Let $X,Y,Z$ be pointed spaces, and $f:X \wedge Y \rightarrow Z$ a map.
Then, $f$ induces a bilinear pairing $\pi_n(X) \times \pi_m(Y) \rightarrow \pi_{n+m}(Z)$ ($n,m \geq 1$).
I see what the pairing is, but I am having difficulty checking the bilinearity. Can someone give a quick argument for that; or cite some textbook which shows it?
Write out the definition of the homotopy group $\pi_n(X) = \mathsf{hTop}_\ast(\Bbb S^n,X)$. Note that $\land:\mathsf{hTop}_\ast\times\mathsf{hTop}_\ast \rightarrow \mathsf{hTop}_\ast$ is a functor and thus in particular gives rise to a map $$\tag{$\star$}\mathsf{hTop}_\ast^2((\Bbb S^m,\Bbb S^n),(X,Y)) = \mathsf{hTop}_\ast(\Bbb S^m,X)\times\mathsf{hTop}_\ast(\Bbb S^n,Y) \rightarrow\mathsf{hTop}_\ast(\Bbb S^m\land\Bbb S^n,X\land Y)$$ Under the homeomorphism $\Bbb{S}^m\land\Bbb{S}^n \cong \Bbb{S}^{m+n}$ it becomes a morphism $\pi_m(X)\times\pi_n(Y) \rightarrow \pi_{m+n}(X\land Y)$. This morphism is already bilinear and your bilinear pairing is just the bilinear map obtained from postcomposing with the group homomorphism $\pi_{m+n}(X\land Y)\rightarrow\pi_{m+n}(Z)$.
The monoid structure on $\pi_m(X)$ can be inferred from the comonoid structure of $\Bbb S^m$ given by pinching ie. contracting the equator to a point. By functoriality of $\land$ pinching of $\Bbb S^m$ on the left hand side of $(\star)$ gives a multiplication on the right hand side of $(\star)$ and the morphism in $(\star)$ is compatible with it. The same holds for pinching $\Bbb S^n$. Finally the three multiplications on $\pi_{m+n}(X\land Y) = \mathsf{hTop}_\ast(\Bbb S^m\land\Bbb S^n,X\land Y)$ given by pinching $\Bbb S^m$, $\Bbb S^n$ or $\Bbb S^{m+n}$, ie. the morphisms $$\begin{align*} \Bbb S^m\land\Bbb S^n &\rightarrow (\Bbb S^m\lor\Bbb S^m) \land \Bbb S^n \cong (\Bbb S^m \land \Bbb S^n)\lor(\Bbb S^m\land \Bbb S^n)\\ \Bbb S^m \land \Bbb S^n &\rightarrow \Bbb S^m\land(\Bbb S^n\lor\Bbb S^n) \cong (\Bbb S^m\land\Bbb S^n)\lor(\Bbb S^m\land\Bbb S^n)\\ \Bbb S^{m}\land\Bbb S^n\cong\Bbb S^{m+n}&\rightarrow\Bbb S^{m+n}\lor\Bbb S^{m+n} \cong (\Bbb S^m\land\Bbb S^n)\lor(\Bbb S^m\land\Bbb S^n) \end{align*}$$ give rise to the same multiplication by an application of the Eckmann-Hilton argument. This shows that the morphism $(\ast)$ is linear in each argument, ie. bilinear.