There is a map $BG \to A(\ast)$ where $BG$ classifies stable spherical fibrations and $A(\ast)$ is
Waldhausen's algebraic $K$-theory of a point. The map is induced by applying Quillen's plus construction to the inclusion 
$$
BGL\_1(S^0) \to BGL\_\infty(S^0)
$$ 
where $BGL\_1(S^0)$ is $BG$. Here $BGL\_\infty(S^0)$ can be defined as the homotopy colimit 
over $k$ and $n$ of $BG(\vee\_k S^n)$, where $G(\vee\_k S^n)$ is the topological monoid of homotopy automorphisms of a $k$-fold wedge of $n$-spheres. (Note: $A(\ast) = \Bbb Z \times BGL\_\infty(S^0)^+$.)


 **Question 1:**  I've heard it mentioned that there can be  **no**  retraction $A(\*) \to BG$ to the map $BG \to A(\ast)$. Can someone please explain to me why there is no such splitting, and if possible, give a reference?


More generally,


 **Question 2:**  If $R$ is a structured ring spectrum, is there a reasonable set of conditions which guarantees that the map $BGL\_1(R)\to K(R)$ admits a retraction? 


(A related question is under what conditions does an $R$-module map
$f: R^n \to R^n$ which is a weak homotopy equivalence 
admit a determinant $\text{det}(f) \in GL\_1(R)$).



Why does the map $BG\to A(*)$  fail to split?
Question 1: There are several arguments.


In degree 2 there is a reference: the proof of corollary 3.7 of Waldhausen's "Algebraic K-theory of spaces, a manifold approach". See <http://www.math.uni-bielefeld.de/~fw/> for a copy. Consider the maps $BG \to A(\ast) \to K(Z)$ and apply $\pi\_2$. Here $\pi\_2 BG = Z/2$, the composite is zero (restrict to $BSG$) and $\pi\_2 A(\ast) \to K\_2(Z)$ is an isomorphism. Hence the first homomorphism cannot be injective.


In degree 3 there is another argument. Right composition with the unstable map $\eta : S^3 \to S^2$ takes the generator of $\pi\_2(BO)$ to zero in $\pi\_3(BO) = 0$, so by naturality with respect to $BO \to BG$, composition with $\eta$ takes the generator of $\pi\_2(BG) = Z/2$ to zero in $\pi\_3(BG)$. On the other hand, composition with $\eta$ takes the generator of $\pi\_2 Q(S^0)$ to a nonzero element in $\pi\_3 Q(S^0)$, so by naturality with respect to $Q(S^0) \to A(\ast)$, composition with $\eta$ is nonzero from $\pi\_2 A(\*) = Z/2$ to $\pi\_3 A(\ast)$. Therefore $BG$ cannot split off $A(\*)$.


In degree 4 there is your argument.


Question 2: So far, I only know that $BGL\_1(R) \to K(R)$ admits a retraction if $R$ is
the realization of a commutative simplicial ring. There is no retraction for $R = ku$ by Corollary 2.3 of "Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere" by Ch. Ausoni, B. I. Dundas and J. Rognes, Documenta Mathematica 13 (2008) 795-801, and ku is already pretty close to a commutative simplicial ring.


