TITLE: $QS^0$ isn't a product of Eilenberg-Mac Lane
QUESTION [12 upvotes]: I am reading Lewis' paper "Is there a convenient category of spectra?". To prove the main result on the non-existence of such a nice category, he shows that otherwise the unit component of $QS^0= \varinjlim \Omega^n S^n$ would have to be weakly equivalent a product of Eilenberg-Mac Lane spaces. So far so good, but it isn't immediately clear to me:
Q: Why can't the unit component of $QS^0$ be a product of Eilenberg-Mac Lane spaces?
This should probably be very easy, given how the paper feels no need to include any justification for it.

REPLY [22 votes]: If $X$ is a product of Eilenberg-MacLane spaces then the map 
$$ \eta^*\colon \pi_2(X) = [S^2,X]\to[S^3,X]=\pi_3(X) $$ is easily seen to be zero (where $\eta\colon S^3\to S^2$ is the Hopf map).  However, standard calculations give: 
\begin{align*}
 \pi_2(QS^0) & =\pi_2^S(S^0)=\mathbb{Z}/2.\eta^2\\
 \pi_3(QS^0) &= \pi_3^S(S^0)=\mathbb{Z}/24.\nu
\end{align*}
with $\eta^3=12\nu$.  Thus, $\eta^*$ is nonzero in this context.