Kunneth Formula: Let M and F are manifolds. If M has a finite good cover then $H^n(M\times F)=\bigoplus _{p+q=n} H^p(M)\bigotimes H^q (F)$
Bott and tu says One can prove Leray-Hirsch theorem by the same argument for Kuneth.
I dont see how does Leray-Hirsch follows via similar argument as Kunneth-Formula. Can someone help?
I saw a similar question Relating the Künneth Formula to the Leray-Hirsch Theorem here. But I guess that post doesn't answer my question.
Best Answer
The idea is arguably the main theme of the first half of the book, "induction on a good cover."
For $U$ a contractible open set, the restricted bundle $E|_U$ over $U$ is homeomorphic to $U \times F$, so its cohomology is just $$H^*(E|_U) = H^*(F) = H^*(F) \otimes H^*(U),$$
since $H^*(U)$ is trivial, and has a basis given by the $e_k$. Now consider what happens on a union $U \cup V$ of two contractible open sets? (We will write $H^*(F) = \mathbb{R}\{e_k\}$ for a vector space, not a ring; the maps involved will typically destroy the multiplicative structure of $H^*(F)$.)
The Mayer–Vietoris sequence yields a triangle
$$H^*(E|_{U \cup V}) \to H^*(E|_U) \times H^*(E|_V) \to H^*(E|_{U \cap V}) \to \cdots,$$
with a graded vector space map from the sequence
$$H^*(F) \otimes H^*({U \cup V}) \to (H^*(F) \otimes H^*(U)) \times (H^*(F) \otimes H^*(V)) \to H^*(F) \otimes H^*({U \cap V}) \to \cdots,$$
given on the basespace factor by $\pi^*$ and on $H^*(F)$ by restriction of the global classes $e_k$ to the appropriate set. This is as in the proof of the Künneth theorem, with the difference that for that theorem, the projection $\rho\colon E = M \times F \to F$ induces the ring map $\rho^* \colon H^*(F) \to H^*(E)$ that we needed; here, although $\rho$ no longer exists, the hypothesis is that $\rho^*$ still exists, at least as a map of vector spaces.
Making this allowance, the map of sequences exists, so because we know the isomorphism for contractible sets, the five lemma gives us a $H^*(U \cup V)$-module isomorphism $$H^*(F) \otimes H^*({U \cup V}) \to H^*(E|_{U \cup V}).$$
Replacing $U$ with $U_1 \cup \cdots \cup U_{n-1}$ and $V$ with $U_n$, this is the induction step, just as in the other proof.
The only further thing I believe you need to check is the commutativity in this case of the analogue of the bottom square from the previous page.