Question about proof of theorem 3.16. (page 219,220 and 221 ) in Hatcher

Question

While i try to solve exercise 3.2.10 i realize that i can't understand where he use the sentence " $H^k(Y,R) $ is a finitely generated free R-module for all $k$" at the proof of theorem 3.16. In addition what about $H^k(X,R) $ Is it also need to be finitely generated for all $k$?

Answer 1

Hatcher's method of proof of the Kunneth formula (in the absolute case, which the relative case depends on) is to show that both sides of the formula are cohomology theories that agree on a point. In particular, we need to show that the functor $$h^n(X,A) := \bigoplus_{i+j=n} H^i(X,A;R) \otimes_R H^j(Y;R)$$ gives the long exact sequence of a pair. We know that $(X,A) \mapsto H^*(X,A;R)$ give long exact sequences, and the condition of being finitely-generated free is used to show that tensoring with $H^j(Y;R)$ preserves this exactness. You don't need $H^i(X;R)$ to be finitely-generated.

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EDIT: (1) Tensoring with any free module preserves exactness. (A module that preserves exactness upon tensoring is called flat, so we're saying free modules are flat.) To show this, we use the fact that tensor products commute with direct sums, so $A \otimes_R R^{\oplus I} \cong A^{\oplus I}$. So we just get $\lvert I \rvert$ copies of the original exact sequence, which is exact.

Here, note that finite generation is not needed - only freeness is. (For finitely generated modules over a nice ring, notions of freeness, projectivity, and flatness coincide.)

(2) I neglected to mention where finite generation of $H^j(Y;R)$ is needed in my original answer. The key is that cohomology theories have to behave well not only with exact sequences, but also with disjoint unions. In particular, we need $$(\prod_\alpha H^i(X_\alpha,A_\alpha;R)) \otimes_R H^j(Y;R) \cong \prod_\alpha (H^i(X_\alpha,A_\alpha;R) \otimes_R H^j(Y;R)).$$ Let $H^j(Y;R) \cong R^{\oplus n}$. Then the left hand side is just $(\prod_\alpha H^i(X_\alpha,A_\alpha;R))^{\oplus n}$ and the right hand side is $\prod_\alpha H^i(X_\alpha,A_\alpha;R)^{\oplus n}$. In general, products don't commute with direct sums, but they do with finite direct sums. One easy way to see this in this case is to note that finite sums are also finite products, and clearly products commute with products.

I hope this clears things up!