On pg. 210 of the hard copy, and on pg 218 on the online copy of Hatcher's Algebraic Topology is defined the notion of relative cross product as follows:
Definition. Let $(X, A)$ and $(Y, B)$ be pairs of topological spaces, and $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$ be the projection maps. Define $\times :H^k(X, A; R)\times H^l(Y, B; R) \to H^{k+l}(X\times Y, A\times Y\cup X\times B; R)$ as $$(a, b)\mapsto p_1^*(a)\cup p_2^*(b)$$ where '$\cup$' denotes the cup product.
I do not understand how this definition makes sense. The map $p_1$ does not give us a map of pairs $(X\times Y, A\times Y\cup X\times B)\to (X, A)$, since the image of $A\times Y\cup X\times B$ under $p_1$ is all of $X$, and not $A$. Similarly for $p_2$.
So it is not clear what is meant by $p_1^*$ and $p_2^*$ here. Can somebody please clarify what is Hatcher trying to say here.
Also, what makes sense is using $p_1:(X\times Y, A\times B)\to (X, A)$ and $p_2:(X\times Y, A\times B)\to (Y, B)$ to get $$\times : H^k(X, A; R)\times H^l(Y, B; R)\to H^{k+l}(X\times Y, A\times B; R)$$ by writing $$(a, b)\mapsto p_1^*(a)\cup p_2^*(b)$$.
Is there a good reason why the second definition cannot be found in literature?
If you have a tripled ($X, A, B)$ with $A$ and $B$ open in $X$ you can define a relative cup product
$$H^*(X, A) \times H^*(X, B) \to H^*(X, A \cup B)$$
Define this on (co)chain level by taking cochains $\alpha, \beta$ from $C^k(X, A)$ and $C^\ell(X, B)$ respectively, cup product of which lives in the subgroup $C^{k+\ell}(X, A + B) \subset C^{k+\ell}(X)$ of cochains vanishing on sums of chains in $A$ and chains in $B$. $\alpha \cup \beta$ does indeed do so (this is just checking the definitions). To push $\alpha \cup \beta$ into something co homologous to a cochain living in the subgroup $C^{k+\ell}(X, A \cup B)$ is all what remains. This can be done by noting that the inclusion $C^*(X, A \cup B) \hookrightarrow C^*(X, A + B)$ is an isomorphism on homology (five lemma + (co)chain-level statement of excision).
The geometric point is that if you think of $\alpha \cup \beta$ as a piecewise-linear function on $X$ (appropriately triangulated), it vanishes on sums of simplices in $A$ and simplices in $B$. It vanishes on any chain on $A \cup B$ (which need not be like that: the chain could contain simplices which goes half into $A$ and half into $B$) because that's what excision says: if you triangulate the chain, refine it enough, it'd consist entirely of simplices of $A$ and simplices of $B$, on which $\alpha \cup \beta$ vanishes. That's all what "upto cohomology" happens.
That being said, $p_1^*(a)$ lives in $H^*(X \times Y, A \times Y)$ and $p_2^*(b)$ lives in $H^*(X \times Y, X \times B)$. Their cup product lives in $H^*(X \times Y, A \times Y \cup X \times B)$.