Verify that $\langle\alpha \cup \beta, u\rangle = \langle\beta, \alpha \cap u\rangle.$

For a commutative ring $R, \alpha \in \tilde{H}^p(X;R), \beta \in \tilde{H}^q(X;R)$ and $ u \in \tilde{H}_{p+q}(X;R),$ Verify $\langle\alpha \cup \beta, u\rangle = \langle\beta, \alpha \cap u\rangle.$

We were asked to solve this after reading section 22.5 of "Modern Classical Homotopy Theory " by Jeffery Strom and 22.3 of Peter May book.

Could anyone help me in solving this please?

Note: this $\langle,\rangle$ represents the cap product.

EDIT: I think the question should be typed as this:

Solution 1:

This question is somewhat poorly posed, because May and Strom use different notations for these operations. In May the symbols $\cap$, $\cup$ and $\langle, \rangle$ denote cup product, cap product, and evaluation pairing respectively, but it appears that Strom uses $\langle , \rangle$ to denote the cap product and $\cdot$ to denote the cup product, and I haven't found a definition of evaluation paring (Strom's definition might just be $\langle u, \alpha \rangle$, i.e. cap product, where $u$ and $\alpha$ have the same degree). This is why it's important to include your definitions in your question, because referring to more than one source can be ambiguous and if it's a problem for a course then they probably expect you to be using a particular definition. Since this is a bountied question I will help you out by including the definitions I'm using for these symbols, these are the definitions from the cited sections of Strom and May, and using Strom's convention of denoting cohomology classes by $u$ and $v$ and denoting homology classes by $\alpha$.

I managed to prove $\langle u \cup v, \alpha \rangle = \langle u, v\cap \alpha \rangle$, which I think differs from what you're asking for by a sign of $(-1)^{|v||u|}$, which may be a slight mistake I made or it might be a consequence of different definitions and conventions. Given the context of the problem it seemed like the idea is to prove it homotopically, rather than in terms of the chain-level definition.

$\cup$

For $u\in \tilde{H}^n(X;G)$, represent it with a map $u\colon X \to K(G,n)$, and similarly let $v\colon X \to K(H,m)$ represent an element of $\tilde{H}^m(X;H)$. Then the cup product $u\cup v \in \tilde{H}^{n+m}(X;G\otimes H)$ is defined by the composition $$ X \stackrel{\bar\Delta}{\to} X \wedge X \stackrel{u \wedge v}{\to} K(G,n)\wedge K(H,m) \to K(G\otimes H,n+m) $$

$\cap$

Let $u\in \tilde{H}^{k}(X;G)$ and $\alpha \in \tilde{H}_n(X;H)$ be represented by maps $u\colon X \to K(G,k)$ and $\alpha\colon S^{n+t} \to X \wedge K(H,t)$ respectively. Then the cap product $u\cap \alpha \in \tilde{H}_{n-k}(X; G\otimes H)$ is represented (in Strom) by

$$\begin{align} S^{n+t} &\stackrel{\alpha}{\to} X \wedge K(H,t)\\ &\stackrel{\bar\Delta \wedge id}{\to} X \wedge X \wedge K(H,t)\\ &\stackrel{id \wedge u\wedge id}{\to} X \wedge K(G,k)\wedge K(H,t)\\ &\to X \wedge K(G\otimes H, k+t)\end{align}$$

$\langle,\rangle$

Finally, for $u\in \tilde{H}^n(X;G)$ and $\alpha\in \tilde{H}_n(X;H)$ the evaluation pairing $\langle u, \alpha \rangle\in G\otimes H$ is defined as $$S^{n+t} \stackrel{\alpha}{\to} X \wedge K(H,t) \stackrel{u\wedge id}{\to} K(G,n)\wedge K(H,t) \to K(G\otimes H, n+t)$$

Now, let $u\in \tilde{H}^p(X;R)$, $v\in \tilde{H}^q(X;R)$, and $\alpha\in \tilde{H}_{p+q}(X;R)$, and we want to prove $\langle u \cup v, \alpha \rangle = \langle u, v\cap \alpha \rangle$. Let us start by writing down the functions that represent these operations.

$\langle u\cup v, \alpha \rangle$: $$\begin{align}S^{p+q+t} \stackrel{\alpha}{\to} X \wedge K(R,t) &\stackrel{\bar\Delta}{\to} X \wedge X \wedge K(R,t)\\ &\stackrel{u\wedge v \wedge id}{\to} K(R, p)\wedge K(R,q)\wedge K(R,t)\\ &\to K(R, p + q + t) \end{align}$$

$\langle u, v\cap\alpha \rangle$: $$\begin{align}S^{p+q+t} \stackrel{\alpha}{\to} X \wedge K(R,t) &\stackrel{\bar\Delta}{\to} X \wedge X \wedge K(R,t)\\ &\stackrel{id \wedge v \wedge id}{\to} X\wedge K(R,q) \wedge K(R,t)\\ &\to X \wedge K(R, q+t) \\ &\stackrel{u \wedge id}{\to} K(R, p) \wedge K(R, q+t) \\ &\to K(R, p + q + t)\end{align}$$

This looks a bit unwieldy, but notice that both operations start with $\bar\Delta \circ \alpha$, so you only have to show that the following commutes up to homotopy:

$\require{AMScd}$ \begin{CD} X\wedge X \wedge K(R,t) @>{id \wedge v \wedge id}>> X \wedge K(R,q) \wedge K(R,t) @>>>X \wedge K(R, q+t)\\ @V{u \wedge v \wedge id}VV && @V{u\wedge id}VV \\ K(R,p)\wedge K(R,q)\wedge K(R,t) @>>> K(R, p + q + t) @<<< K(R,p) \wedge K(R, q+ t) \end{CD}

Here you have to use associativity of $\wedge$ and of the maps $\phi_{n,m}\colon K(G,n) \wedge K(H,m) \to K(G\otimes H, n+m)$.