# Cohomology groups of a homotopy fiber

I am reading the following: http://www.indiana.edu/~jfdavis/teaching/m623/book.pdf and on page 316 there is a thing that gets me confused:

Consider the following situation: Assume that we know that $H^\ast(K(\mathbb{Z},n),\mathbb{Q}) = \mathbb{Q}[x]$ (the rational cohomology ring) and $\Lambda(x)$ (i.e the exterior algebra on one generator over $\mathbb{Q})$.

With this, let n be even. We have a map $f:K(\mathbb{Z},n) \rightarrow K(\mathbb{Z},2n)$ corresponding to $x^2 \in H^{2n}(K(\mathbb{Z},n))$. One can form the homotopy fiber of f, to obtain a fibration $F \rightarrow K(\mathbb{Z},n) \rightarrow K(\mathbb{Z},2n)$ and one can show that $\pi_k(F)=0$ if $k \neq n, 2n-1$ and $\pi_n(F) \cong \pi_{2n-1}(F) \cong Z$ from the long exact sequence of a fibration. Now, form the homotopy fiber of $F \rightarrow K(\mathbb{Z},n)$ , to get a fibration

$$K \rightarrow F \rightarrow K(\mathbb{Z},n).$$

Now, $K$ is homotopy equivalent to $K(\mathbb{Z},2n-1)$, so we can rewrite this as $$K(\mathbb{Z},2n-1) \rightarrow F \rightarrow K(\mathbb{Z},n).$$ Now, In the lecture notes I posted, it says that from an easy application of the Serre spectral Sequence, and the above knowledge of $H^\ast(K(\mathbb{Z},n),\mathbb{Q})$ one can conclude that $H^\ast(F,\mathbb{Q}) = \Lambda(x)$, where x is a generator in degree n. I however, don't see why this is true. Could anyone help me?

Update

We have come that the crucial problem is to see that the transgression is surjective, but it is not obvious (to me atleast) how this is done. Does anyone see how this can be done?

• Did you get this figured out? – Joe Johnson 126 Oct 16 '13 at 0:40
• Unfortunately no! – user101036 Oct 16 '13 at 7:59
• I worked with it a bit and found that on the $E_{2n}$ page of the spectral sequence, you have a bunch of maps. All you would need is for them to be onto. – Joe Johnson 126 Oct 16 '13 at 11:58
• @Joe Johnson: Right, that is what I concluded too, but I am not sure what these maps really are... Something about the transgression I think - do you know this? – user101036 Oct 16 '13 at 15:56
• @JoeJohnson126: Have you made any progress? I see thatyou have asked some questions regarding the question. Unfortunately, I don't have any answers, but the transgression has another form (as you probably know) , see for example pg. 186 of "User's guide to spectral sequences" . But I don't see how to calculate with that definition at all. – user101036 Oct 24 '13 at 18:38

Maybe this is it? I will write $K_i$ for the $i$th Eilenberg-Maclane space. Since $$F\longrightarrow K_n\stackrel{f}{\longrightarrow} K_{2n}$$ is the null map, we get a diagram: $$\require{AMScd} \begin{CD} \Omega K_{2n} @>>> PK_{2n} @>>> K_{2n}\\ @| @AAA @AA{f}A \\ \Omega K_{2n} @>>> F @>>> K_n \end{CD}$$ where the right square strictly commutes, and the left square commutes up to homotopy. This induces a map on spectral sequences that is an isomorphism for $(p,q)=(2n,0)$ and $(p,q)=(0,2n-1)$. We know that the map (transgression) $$E_{2n}^{0,2n-1}\longrightarrow E_{2n}^{2n,0}$$ is an isomorphism for the path-loop fibration. Thus the transgression must be an isomorphism for the other.
• @Tedar The path space is contractible. So, the only nontrivial group in $E_\infty$ is at $(p,q)=(0,0)$. This causes the transgression to be an isomorphism. – Joe Johnson 126 Oct 26 '13 at 13:38