I have been working on this question here. Here is the setup: First, all cohomology groups are assume to be with $\mathbb{Q}$ coefficients. We assume that $H^*(K(\mathbb{Q},n))=\mathbb{Q}[x]$, with $|x|=n$ (here $n$ is even). Thus $x^2$ generates $H^{2n}$. This means that there is a map $f:K(\mathbb{Q},n)\longrightarrow K(\mathbb{Q},2n)$ that classifies $x^2$. Let $E$ be the homotopy fiber of $f$. Moreover one can show that the homotopy fiber of $E\longrightarrow K(\mathbb{Q},n)$ is $K(\mathbb{Q},2n-1)$.
Now, in the Serre spectral sequence, we have a map $$E_{2n}^{0,2n-1}=H^{2n-1}(K(\mathbb{Q},2n-1))\longrightarrow H^{2n}(K(\mathbb{Q},n))=E_{2n}^{2n,0}$$ given by the transgression map. I'd like to show that the transgression is onto.
To this end all I need to show is that $x^2$ is in the image. We consider the diagram $$\require{AMScd} \begin{CD} E\cup CF @>{\delta}>> \Sigma F\\ @V{p}VV \\ K(\mathbb{Q},n) @>{f}>> K(\mathbb{Q},2n) \end{CD}$$ If I can fill in a map $\Sigma F\longrightarrow K(\mathbb{Q},2n)$ to make the square homotopy commutative, I will have proved that $x^2$ is in the image of the transgression.
My main confusion is that I believe the composition $f\circ p$ is nullhomotopic, since this is just the fibration listed above. This would suggest I can fill in the diagram with the null map. But, $f$ is certainly not null (in fact it is an isomorphism) and $p$ induces an injective map on $H^{2n}$. If not, it would have nontrivial kernel. But, $E_{2n}^{2n,0}\cong H^{2n}(K(\mathbb{Q},n))/\ker(p^*)\cong H(K(\mathbb{Q},n))$ due to lack of nonzero differentials until the $2n$ page.
What is it that I am missing? My guess would be that $\Sigma\operatorname{Id}$ is the map I should be using to fill in the diagram. Is this correct?
• Just because the composite $fp$ is nullhomotopic in $E$ canonically, from the fibration sequence doesn't mean that it's nullhomotopic: you have to make a choice of an extension of $p$ to the cone. – Tyler Lawson Oct 19 '13 at 2:06