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?


1 Answer 1


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 ...

  • $\begingroup$ Thanks. If I use the extra point as basepoint, then $J(X) = * \sqcup (S^1)^\infty$, and $J_0(X) = * \neq \Omega_0 \Sigma X$. I guess I need to reconsider those higher homotopy groups. $\endgroup$
    – Jeff Strom
    Oct 7, 2010 at 14:06

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