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

Question

(This problem is from chapter 9 of J. P. May's A Concise Course in Algebraic Topology.)

At first I thought of showing that the exact sequence $$ \cdots \to \pi_{n+1}(X \times Y) \to \pi_{n+1}(X \times Y, X \vee Y) \overset{\partial}{\to} \pi_n(X \vee Y) \\ \to \pi_n(X \times Y) \to \pi_n(X \times Y, X \vee Y) \to \cdots $$ splits; the result would then follow as $\pi_n(X \times Y) \cong \pi_n(X) \oplus \pi_n(Y)$ for $n \geq 2$. Since $\partial$ is defined by restricting maps $$(I^n, \partial I^n, J^n) \to (X \times Y, X \vee Y, *)$$ to maps $$(I^{n-1} \times \{1\}, \partial I^{n-1} \times \{1\}) \to (X \vee Y, *)$$ where $J^n := \partial I^{n-1} \times I \cup I^{n-1} \times \{0\} \subset I^n$ for $n \geq 2$, perhaps I could find some retract $r$ with $r \circ \partial = \text{id}$, but it doesn't seem possible since information is lost by restricting the maps.

Also, $\pi_{n+1}(X \times Y)$ and $\pi_n(X \times Y, X \vee Y)$ need not be trivial, and I seem to need an exact sequence of the form $$0 \to A \to B \to C \to 0$$ to talk about splitting meaningfully.

I don't know what I'm doing actually. Please send help.

EDIT: See also the proof of Theorem 6.10.5 in tom Dieck's Algebraic Topology.

Answer 1

You can get the answer by directly analyzing this LES.

First, the LES does split into SES's because the map $\pi_n(X \times Y) \rightarrow \pi_n(X \times Y,X \vee Y)$ has image 0. This is because any element of $\pi_n(X\times Y)$ can be written as $[(f,g)]$ which equals $[(f+0,0+g)]$ which gets sent to $0$ under the map.

The SES $0 \rightarrow \pi_n(X \times Y , X\vee Y) \rightarrow \pi_{n-1}( X \vee Y) \rightarrow \pi_{n-1}(X \times Y) \rightarrow 0$ splits by the exact same idea by sending $[(f,g)] \in \pi_{n-1}(X\times Y)$ to $[f]+[g] \in \pi_{n-1}(X\vee Y)$ where this is well defined if $n-1>1$ since the homotopy groups will be abelian.