When I work out the James construction for a discrete pointed space, it appears that the 
induced map $\pi\_0 (J(X)) \to \pi\_0( \Omega\Sigma X)$ is the inclusion of the free monoid on $\pi\_0(X)$ into the free group on $\pi\_0(X)$, so $J(X) \to \Omega\Sigma X$
 is not a homotopy equivalence; and it seems clear that the same phenomenon holds
 for other disconnected spaces (which is why I made the distinction between $X$ and $\pi\_0(X)$).


I also think that $J(X) \to \Omega \Sigma X$ will induce isomorphisms on $\pi\_n$ for $n \geq 1$, which would mean that the restriction to the basepoint components, which I'll denote
$J\_0(X) \to \Omega\_0 \Sigma X$, should be a (weak) equivalence. But I don't recall having seen this claim anywhere, so: is this written up somewhere? 



James Construction for Disconnected Spaces
No. Let $X$ be the disjoint union of a point and a circle, and compare the two different $J(X)$ that you get according to what point you call the base point. Now think about $\pi\_1(\Omega \Sigma X)=\pi\_2(S^1\vee S^2)$. Same as $\pi\_2$ of the universal covering space ...


