# Is it ever true that $\Omega^n(\bigvee_I X)\simeq\bigvee_I\Omega^n X$?

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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Depending on the response I might ask this on MathOverflow as well. Please stop me with a comment if you think this question is too broad for that page :D –  Eivind Dahl Mar 24 '11 at 20:26
Is the notation you use standard? –  Raphael Mar 24 '11 at 22:30
I would think so. $\Omega X:=\hom_*(S^1,X)$ and $\Omega^n X:=\hom_*(S^1,\Omega^{n-1}(X)$ is the $n$-th loopspace of pointed maps. Was that what you were thinking about? –  Eivind Dahl Mar 24 '11 at 22:37
Notice there is a retraction $\Omega^n \bigvee_I X \to \prod_I \Omega^n X$. This is almost never a homotopy-equivalence though. Have you read the Hilton-Milnor Theorem? It's the theorem that tells you about loop-spaces of wedges of things. –  Ryan Budney Mar 24 '11 at 22:44
Anyhow, I think the answer is no provided $I$ has more than one element and $X$ has any non-trivial homotopy-groups in dimension $k \geq n$. This is because the space $\vee_IX$ has non-trivial Whitehead products, but $\prod_I X$ does not. –  Ryan Budney Mar 25 '11 at 5:07