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This is an exercise from May's Concise Course In Algebraic Topology

Show for $n\geq 2$ that $\pi_n(X\vee Y)\cong \pi_n(X)\oplus\pi_n(Y)\oplus\pi_{n+1}(X\times Y,X\vee Y)$

I feel like I'm missing something obvious. From the relative long exact sequence, we probably want to show that the maps $\pi_n(X\times Y)\to \pi_n(X\times Y,X\vee Y)$ are trivial and that things split.

One thing i've tried is looking at the inclusion $\Omega(X\times Y)\to P(X\times Y,X\vee Y)$. If I'm not mistaken then if we can show the image is contractible, this gives the above map being trivial.

Saying that, I'm super confused about everything right now and any help would be appreciated.

Edit: let $h_X:P(X) \wedge I_+\to P(X)$ be contraction on the path space and $i:\Omega(X\times Y)\to P(X\times Y,X\vee Y)$ the inclusion. Then let $H=((h_X\times id_Y)\star(id_X\times h_Y))\circ i$. i,e. contract on paths on $X$ and then $Y$ and compose this with the inclusion. This shows $i$ is homotopic to the constant map and so the maps on homotopy groups are trivial.

Actually I think I can use this homotopy to construct a splitting.

Suppose I have a map $f:S^{n-1}\to \Omega(X\times Y)$. Then I can use the homotopy to get an extension $f^\prime:S^{n}\to P(X\times Y,X\vee Y)$ which i then compose with $p_1$ to get a map $S^n\to X\vee Y$ which I think should be what I'm after.

Leon Sot
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  • The maps in the relative sequence aren’t trivial but they do admit splittings. – Connor Malin Dec 24 '19 at 01:10
  • What are the splittings? Also, while the maps in general aren't trivial, it seems to me that they are in this case. I'll edit my question with my current idea for that. – Leon Sot Dec 24 '19 at 04:17
  • An element of $\pi_n(X \times Y)$ is a pair of elements of $\pi_n(X)$ and $\pi_n(Y)$. We can always map this to the element of $\pi_n(X \vee Y)$ that is the first concatenated by the second. Show if $n>1$ this is a homomorphism. – Connor Malin Dec 24 '19 at 05:09
  • Oh of course! Thanks a lot. – Leon Sot Dec 24 '19 at 05:34

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