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.

enter image description here

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.


1 Answer 1


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.

  • $\begingroup$ Why on a contractible open set $U$, $E|_U$ is homeomorphic to $U\times F$. By the definition of fiber bundle the only thing we know is there is a open cover of $U_\alpha$ of $M$ s.t restricted to the open cover it is trivial. $\endgroup$
    – Babai
    Jan 18, 2015 at 7:55
  • $\begingroup$ The book is assuming your "good cover," where the sets $U_\alpha$ and intersections of those sets are all diffeomorphic to $\mathbb R^n$, is also a trivializing cover. That is, the fiber-preserving homeomorphisms $E|_U \approx U \times F$ are assumed. It is actually a theorem that all bundles over paracompact bases are trivial, so this is a reasonable thing to do, but this might not be something that Bott & Tu have invoked so far. $\endgroup$
    – jdc
    Jan 18, 2015 at 18:13
  • $\begingroup$ @jdc I think you’ve made a mistake in your comment — it is certainly not the case that all bundles over paracompact bases are trivial (e.g. take the Mobius bundle over the circle). Perhaps you meant to say contractible? $\endgroup$ Apr 20, 2019 at 8:56
  • $\begingroup$ Yes, that's definitely what I meant. Paracompact (maybe paracompact Hausdorff? I'll check Steenrod) and contractible is what you want. $\endgroup$
    – jdc
    Apr 20, 2019 at 20:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .