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?](https://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent/53409#53409)
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:


5. What are good counterexamples to the ''group-completion'' theorem in general that explain why the hypothesis is essential?



Is every ''group-completion'' map an acyclic map?
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.
> 
> 
> 


