (\coprod_{P} B Aut (P))$ ($P$ ranges over all finitely generated projective $R$-modules). The commutator subgroup is perfect due to the Whitehead lemma.
  2. $X=\mathbb{Z} \times B \Sigma_{\infty}$; $Y=QS^0$. The alternating groups are perfect.
  3. $X=\mathbb{Z} \times B \Gamma_{\infty}$ (the stable mapping class group); $Y$ the Madsen-Weiss infinite loop space. Here there is no problem, $\Gamma_g$ is perfect for large $g$.
  4. $X=\mathbb{Z} \times B Out(F_{\infty})$ (outer automorphisms of the free group), $Y=Q S^0$. Galatius proves that this is a weak homology equivalence and he states implicitly this map is a strong homology equivalence.

I explain why I am interested: if you only look at the homology of $X$ and $Y$, this is only an aesthetical question. I want to take homotopy fibres and as explained above, the distinction is essential and a mistake here can ruin any argument.

In all these cases, there is a topological monoid $M$ (for example $\coprod_{P} B Aut (P)$) and $X$ is the limit $M_{\infty}$ obtained by multiplying with a fixed element. There is an identification $\Omega BM$ with $Y$ that results from geometric arguments and does not play a role in this discussion.

There is a map $\phi:M_{\infty} \to \Omega BM$, which is the subject of the ''group-completion theorem'', see the paper "Homology fibrations and the ''group completion'' theorem" by McDuff-Segal. The map arises from letting $M$ act on $M_{\infty}$ and forming the Borel construction $EM \times_M M_{\infty} \to BM$. The point-preimage is $M_{\infty}$, the space $EM \times_M M_{\infty}$ is contractible and so the homotopy fibre is $\Omega BM$. $\phi$ is the ''geometric-fibre-to-homotopy-fibre'' map.

What Segal and McDuff prove is that if the action is by weak homology equivalences, then $\phi$ is a weak homology equivalence. This is what is typically used to established the above results. To prove that 1,2,3 are strong homology equivalences, one can invoke an extra argument which is specific to each case.

Now, in McDuff-Segal, I find the claim (Remark 2) that their methods give that $M_{\infty} \to \Omega BM$ is a strong homology equivalence and I want to understand this.

I convinced myself that the whole argument goes through with strong homology equivalences (and the corresponding notion of "strong homology fibration"). Proposition 2 loc.cit. then has the assumption that $M$ acts on $M_{\infty}$ by strong homology equivalences (one needs the notion of homology equivalences one wants to prove in the end - which I find plausible).

This amounts, say in example 4, to prove that the stable stabilization map $B Out(F_{\infty}) \to B Out(F_{\infty})$ is a homology equivalence in the strong sense. For "weak homology equivalence", one invokes the usual homology stability theorem (Hatcher-Vogtmann-Wahl). But it seems that for the map being a strong homology equivalence, one needs a stronger homological stability result. I can imagine how the homological stability arguments can be modified to include abelian coefficient system, but that is not a satisfying solution.

Here are, finally, some questions:

  1. McDuff and Segal refer to ''argument by Wagoner'' in his paper ''Delooping classifying spaces in algebraic K-Theory''. I am unable to find an argument in Wagoners paper that proves under general assumptions quasiperfectness. What argument do McDuff and Segal refer to?

  2. If $M$ is a topological monoid and if $M_{\infty} \to \Omega BM$ is a weak homology equivalence, is it always a strong homology equivalence?

  3. If not, do you know a counterexample?

  4. If 2 is not true, is there a useful general criterion to prove that the group completion map is acyclic (besides the trivial case $H_1 (M_{\infty})=0$ and besides proving quasiperfectness of $\pi_1 (M_{\infty})$ by hands).

A related, but not central question:

