Obstruction to a cohomology class on total space being a pullback of a class on the base space is the restriction to the fiber

Question

Let $\pi \colon E \to X$ be a fiber bundle with fiber $F$ and suppose that $\tilde H^i(F) = 0$ for $0 \leq i \leq k-1$.

Using the Leray-Serre spectral sequence, we get an exact sequence $$ 0 \to H^k(X) \to H^k(E) \to H^k(F) \to H^{k+1}(X) \to 0. $$ Therefore, a class in $H^k(E)$ is pulled back from a class in $H^k(X)$ if and only if it is zero when restricted to the fiber.

I want to show this using explicit representatives of these cohomology classes as homotopy classes of maps into Eilenberg-Maclane spaces.

That is, given a map $f \colon E \to K(\mathbb Z,k)$ such that the composition $F \to E \to K(\mathbb Z, k)$ is null-homotopic, I want to explicitly construct a map $g \colon X \to K(\mathbb Z, k)$ such that $f \simeq g \circ \pi$.

How could I go about this? I've read some obstruction theory, but haven't found anything applicable. If it helps, the specific case I'm interested in is when $F = T^n * T^n$ is the join of two tori and $k=3$.

Answer 1

I don't know if this is explicit enough for your purpose, but the existence of this map follows from the (dual) Blakers Massey theorem, at least with mild assumptions on $\pi$.

Let's start with the best case scenario: when the fiber sequence $F\to E\to X$ is also a cofiber sequence. In this case, it is clear that we can extend $E\to K(\mathbb{Z},k)$ to $X$, which doesn't require connectivity of $F$..

In your case, we have $F$ is $(k-1)$-connected, i.e. the map $F\to *$ is $k$-connected. If $E$ is $0$-connected, then the Blakers Massey theorem (see Theorem 4.4.2 in Munson-Volic) tells us that the map $Y:=\text{cof}(F\to E)\to X$ is $(k+1)$-connected. In particular, $H^{k+1}(X,Y;\pi_k(K(\mathbb{Z},k))=0$, so the obstruction to extending the map $Y\to K(\mathbb{Z},k)$ to $X$ vanishes. Thus, a map $g:X\to K(\mathbb{Z},n)$ such that $f\simeq g\circ \pi$ exists.