# The proof of Duality theorem ( Milnor & Stasheff )

At the beginning of the proof(theorem 11.10 page 128), it says that it follows easily that the diagonal class can be expressed as r-fold sum \begin{align*} u''=b_i\times c_i+\ldots+b_r\times c_r \end{align*} by using the Kunneth formula \begin{align*} H^*(X\times Y)\cong H^*(X)\otimes H^*(Y) \end{align*} where $$b_i,c_i$$ are basis for $$H^*(M)$$ and $$|b_i|+|c_i|=n=\dim(M)$$. $$u'':=u'|_{M\times M}$$ with

$$u'\in H^{n}(M\times M,M\times M-\Delta(M))\cong H^{n}(TM,TM-M)$$

Here the isomorphism is derived by the diagonal embedding $$\Delta:M \rightarrow M\times M\quad x\mapsto (x,x)$$ and Tubular Neighborhood Theorem

I think I miss something here, I only know that $$H^n(M \times M)=\oplus_{i+j=n}H^{i}(M)\otimes H^{j}(M)$$. So i don't know why $$u''$$ has this explicit form instead of $$u''=b_1\times c_1$$ or some other formula.

The answer below pointed out my flow. Thanks a lot.

• What ring are you working with for your coefficients? Künneth gives you a short exact sequence $$0 \to H^∗(X,L) \otimes H^∗ (Y,M) \to H^∗ (X\times Y,L\otimes M) \to \text{Tor}^1 (H^∗ (X,L), H^∗ (Y,M)) \to 0$$ So you only have an iso in specific cases such as $L = M$ and cohomologies are finite dimensional. I assume here $L = M = \mathbb{Z}$. Ok I did not see that you had finite basis my bad :) Jun 24, 2023 at 14:26
• I think this has to do with the fact that $u''$ is the diagonal class, that's why you have each coefficient with scalar $1$ Jun 24, 2023 at 14:33

They don't claim that $$\{c_1, \dots, c_r\}$$ is a basis for $$H^*(M)$$, they only assume $$\{b_1, \dots, b_r\}$$ is a basis for $$H^*(M)$$.

Any element of $$H^*(X)\otimes H^*(Y)$$ can be written as $$\sum_{i=1}^pu_i\otimes v_i$$ where $$u_1, \dots, u_p \in H^*(X)$$ and $$v_1,\dots, v_p \in H^*(Y)$$. The isomorphism $$H^*(X)\otimes H^*(Y) \to H^*(X\times Y)$$ is generated by $$x\otimes y \mapsto \pi_1^*x\cup\pi_2^*y = x\times y$$, so $$\sum_{i=1}^pu_i\otimes v_i \mapsto \sum_{i=1}^pu_i\times v_i$$. That is, every element of $$H^*(X\times Y)$$ can be written as $$\sum_{i=1}^pu_i\times v_i$$ for some $$u_1, \dots, u_p \in H^*(X)$$ and $$v_1,\dots, v_p \in H^*(Y)$$.

Now consider the case $$X = Y = M$$ and suppose $$\{b_1, \dots, b_r\}$$ is a basis for $$H^*(M)$$. Then for any $$u_i \in H^*(M)$$, we have $$u_i = \sum_{j=1}^rm_{ij}b_j$$ where $$m_{ij} \in \Lambda$$, the field of coefficients. Therefore

\begin{align*} \sum_{i=1}^pu_i\times v_i &= \sum_{i=1}^p\left(\sum_{j=1}^rm_{ij}b_j\right)\times v_i\\ &= \sum_{i=1}^p\sum_{j=1}^rm_{ij}b_j\times v_i\\ &= \sum_{j=1}^r\sum_{i=1}^pm_{ij}b_j\times v_i\\ &= \sum_{j=1}^r\sum_{i=1}^pb_j\times m_{ij}v_i\\ &= \sum_{j=1}^rb_j\times\left(\sum_{i=1}^pm_{ij}v_i\right)\\ &= \sum_{j=1}^rb_j\times c_j \end{align*}

where $$c_j := \sum_{i=1}^pm_{ij}v_i$$.

• So at this moment we don't know if $c_j$ is zero or not? We know it at the end of the proof?
– Jino
Jun 24, 2023 at 14:55
• At this stage of the proof, they are just using the fact that the diagonal class, like any class in $H^*(M\times M)$, can be written as $b_1\times c_1 + \dots + b_r\times c_r$ for some $c_1, \dots, c_r \in H^*(M)$. Using the properties of the diagonal class, they then determine what the classes $c_i$ are. Jun 24, 2023 at 14:57
• thanks. i assumed $c_j$ are also basis.
– Jino
Jun 24, 2023 at 15:00
• @Jino You do not need to assume that. The classes $c_i$'s are just undetermined elements in $H^*(M)$. In the course of the proof they determine that $c_j = (-1)^{\text{deg} b_j}b_j^\#$ but you do not and in fact, cannot assume that $c_j$'s form a basis because not every cohomology class in the product will be cross product of basis elements. Jun 24, 2023 at 23:04
• yes, this is the reason i got confused
– Jino
Jun 24, 2023 at 23:07