After Theorem 3 in Segal's Configuration Spaces and Iterated Loop Spaces, he gives some special cases.I do not understand $n = 2$ case, i.e. how is $B(\coprod_{k\geq 0} B(Br_k)) \simeq \Omega S^2$?
Reason why I'm confused :
$C_2 = \coprod_{k \geq 0} C_{2,k}$ by definition. $C_{2,k} \simeq BB_k$, that is clear.And we know that $C_2 \simeq C_2'$ and Theorem 1 gives us :
$BC'_2 \simeq \Omega S^2$.So, if we can show that $BC_2 \simeq BC_2'$, then we are done. But I can not see this [Please see below, for more details].
The entire Segal's paper is trying to prove Theorem 2, which follows from Proposition 2.1, which in turn follows from Proposition 2.4, along with the fact that $BC_n''(X) \rightarrow BC_{n-1}(X)$ is a homotopy equivalence. Summing up, what Segal proves in the paper is, that $BC_{n-1}(X) \simeq BC_n'(X)$.
In other words, we have $BC_{n-1} \simeq BC_n'$.
Combining with the above $n = 2$, we need to show that $BC_2 \simeq BC_2' \simeq BC_1$.
So my question is, how is $BC_1 \simeq BC_2$?
PS) Segal remarks that Theorem 1 is Theorem 2 for the case $X = S^{0}$.I can't see this clearly, that is, why is $BC_{n}'(S^0) \simeq BC_n'$ ?
