Maps to the group completion

Question

Let $M$ be an H-space, topological monoid (homotopy-commutative if necessary):

What does the group comletion $\Omega BM$ represent in homotopy category? Is $[X,\Omega B M]$ always equal to the universal group associated to $[X,M]$?

Thanks!

Answer 1

Section 1 of ``James maps, Segal maps, and the Kahn-Priddy theorem, by J. Caruso, F.R. Cohen, L.R. Taylor, and myself (http://www.math.uchicago.edu/~may/PAPERS/48.pdf) is entitled "Universal properties of group completions". With an appropriate and standard definition of a group completion $g\colon X\to Y$ of $H$-spaces, Lemma 1.1 there observes that if $Y$ is grouplike, then the functor $[-,Y]$ takes homology isomorphisms to isomorphisms. At least if $\pi_0(X)$ contains a countable cofinal sequence, Proposition 1.2 then gives a universal property of a group completion $g$ in terms of weak homotopy classes of maps: for a weak $H$-map $f\colon X \to Z$, where $Z$ is grouplike, there is a weak $H$-map $\tilde{f}\colon Y \to Z$ such that $\tilde f\circ g \simeq f$, and $\tilde{f}$ is unique up to weak homotopy. Corollary 1.3 then interprets this to say that on finite CW-complexes $A$, $g_{*}\colon [A,X]\to [A,Y]$ is universal in a reasonable sense of abelian monoid valued represented functors. The section was meant to clarify observations of Segal in "Operations in stable homotopy theory". I was never satisfied with this, but it was the best I could do.

Answer 2

The answer to your second question is usually no. For example, take $M = \coprod_P BAut(P)$, where $P$ runs over a set of representatives for the isomorphism classes of finitely generated projective modules over a (discrete) ring $R$. Then $\pi_m BAut(P) = 0$ for $m\neq 1$ (this is just the classifying space of a discrete group), so every map $S^m \to BAut(P)$ ($m>1$) is nullhomotopic and we find that the monoid of (unbased) homotopy classes of (unbased) maps $[S^m, M]$ is just the monoid of isomorphism classes of finitely generated projective $R$-modules. So the universal group associated to this is just $K_0 (R)$. On the other hand, $\pi_m \Omega BM = K_{m} (R)$, Quillen's higher K-theory groups of $R$, which are much more complicated. As Neil points out in the comments, this implies that $[S^m, \Omega BM] = \pi_0 \Omega BM \times \pi_m \Omega BM = K_0 (R) \times K_m (R)$.

(One can use the McDuff-Segal version of the group completion theorem here, which identifies $\Omega BM$ up to homotopy as the plus construction applied to $K_0 (R) \times BGL(\infty, R)$. This is discussed in McDuff-Segal; see Example 2 on p. 283 and Remark 2 on p. 280.)