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Suppose $X$ and $Y$ are finite, simply connected, based CW-complexes and $m,n\geq 2$. If $a\in \pi_m(X)$ and $b\in \pi_n(Y)$, then one can regard these as elements of the homotopy groups of $X\vee Y$. If $a$ and $b$ are non-trivial, must the corresponding Whitehead product $[a,b]\in \pi_{m+n-1}(X\vee Y)$ be non-trivial?

I understand some cases using homotopy excision and the smash product but not this general situation.

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Suppose $\pi_m(X)\cong \mathbb Z/p$ and $\pi_n(Y)\cong \mathbb Z/q$, where $p$ and $q$ are distinct primes. Then $[a, b]=0$, by bilinearity of the Whitehead product.

On the other hand, if $a$ and $b$ are non-zero in $\pi_*\otimes \mathbb Q$, then I suspect that you can use rational homotopy theory to prove that $[a, b]\ne 0$.

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  • $\begingroup$ Suspicion in rational case is right: homotopy Lie algebra of a wedge of two simply connected spaces of finite rational type is the free product of their homotopy Lie algebras. This result was first published in THE RATIONAL HOMOTOPY OF A WEDGE. ALLAN CLARK AND LARRY SMITH. Pacific, volume 24, No. 2, 1968 $\endgroup$
    – Denis T
    Commented Aug 26 at 14:58

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