# proof for Kunneth formula using the double complex

prove: When a homomorphism $f: K \rightarrow K'$ of double complexes induces $H_d$-isomorphism, it also induces $H_D$-isomorphism.

This is in the context of proving Kunneth formula using the double complex. The map is the pullback of wedge products:

Let $U=\{U_{\alpha}\}$ be a good cover for $M$ and $\pi: M\times F\rightarrow M$ projection into $M$ and $\rho$ likewise into $F$. Then $\{\pi^{-1}U_{\alpha}\}$ is some cover for $M\times F$. Assume cohomology $H^*(F)$ is finite dimensional with closed forms basis $\{[\omega_{\alpha}]\}$ Then define

$$\pi^*: H^*(F) \otimes C^*(U,\Omega^*)\rightarrow C^*(\pi^{-1}U,\Omega^*)$$

$$\pi^*([\omega_{\alpha}]\otimes\phi)=\rho^*\omega_{\alpha}\wedge \pi^*\phi$$

This commutes with $d$ and $\delta$ and induces an isomorphism in $d$-cohomology.

*without use of spectral sequences which comes later in the book.

book bott and Tu: 108 and 107, http://www.maths.ed.ac.uk/~aar/papers/botttu.pdf

Edit: I do not understand the answer below. The cohomolgy classes for me are given by "diagonals, see text. The answer below does not seem correct to me.

• FWIW, all that follows "This is the context..." is not really relevant to your question. Feb 1, 2017 at 1:40

Let $K$ and $K'$ be two first-quadrant cohomological complexes, let $d$ be the vertical differentials and $D$ the total one. Let $f:K\to K'$ be a morphism which induces an isomorphism in cohomology with respect to $d$.

Let us suppose that both $K$ and $K'$ have finitely many nonzero columns, and let us do induction on the number of columns. The hypothesis holds in your case, since the de Rham complex of a manifold has finite length.

Let $K_1$ and $K'_1$ be the subcomplexes of $K$ and $K'$ of all columns of positive degree and let $C$ and $C'$ be the first columns of $K$ and of $K'$. There is a commutatvie diagram of complexes

0 —> K_1  —> K  —> C  —> 0
|       |     |
V       V     V
0 —> K_1' —> K' —> C' —> 0


with exact rows and vertical arrows induced by $f$, which induces an infinite ladder of the form

H_{i-1}(C)  —> H_i(K_1)  —> H_i(K)  —> H_i(C)  —> H+i(K_1)
|         |            |          |          |
V         V            V          V          V
H_{i-1}(C') —> H_i(K_1') —> H_i(K') —> H_i(C') —> H+i(K_1')


Here the vertical arrows $H_{i-1}(C)\to H_{i-1}(C')$ and $H_{i}(C)\to H_{i}(C')$ are isomorphism by hypothesis, and the arrows $H_i(K_1)\to H_i(K_1')$ and $H_i(K_1)\to H_i(K_1')$ are isomorphisms inductively, as $K_1$ and $K_1$ have one less nonzero column than $K$ and $K'$. The arrow $H_i(K)\to H_i(K')$ is an isomorphism by the $5$-lemma.

The result is true even for complexes with infinitely many nonzero arrows. I'll leave it for someone with the energy to do so to write a proof :-)

• One way to handle the general case is to notice that the $i$th cohomology group of a first-quadrant double complex $K$ depends only on the quotient of $K$ by the subcomplex spanned by the columns with index greater than $i+2$, and reduce the general case to the finitely-columned one. Feb 1, 2017 at 2:01
• I think I nearly get it. Except that my "D" is an operator between "diagonals" not columns of the complex. Can I go from this argument to showing that the H_D cohomology is an isomporphism? (see the page in the book 107-108 if need be) Feb 1, 2017 at 3:18
• My D is not the horizontal differential but the total one, just as I wrote. In other worda, it is the sum of the vertical and the horizontal differentials, as in the book. Feb 1, 2017 at 3:28
• (Do not accept answers unless you understand them and see they actually solve your problem, by the way. If you accept something which is not quite what you want, it is much less probable that some ele will come and provide another, better answer) Feb 1, 2017 at 3:29
• @ Mariano Suárez-Álvarez oh ok I wanted to say thank you quickly :) Is it because your D is between sub-complexes so vertical and horizontal "sum " takes us from one subcomplex to the "bigger one". The columns are infinite so in effect my diagonals induce your "total" D between sub-complexes. If I am not clear let me get back to you with nice drawings later as to what I mean. Feb 1, 2017 at 3:40