Corollary 14. Let $\{X\times B,A\times Y\}$ be an excisive couple in $X\times Y$ and let $p_1:(X,A)\times Y\to(X,A)$ and $p_2:X\times(Y,B)\to (Y,B)$ be the projections. Given $u\in H^p(X,A;G)$ and $v\in H^q(Y,B;G)$, then in $H^{p+q}((X,A)\times(Y,B);G\otimes G')$, we have $$u\times v = p_1^*(u)\smile p^*_2(v)$$

I have a question in Spanier AT Example 5.6.15:

Let $p$ and $q$ be integers $\geq 1$ and let $X$ be the union of $S^p,S^q$ and $S^{p+q}$ all identified at one point. If $i:S^p\hookrightarrow X, j:S^q\hookrightarrow X$ and $k:S^{p+q}\hookrightarrow X$ then $i_*\tilde{H}(S^p)\oplus j_*\tilde{H}(S^p)\oplus k_*\tilde{H}(S^{p+q})\simeq\tilde{H}(X)$. Computing $H(S^p\times S^q)$ by the Kunneth formula, we see that $H(X)\simeq H(S^p\times S^q)$. By the universal coefficient theorem, $X$ and $S^p\times S^q$ have isomorphic homology and cohomology groups for any coefficient group. Since $k^*:H^{p+q}(X;\Bbb Z)\simeq H^{p+q}(S^{p+q};\Bbb Z)$ and $k^*$ commutes with the cup product, it follows that the cup product of integral cohomology classes of degree $p$ and $q$, respectively, in $X$ is zero. However, it follows from corollary 14 that there are integral cohomology classes of $S^p\times S^q$ of degrees $p$ and $q$, respectively, whose cup product is nonzero....

There are two parts that I can't understand:

  1. ...it follows that the cup product of integral cohomology classes of degree $p$ and $q$, respectively, in $X$ is zero.
  2. ...there are integral cohomology classes of $S^p\times S^q$ of degrees $p$ and $q$, respectively, whose cup product is nonzero.

Could you explain why these are true?


1 Answer 1


$\textbf{Question 1:}$ Here, I show all cup products in the positive degrees of the wedge of finitely many spheres$($of different dimensions$)$ vanishes.

Let $p_i\colon \Bbb S^{r_1}\lor \cdots\lor \Bbb S^{r_k}\to \Bbb S^{r_i}$ be the projections for $i=1,...,k$. We have the isomorphism $p^*_1\oplus \cdots\oplus p_k^*\colon H^n(\Bbb S^{r_1};R)\oplus\cdots \oplus H^n(\Bbb S^{r_k};R)\to H^n(\Bbb S^{r_1}\lor\cdots\lor \Bbb S^{r_k})$.

Consider the projections $g\colon \Bbb S^{r_1}\lor\cdots\lor\Bbb S^{r_k}\to \Bbb S^{r_i}\lor \Bbb S^{r_j}$ and $q_i,q_j\colon \Bbb S^{r_i}\lor \Bbb S^{r_j}\to \Bbb S^{r_i},\Bbb S^{r_j}$. So, $q_ig=p_i,q_jg=p_j$. Hence, for $r_i+r_j>0$ we have

$$p_i^*[\Bbb S^{r_i}]^*\smile p_j^*[\Bbb S^{r_j}]^*=g^*\big(q_i^*[\Bbb S^{r_i}]^*\smile q_j^*[\Bbb S^{r_j}]\big)$$$$\in g^*\big(H^{r_i+r_j}(\Bbb S^{r_i}\lor\Bbb S^{r_j})\big)=g^*\big(H^{r_i+r_j}(\Bbb S^{r_i})\oplus H^{r_i+r_j}(\Bbb S^{r_j})\big)=g^*(0\oplus 0)=0.$$

$\textbf{Product Theorem:}$ Let $(X,A)$ and $(Y,B)$ be pairs of topological spaces. Let $A_1,A_2\in \{\varnothing ,A\}$ and $B_1,B_2\in \{\varnothing, B\}$. Consider the projections $p,q\colon X\times Y\to X,Y$. Suppose, $(X\times Y,A\times Y,X\times B)$ is an excisive triad. Let $\alpha\in H^k(X,A_1),\ \beta\in H^l(Y,B_1), \mu\in H_{m-k}(X,A_2),\ \nu\in H_n(Y,B_1\cup B_2)$. Then, $$\big(p^*(\alpha)\smile q^*(\beta)\big)\frown (\mu\times \nu)=(-1)^{\deg \alpha\cdot \deg \beta\ +\ \deg \beta \cdot \deg \mu}(\alpha\frown \mu)\times (\beta\frown \nu)$$ in $H_{m+n-k-l}(X\times Y,X\times B_2\cup A_2\times Y)$.

$\textbf{Note:}$ Here $\times$ is the cross product, and the power of $-1$ may be different in different literature.

$\textbf{Fact:}$ Let $M,N$ be two connected compact oriented manifolds with the fundamental classes $[M]\in H_m(M,\partial M),\ [N]\in H_n(N,\partial N)$. Let $p,q\colon M\times N\to M,N$ be the projections. So, $p^*[M]^*\in H^m(M\times N,\partial M\times N),\ q^*[N]^*\in H^n(M\times N,M\times \partial N)$.

