I am trying to prove the following for singular cohomology with coefficients in a commutative ring $R$:
Let there be excisive triad $(U\cup V;U,V)$, a subspace $W\subseteq U\cup V$ and $i:(U\cap V, W\cap A\cap B)\hookrightarrow (U\cup V, W)$ the canonical inclusion. Let $x\in H^{p}(U\cap V)$ and $y\in H^{q}(U\cup V, W)$. Let $\varDelta$ denote the connecting homomorphism of the Mayer-Vietoris sequence of the triad $(U\cup V;U,V)$ and $\varDelta'$ the connecting homomorphism of the relative Mayer-Vietoris sequence of the excisive triads $(U\cup V; U,V)$ and $(W, W\cap U, W\cap V)$.
Then in $H^{p+q+1}(U\cup V,W)$ the equation
$$\varDelta'(x\cup i^{*}y)=\varDelta x\cup y$$
holds true.
Here are some of my thoughts:
The definition of the cup product I am working with is via the cross product: $x\cup y:=\delta^{*}(x\times y)$ where $\delta^{*}$ is the homomorphism induced by the diagonal map. So it seemed reasonable to me to show the analogue for the cross product. If I consider the absolute cross product I obtain a statement of the form $\varDelta'(x\times y)= (\varDelta x)\times y$ by comparing two short exact sequences from which the Mayer-Vietoris sequences arise and applying the five lemma:
$$\require{AMScd} \begin{CD} 0 @>>> hom(C_{*}(U)+C_{*}(V),R) @>>> C^{*}(U)\oplus C^{*}(V) @>>> C^{*}(U\cap V) @>>> 0 \\ @| @VV\times zV @VV\times zV @VV\times zV @| \\ 0 @>>> hom(C_{*}(U\times Z)+C_{*}(V\times Z),R) @>>> C^{*}(U\times Z)\oplus C^{*}(V\times Z) @>>> C^{*}(U\cap V) @>>> 0 \\ \end{CD} $$ for some $z\in C^{*}(Z)$.
This is where I get stuck. I keep failing to generalize this proof strategy to the case of relative cross products.
Any idea? Or is there a different way to deal with the problem? Thanks in advance.