Suppose that $A$ is a Banach algebra with unit. We can consider $GL(A)$ as a topological group in either the discrete topology or the topology that it inherits from the norm topology of $A$, and the identity map $i$ determines a continuous map $Bi: BGL(A) \longrightarrow B^{top}GL(A)$. If we apply Quillen's plus construction, we then obtain a continuous map $(Bi)^+: BGL(A)^+ \longrightarrow B^{top}GL(A)^+$. Because $\pi_1(B^{top}GL(A))$ is abelian, we know that $B^{top}GL(A)^+$ is (homotopy equivalent to) $B^{top}GL(A)$. Therefore we have an induced homomorphism $(Bi)_1^+: \pi_1(BGL(A)^+) \longrightarrow \pi_1(B^{top}GL(A))$. Of course, we know that $\pi_1(BGL(A)^+)$ is isomorphic to $K_1^{alg}(A) = GL(A)/E(A)$, while $\pi_1(B^{top}GL(A))$ is the topological (or operator algebraic) group $K_1^{top}(A) = GL(A)/GL(A)_0$. Now, there is an obvious map from $K_1^{alg}(A)$ to $K_1^{top}(A)$; namely, the quotient map, but it not obvious (to me) that $(Bi)_1^+$ is that map. Furthermore, in Loday's discussion of relative Banach algebra $K$-theory in his book "Cyclic Homology" (Section 11.5.2, p. 372), he made an offhand comment that suggests that $(Bi)_1^+$ can have a nontrivial cokernel in a group $K_0^{rel}(A)$ (that he did not define). Can anyone tell me what is known about this map $(Bi)_1^+$?
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The $(Bi)^+_1$ is obviously surjective, since it is induced by the identity on $GL(R)$. The only reasonable definition for $K_0^{rel}(A)$ for a Banach algebra is the zero abelian group, since also $K^{alg}_0(A)=K^{top}_0(A)$.
answered Mar 18, 2015 at 9:04
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1$\begingroup$ It is hard to say more. Any map $G \to H$ induces a map $BG \to BH$ with $\pi_1(BG) \to \pi_1(BH)$ being induced by $G \to H$. The same also if you perform the $+$-construction. Thus, $K_1^{alg}(A) \to K_1^{top}(A)$ is always surjective. Since, $K_0$ makes no distinction between "algebraic" and "topological", the only choice for $K_0^{rel}(A)$ is the zero abelian group. I think, that Loday did not mean to say that $K_0^{rel}(A)\neq 0$ in his comment - only that one could define it somehow, namely as the zero group. $\endgroup$ Commented Mar 19, 2015 at 21:58
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$\begingroup$ It was Loday's comment that has always made me wonder about this. It is sad that he is not with us so that we could ask him about it. $\endgroup$ Commented Mar 20, 2015 at 15:46