Now, for $x\in M,\ y\in N$ in $H_0(M\times N)$ we have$$\big(p^*[M]\smile q^*[N]\big)\frown [M\times N]=\big(p^*[M]^*\smile q^*[N]^*\big)\frown\big([M]\times [N]\big)$$$$=(-1)^{mn+mn}\big([M]^*\frown [M]\big)\times ([N]^*\frown [N]\big)=[x]\times [y]=\big[(x,y)\big]$$$$\implies p^*[M]^*\smile q^*[N]^*=[M\times N]^*.$$ Note that the fundamental class $[M\times N]\in H_{m+n}\big(M\times N, \partial(M\times N)\big)$ is same as the $[M]\times [N]$, this follows once you consider the isomorphism $$H_m(M,M\backslash x)\otimes H_n(N, N\backslash y)\xrightarrow[\cong]{\text{cross product}} H_{m+n}\big(M\times N,M\times N\backslash (x,y)\big)$$

$\textbf{Question 2:}$ Consider pojections $p,q\colon \Bbb S^p\times \Bbb S^q\to\Bbb S^p,\ \Bbb S^q$ to see $p^*[\Bbb S^p]^*\smile q^*[\Bbb S^q]^*=[\Bbb S^p\times \Bbb S^q]^*=$the dual fundamental class of the oriented $(p+q)$-dimensional manifold $\Bbb S^p\times \Bbb S^q$.

$\textbf{The Künneth Theorem for cohomology groups:}$ Let $X$ and $Y$ be two topological spaces. We denote by $p,q \colon X \times Y \to X,Y$ the projection maps. If all the homology groups of $X$ are finitely generated, then given any $n\geq 0$, there exists a short exact sequence $$0\to \bigoplus_{k+l=n}H^k(X;\Bbb Z)\otimes H^l(Y;\Bbb Z)\xrightarrow{\times} H^n(X\times Y;\Bbb Z)$$$$\to \bigoplus_{k+l=n+1}\text{Tor }\big(H^k(X;\Bbb Z),H^l(Y;\Bbb Z)\big)\to 0 $$ which is natural with respect to the topological spaces $X$ and $Y$. If all the homology groups of $Y$ are also finitely generated, then the short exact sequence splits. The map $\times$ can be described considering $H^k(X;\Bbb Z)\otimes H^l(Y;\Bbb Z)\ni \varphi\otimes \psi\longmapsto p^*\varphi\smile q^*\psi\in H^{k+l}(X\times Y;\Bbb Z)$

$\textbf{Question 2:}$ One can also prove $p^*[\Bbb S^p]^*\smile q^*[\Bbb S^q]^*\neq 0$ considering the above, as for this particular case we have the isomorphism $$\Bbb Z\cong\Bbb Z\otimes \Bbb Z\cong H^p(\Bbb S^p;\Bbb Z)\otimes H^q(\Bbb S^q;\Bbb Z)\ni \varphi\otimes \psi\longmapsto p^*\varphi\smile q^*\psi\in H^{p+q}(\Bbb S^p\times \Bbb S^q;\Bbb Z)\cong \Bbb Z$$

  • $\begingroup$ Thank you for your answer. 1. I understood the most of the part except the isomorphism. I only know the isomorphism $H^n(\Bbb S^{r_1};R)\oplus\cdots\oplus H^n(\Bbb S^{r_k};R)\simeq H^n(\Bbb S^{r_1}\vee\cdots\vee S^{r_k};R)$ from M-V sequence. How do you know the map $p_1^*\oplus\cdots\oplus p_k^*$ gives an isomorphism?. 2. Actually I don't know the cap product. That topic appears after the example. Could you explain this without that concept? $\endgroup$ Jul 17, 2021 at 10:06
  • $\begingroup$ For a good wedge, $A\lor B$ with inclusions $i,j\colon A, B\hookrightarrow A\lor B$ and projections $p,q\colon A\lor B\to A,B$ use Mayer-Vietoris Theorem to show, $i^*\oplus j^*\colon H^n(A\lor B)\to H^n(A)\oplus H^n(B)$ is the inverse of $p^*\oplus q^*\colon H^n(A)\oplus H^n(B)\to H^n(A\lor B)$. $\endgroup$
    – Sumanta
    Jul 17, 2021 at 10:14
  • $\begingroup$ Ah, just an inverse! Thanks. Could you answer 2? I mean without cap product concept. The context makes me feel like I can make a conclusion without using cap product. $\endgroup$ Jul 17, 2021 at 10:17
  • $\begingroup$ From Corollary $14$ we have $[\Bbb S^p]^*\times [\Bbb S^q]^*=p^*[\Bbb S^p]\smile q^*[\Bbb S^q]^*$. Now, use Theorem $1$ on page $249$ to show $[\Bbb S^p]^*\times [\Bbb S^q]^*=[\Bbb S^p\times \Bbb S^q]^*=$the dual fundamental class of $(p+q)$-dimensional manifold $\Bbb S^p\times \Bbb S^q$. Let me know if you want more details. $\endgroup$
    – Sumanta
    Jul 17, 2021 at 10:30
  • $\begingroup$ Thanks. Just one more thing. $[\Bbb S^p]^*$ stands for a generator? $\endgroup$ Jul 17, 2021 at 10:36

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