Let $X$ and $Y$ be finite CW-complexes and $p,q\geq 2$. The Whitehead bracket induces a homomorphism $\pi_{p}(X)\otimes \pi_{q}(Y)\to \pi_{p+q-1}(X\vee Y)$, $\alpha\otimes \beta\mapsto [\alpha,\beta]$. Is this homomorphism always injective?
Here, I'm not assuming any higher connectedness assumptions on the spaces involved except for maybe $1$-connectedness.
The answer is (still) no. It comes down to the assertion that the homomorphism $\pi_p(A)\otimes \pi_q(B)\to \pi_{p+q}(A\wedge B)$ is not always injective.
To see this, let us rewrite the Whitehead product as a Samelson product. There is a natural map $\Omega X \wedge \Omega Y\to \Omega(X\vee Y).$ It is essentially the commutator of the two inclusion maps. From this we obtain a composition of homomorphisms $$ \pi_{p}(\Omega X)\otimes \pi_{q}(\Omega Y) \to \pi_{p+q}(\Omega X\wedge \Omega Y)\to \pi_{p+q}(\Omega(X\vee Y)).$$ This homomorphism is equivalent to the Whitehead bracket as defined in the OP. So to find an example when the Whitehead bracket is not injective it is enough to find an example where the homomorphism $ \pi_{p}(\Omega X)\otimes \pi_{q}(\Omega Y) \to \pi_{p+q}(\Omega X\wedge \Omega Y)$ is not injective. Let's take $X=S^{m+1}, Y=S^{n+1}$. Recall that there is a $2m-1$-connected map $S^m\to \Omega S^{m+1}$. From this, we obtain a diagram of groups $$ \begin{array}{ccc} \pi_p(S^m) \otimes \pi_q(S^n) & \to & \pi_{p+q}(S^{m}\wedge S^n) \\ \downarrow & & \downarrow \\ \pi_{p}(\Omega S^{m+1})\otimes \pi_q(\Omega S^{n+1}) & \to & \pi_{p+q}(\Omega S^{m+1}\wedge \Omega S^{n+1}). \end{array} $$ When $p< 2m-1$ and $q<2n-1$, the left vertical homomorphism is an isomorphism, and the homotopy groups are in the stable range. To find an element in the kernel of the bottom homomorphism, choose two elements in the stable homotopy groups of spheres whose tensor product is not zero, but whose smash product is zero. For example $\eta$ and $\nu$. Then realise them as unstable homotopy groups of spheres of high enough dimension.
