TITLE: Plus construction considerations.
QUESTION [21 upvotes]: In order to realise the K-groups of a ring as the homotopy groups of some space associated to that ring, Quillen proposed the following (roughly-sketched) construction:
Recall that $K_1(R) = GL(R)/E(R)$, so we're, at least, looking for a space $X$ with $\pi_1(X) = K_1(R)$. The classifying space of $K_1(R)$ is obviously not a serious candidate, but we can start with the classifying space of $GL(R)$ (given the discrete topology), $BGL(R)$. Then, choosing representative loops whose classes generate $E(R) \subset \pi_1(BGL(R))$, and then attaching 2-cells using these loops on the boundaries, we end up with something that has a fundamental group of $K_1(R)$. Furthermore, now Quillen adjoins 3-cells essentially to correct the homology back to that of $BGL(R)$, which was messed up in the addition of those 2-cells. We end up with a space denoted $BGL(R)^+$, on which we define $K_i(R) := \pi_i(BGL(R)^+)$.
More specifically, Quillen sought to find a space $BGL(R)^+$ for which $(BGL(R), BGL(R)^+)$ was an acyclic pair (that is, the induced map $H_*(BGL(R), M) \to H_{*}(BGL(R)^+,M)$ is an isomorphism for all $K_1(R)$-modules $M$). My question is

In search of a space on which to define K-groups, why was it desirable to find something satisfying the above condition on homology?

My best guess is that it was observed that $K_1(R) = GL(R)/E(R) = GL(R)_{ab} = H_1(GL(R), \mathbb{Z})$, and $K_2(R) = H_2(E(R), \mathbb{Z}) = H_2([GL(R), GL(R)], \mathbb{Z})$, and so it seemed reasonable that all K-groups should be related to the homology of $GL(R)$ - and so the above is a stab at preserving that homology.
I am trying to learn about K-theory, but, as with most presentations of math, I'm sure, I'm coming across too many cleanly-shaven formulas and propositions, entirely divorced from any sorts of thought-processes, big-pictures, or mentions of what is trying to be done and to what ends. Please share with me what you think is going on; here, with the +-construction, that is. (And, if and only if it is not a completely unrelated question, what sort of K-theoretical phenomena suggested that these groups should be homotopy groups?)

REPLY [12 votes]: For convenience (at least my own) and completeness, I want to give an explanation of Tom Goodwillie's answer, as it was not obvious to me how to prove the statement he makes. I wanted to leave it as a comment to his answer, but it became way too long. In summary, the point is that applying the plus construction to the spaces defined by Tom gives the connective covers (i.e. the Whitehead tower) of $BGL(R)^+$. This follows from the fact that the plus construction preserves certain fibre sequences.
Abbreviate $X=BGL(R)$. Let $F_1=X=BGL(R)$, and denote by $F_{i+1}$ the $i$-th space constructed by Tom, so that his claim becomes $H_i(F_i) = \pi_i(X^+) = K_i R$. Explicitly, the spaces $F_k$ are defined inductively by a fibre sequence
$$ F_{i+1}\longrightarrow F_i\longrightarrow K(H_i(F_i),i) $$
for each $i\geq 1$, where the map on the right induces the "identity" on $H_i$. Applying the plus construction to this fibre sequence, we get a new sequence
$$ (FS^+): \qquad\qquad (F_{i+1})^+ \longrightarrow (F_i)^+ \longrightarrow K(H_i(F_i),i)^+ \simeq K(H_i(F_i),i) \hphantom{\qquad\qquad\qquad} $$
where we have noted that the natural map $K(H_i(F_i),i) \to K(H_i(F_i),i)^+$ is a weak equivalence. Importantly, the new sequence $(FS^+)$ is again a fibre sequence. This result is an instance of proposition 3.D.3-(2) in page 74 of Dror Farjoun's book "Cellular spaces, null spaces and homotopy localization": applying the plus construction to a fibre sequence of path connected spaces gives a fibre sequence as long as the homotopy type of the base space is unchanged by the plus construction. In fact, the book states this for any nullification functor, and the plus construction is one such functor.
The fundamental group of $(F_1)^+=X^+$ is abelian (it is $K_1 R$, after all). From the fibration sequence $(FS^+)$ for $i=1$ we then conclude that $(F_2)^+$ is simply connected. Moreover, that fibre sequence is just $(F_2)^+\to X^+\to B(\pi_1(X^+))$, and it realizes $(F_2)^+$ as the universal cover of $X^+$.
We can proceed inductively in this manner. More precisely, knowing that $\pi_1((F_1)^+)=\pi_1(X^+)$ is abelian, we can show by induction that for all $i\geq 1$:

The space $(F_i)^+$ is $(i-1)$-connected, and the Hurewicz theorem implies $\pi_i((F_i)^+)=H_i((F_i)^+)=H_i(F_i)$.
Hence, the second map in the fibre sequence $(FS^+)$ above is an isomorphism on $\pi_i$. In other words, that fibre sequence is simply killing $\pi_i$ of $(F_i)^+$.
Finally, there is a canonical equivalence $(F_i)^+ \simeq (X^+)^{\geq i}$ from $(F_i)^+$ to the $(i-1)$-connected cover of $X^+$ (also known as the $i$-th stage of the Whitehead tower of $X^+$).

Consequently, $H_i(F_i)=H_i((F_i)^+)=H_i((X^+)^{\geq i})=\pi_i((X^+)^{\geq i})=\pi_i(X^+)$ as desired.
Obviously, the above argument is not specific to $X=BGL(R)$. In fact, we only require that the pointed space $X$ is path connected, and $\pi_1(X^+)$ is abelian.