Timeline for What is the intuition behind the Freudenthal suspension theorem?

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Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 31, 2017 at 15:58	comment	added	John Klein		The proof uses a result: if $Y_\alpha$ is a diagram of spaces over $X$, then hocolimit commutes with homotopy fibers. Apply the result to the diagram $C-\Omega X \leftarrow \Omega X \to C_+\Omega X$ ($C_\pm$ are cones) over $X$. This will give that the homotopy fiber of the map $\Sigma \Omega X \to X$ is identified with the join $\Omega X\ast \Omega X$. The join has the homotopy type of $\Sigma \Omega X \wedge \Sigma \Omega X$.
Jan 31, 2017 at 14:13	comment	added	Saal Hardali		How do you see so quickly that the homotopy fiber is $\Sigma \Omega X \wedge \Omega X$? I managed to get it to the form $colim (lim (\Omega X \wedge \Omega X \rightrightarrows X) \rightrightarrows \Omega X)$ using that homotopy colimits are universal. But now i'm stuck. Of course once I know this is true I can prove this by choosing a model and computing but i'm trying to understand why.
Feb 24, 2011 at 12:28	history	edited	John Klein	CC BY-SA 2.5	added 80 characters in body
Feb 24, 2011 at 0:53	history	edited	John Klein	CC BY-SA 2.5	deleted 6 characters in body
Feb 23, 2011 at 23:46	history	answered	John Klein	CC BY-SA 2.5