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Let $X$ be an $n$-connected space and $Y$ be an $m$-connected space. How can I prove that the join $X*Y$ is $(n+m+1)$-connected?

I thought that homotopy excision would do the trick, but it does not seem so.

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  • $\begingroup$ If $X$, $Y$ are CW, then there is an obvious CW structure on $X\ast Y$, and oce you have determined this you can use cellular methods to decide why the connectivity statement is true. $\endgroup$ – Tyrone Nov 15 '18 at 11:26
  • $\begingroup$ I'm afraid I don't quite get your point. Wouldn't the CW structure on the join contain $X$ and $Y$ as subcomplexes? What results are you refering to? $\endgroup$ – user09127 Nov 20 '18 at 13:22
  • $\begingroup$ The join of two CW complexes $X,Y$ is a quotient of $X\times I\times Y$ by a certain relation. Take the product CW structure on $X\times I\times Y$ and then figure out which cells you need to quotient out. You can also use the fact that $X\ast Y$ is the pushout of the inclusions $X\times CY\leftarrow X\times Y\rightarrow CX\times Y$ to get a CW structure. The end result is that the cells of $X\ast Y$ are the joins of the cells of $X$ and $Y$. I'll leave you to figure out what the joins $D^n\ast D^m$ and $S^{n-1}\ast S^{m-1}$ are. $\endgroup$ – Tyrone Nov 20 '18 at 13:47
  • $\begingroup$ If $X$ is $n$ connected, and $Y$ is $m$ connected, then you'll see that the first cell of $X\ast Y$ above dimension $0$ that you need to worry about is $e^n\ast e^m$, so you can figure out the connectivity of $X\ast Y$ from, say, cellular homology, depending on what you are confident with. $\endgroup$ – Tyrone Nov 20 '18 at 13:49
  • $\begingroup$ To move to the general case use CW approximation and the functorality of the join construction. $\endgroup$ – Tyrone Nov 20 '18 at 13:50

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