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I'm curious about the following: Given a countable index set $I$, is it ever true that $\Omega^n(\bigvee_I X)\simeq\bigvee_I\Omega^n X$? What would we have to assume about $X$ (if possible) to make it true?

My motivation (wishful thinking) is that this would allow the following computation: $\pi_n\bigvee_I X\cong\pi_1\Omega^{n-1}(\bigvee_I X)\cong\pi_1\bigvee_I\Omega^{n-1}X\cong\coprod_I\pi_1\Omega^{n-1} X\cong\coprod_I\pi_n X$. Here $\coprod$ is the free product of groups.

My initial space was $X=S^1$. I don't think it's true for that particular case, but I have not proved that there is no such homotopy equivalence. (Edit: My initial intuition for thinking it might be true for that particular $X$ was for some reason its $H$-cogroup structure, i.e. the pinch map $S^1\to S^1\vee S^1$.)

Thanks!

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    @Ryan Sounds good to me. Thank you!2011-04-03

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