# $H^*(X;\Bbb Z)$ and $H^*(S^p\times S^q;\Bbb Z)$ are not isomorphic by an isomorphism preserving the cup product

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?

$$\textbf{Question 1:}$$ 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 $$\Bbb Z\cong\Bbb Z\otimes \Bbb Z\cong H^p(\Bbb S^p;\Bbb Z)\otimes H^q(\Bbb S^q;\Bbb Z)\ni \varphi\otimes \psi\longmapsto p^*\varphi\smile q^*\psi\in H^{p+q}(\Bbb S^p\times \Bbb S^q;\Bbb Z)\cong \Bbb Z$$

• 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? Jul 17, 2021 at 10:06
• 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 $p^*\oplus q^*\colon H^n(A)\oplus H^n(B)\to H^n(A\lor B)$. Jul 17, 2021 at 10:14
• 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. Jul 17, 2021 at 10:17
• 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. Jul 17, 2021 at 10:30
• Thanks. Just one more thing. $[\Bbb S^p]^*$ stands for a generator? Jul 17, 2021 at 10:36