Rational homotopy groups of spectra

Question

It is claimed in a paper of Adams, Harris and Switzer that $$\pi_*E \otimes \pi_*F \otimes \mathbb{Q} \to E_*F \otimes \mathbb{Q}$$ is an isomorphism. This map is constructed by taking the map $\pi_*E \otimes \pi_*F \to E_*F$ which smashes $f:\Sigma^{|f|}\mathbb{S} \to E$ and $g:\Sigma^{|g|}\mathbb{S} \to F$ and together to get $f \wedge g:\Sigma^{|f|+|g|}\mathbb{S} \to E \wedge F$, then tensoring with $\mathbb{Q}$. It is sais that the isomorphism in question falls out of exactness of tensoring with $\mathbb{Q}$, but I do not see it. Any help would be appreciated.

Answer 1

Fix a spectrum $E$.

The functor $F \mapsto E_*(F) \otimes \mathbb{Q}$ is a homology theory because it is exact. (Clearly $E_*(-)$ is a homology theory, and tensoring with $\mathbb{Q}$ is exact.)

The functor $F \mapsto \pi_*E \otimes \pi_*F$ need not be exact, since tensoring with $\pi_*E$ need not be exact. But $\pi_*E \otimes \mathbb{Q}$ is a rational vector space, so tensoring with it is exact, and we get a homology theory $F \mapsto \pi_*E \otimes \pi_*F \otimes \mathbb{Q}$.

These two homology theories agree when $F=S^0$, so they agree on all spectra. Therefore $E_*(F) \otimes \mathbb{Q} = \pi_*E \otimes \pi_*F \otimes \mathbb{Q}$ for all $E$ and $F$.