# Why is $BGL(R)^+$ an $H$-space?

Of course, an easy answer is to use the fact that $BGL(R)^+$ is the identity component of $K(R)$ and construct the $H$-space structure using a more convenient model for $K(R)$, such as the group completion of $S = \amalg_n BGL_n(R)$. But the only proof I know that $BGL(R)^+$ coincides with $|S^{-1}S|_0$ actually uses the fact that $BGL(R)^+$ is already known to be an $H$-space (this is the proof found in Grayson's Higher Algebraic K-Theory II or Weibel's K-Book -- the $H$-space structure is used to show that a homology equivalence implies a homotopy equivalence. My question is actually Exercise 1.11 in Weibel.)

A direct proof that $BGL(R)^+$ is an $H$-space can be found in Berrick's An Approach to Algebraic K-Theory, but it's quite technical, involving a detailed study of acyclic maps, and can't possibly be the method Weibel has in mind in his exercise. I was surpised that I couldn't find any discussion or reference for this fact in Grayson's paper or in Quillen's Higher Higher Algebraic K-Theory I. So more precisely, my question is:

Question: Is there some way to show that $BGL(R)^+$ has an $H$-space structure induced by $\oplus$ which doesn't presuppose the equivalence with the group completion model for $K(R)$ and which is less technical than Berrick's approach (and preferably is elementary enough to be what Weibel has in mind as the solution to the exercise in his book)? Alternatively, what reference for this fact does Grayson have in mind in Higher Algebraic K-theory II?

On the face of it, it sounds plausible that one should be able to extend the $H$-space structure $\oplus$ on $\amalg_n BGL_n(R)$ in a straightforward way by following one's nose. But all the paths my nose is taking me down would require me to first show that $BGL(R)$ is an $H$-space, which is simply false -- for instance $\pi_1(BGL(R))$ is not abelian (unless $R$ is the trivial ring :)), so it can't be an $H$-space.

The best I can figure is that $\oplus$ does induce a multiplication on $BGL(R)$, but it's not homotopy unital. In order to construct homotopies witnessing unitality, one needs to use the new cells added in the plus construction. But writing down such a homotopy by hand seems like a daunting task. Is there a slicker way to do this? Or is it maybe not so daunting as I'm making it out to be?

• Thank you, Professor Grayson! On a second look through your references in HAK II, I did find the proof of the $H$-space structure in a paper of Wagoner. It's a little awkward because he has to modify the natural multiplication $\mu$ to make it an $H$-space structure. However the first part of the proof is a self-contained proof that $BGL(R)^+$ is a weakly simple space, which is enough to complete the proof that $BGL(R)^+ \simeq |S^{-1}S|_0$, showing that this modification of $\mu$ didn't actually change it after all. – tcamps Dec 18 '17 at 19:03