  1. What are good counterexamples to the ''group-completion'' theorem in general that explain why the hypothesis is essential?
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I don't have time for a long answer, but the standard notion of group completion in infinite loop space theory is an H-map X >--> Y, where \pi_0(Y) is a group and (to avoid an unconvincing morasse in the literature) X and Y are homotopy associative and commutative such that \pi_0(X) >--> \pi_0(Y) is a group completion in the obvious sense and for every commutative ring (not just Z) the map H_*(X;R) >--> H_*(Y;R) is a localization of graded rings obtained by inverting the elements of \pi_0(X) [these elements being an R-basis for H_0(X;R)]. McDuff somewhere published corrections to M-Segal. –  Peter May Oct 14 '12 at 16:15
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@Peter May: Am I overlooking something here, or is it just a reformulation of the problem? I guess the important case is where $R$ is the group ring of the fundamental group of $Y$. But verifying the hypothesis that $\pi_0$ is central in this ring needs (say for $Out (F_{\infty})$) homological stability with abelian coefficients, just in the same way the argument works for constant coefficients. With the ''standard notion of group completion'', the question becomes: how do I see that the above maps are group completions? –  Johannes Ebert Oct 14 '12 at 16:59
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Sorry, I was just explaining the term ``group completion'' for those readers who might not know. Your question is all about isomorphisms, so perhaps doesn't make that clear. From the point of view of group completion as I defined it, introducing $M_{\infty}$ is irrelevant: if $M$ is a topogical monoid, $\pi_0(M)$ is homotopically central in $M$, and $\pi_0(\Omega BM)$ is homotopically central in $\Omega BM$, then the natural map $M\to \Omega BM$ is a group completion. In that generality, $M_{\infty}$ as you define it using just one element can be misleading. I doubt this is helpful to you. –  Peter May Oct 14 '12 at 23:06
    
"In that generality, M∞ as you define it using just one element can be misleading." I agree, and taking another element could a priori lead to a different answer to the question. –  Johannes Ebert Oct 15 '12 at 15:18
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I have written up what I know about this problem, and it is available at dpmms.cam.ac.uk/~or257/GCrem.pdf. –  Oscar Randal-Williams Nov 21 '12 at 10:06

1 Answer 1

up vote 12 down vote accepted

I think I have been able to reproduce the "argument by Wagoner" (perhaps it was removed from the published version?). It certainly holds in more generality that what I have written below, using the notion of "direct sum group" in Wagoner's paper (which unfortunately seems to be a little mangled).

Let $M$ be a homotopy commutative topological monoid with $\pi_0(M)=\mathbb{N}$. Choose a point $1 \in M$ in the correct component and let $n \in M$ be the $n$-fold product of 1 with itself, and define $G_n = \pi_1(M,n)$. The monoid structure defines homomorphisms $$\mu_{n,m} : G_n \times G_m \longrightarrow G_{n+m}$$ which satisfy the obvious associativity condition. Let $\tau : G_n \times G_m \to G_m \times G_n$ be the flip, and $$\mu_{m,n} \circ \tau : G_n \times G_m \longrightarrow G_{n+m}$$ be the opposite multiplication. Homotopy commutativity of the monoid $M$ not not ensure that these two multiplications are equal, but it ensures that there exists an element $c_{n,m} \in G_{n,m}$ such that $$c_{n,m}^{-1} \cdot \mu_{n,m}(-) \cdot c_{n,m} = \mu_{m,n} \circ \tau(-).$$

Let $G_\infty$ be the direct limit of the system $\cdots \to G_n \overset{\mu_{n,1}(-,e)}\to G_{n+1} \overset{\mu_{n+1,1}(-,e)}\to G_{n+2} \to \cdots$.

Theorem: the derived subgroup of $G_\infty$ is perfect.

Proof: Let $a, b \in G_n$ and consider $[a,b] \in G'_\infty$. Let me write $a \otimes b$ for $\mu_{n,m}(a, b)$ when $a \in G_n$ and $b \in G_m$, for ease of notation, and $e_n$ for the unit of $G_n$.

In the direct limit we identify $a$ with $a \otimes e_n$ and $b$ with $b \otimes e_n$, and we have $$b \otimes e_n = c_{n,n}^{-1} (e_n \otimes b) c_{n,n}$$ so $b \otimes e_n = [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)$. Thus $$[a \otimes e_n, b \otimes e_n] = [a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)]$$ and because $e_n \otimes b$ commutes with $a \otimes e_n$ this simplifies to $$[a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)]].$$ We now identify this with $$[a \otimes e_{3n}, [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}]$$ and note that $a \otimes e_{3n} = c_{2n,2n}^{-1}(e_{2n} \otimes a \otimes e_{n})c_{2n,2n} = [c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})]\cdot (e_{2n} \otimes a \otimes e_{n})$. Again, as $(e_{2n} \otimes a \otimes e_{n})$ commutes with $[c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}$ the whole thing becomes $$[a,b]=[[c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})], [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}],$$ a commutator of commutators.

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Great! Thanks! Do you see whether you can generalize it to the ''multi-object-case'' as in GMTW or the noncommutative case? I don't. –  Johannes Ebert Oct 16 '12 at 18:28
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No, I don't, and I invested considerable fruitless effort in trying to find a counterexample. –  Oscar Randal-Williams Oct 16 '12 at 19:48

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