# Relating the Künneth Formula to the Leray-Hirsch Theorem

I am reading through Bott & Tu's Differential Forms in Algebraic Topology, which very early on discusses the Künneth formula and the Leray-Hirsch theorem for smooth principal bundles. The proof of Künneth is given explicitly, while Leray-Hirsch is said to follow by the same kind of argument but by adding the appropriate twists (no pun intended!).

It seems to me that the proofs Bott & Tu offer are independent of one another. Thus, in the presence of the Künneth formula, one might boldly restate Leray-Hirsch as

If a principal bundle contains global cohomology classes which freely generate the cohomology of the fibres, the bundle "looks" trivial in cohomology.

But this makes me wonder if I am putting Descartes before the horse (bad pun now intended); that is, perhaps Künneth is really just a simple corollary of Leray-Hirsch. Indeed, this would be the case if someone could prove something along the lines of the following:

Conjecture: Any smooth trivial principal bundle $\pi: G \times B \to B$ admits global cohomology classes on $G \times B$ which freely generate $H^*(G)$.

So here is where I'm not certain how to proceed. Triviality of the bundle guarantees we can actually define a projection onto the fibres: $\rho: G \times B \to G$, which I could then use to pullback cohomology classes from $G$ into the total space. Are these global?

• +1 for twisted humor. I'll think about the math in a whale. May 24, 2013 at 23:03

I'm not sure what 'global cohomology class' means, but for the Leray-Hirsch theorem I know of it's enough to ask that the inclusion of the fiber $G \rightarrow G \times B$ induces a surjection $H_{dR}^*(G \times B) \rightarrow H^*(G)$.
But this is clear since $G \rightarrow G \times B \rightarrow G$ is the identity, and so $G$ is a retract of $G \times B$. Since $H_{dR}^*$ is functorial it takes retracts to retracts (but flipping arrows), and so the hypotheses of the Leray-Hirsch theorem are satisfied, whence we have $H_{dR}^*(G) \otimes H_{dR}^*(B) \cong H_{dR}^*(G \times B)$.