Cup products in the Mayer-Vietoris sequence Let $(X;U,V)$ be an excisive triad and consider the corresponding part of the Mayer-Vietoris sequence $H^{\bullet-1}(U\cap V)\stackrel{\delta^*}{\to} H^\bullet(X)\to H^\bullet(U)\oplus H^\bullet(V)$. Now let $\alpha,\beta\in H^{\bullet-1}(U\cap V)$. Can we say anything about $\delta^*(\alpha\smile\beta)$?
 A: Here are some comments. 
The map $\delta^\ast$ is a composition of the form
$\require{AMScd}$
\begin{CD}
H^{\ast-1}(U\cap V) @> \sigma >\cong > H^{\ast}(\Sigma (U\cap V)) \to H^\ast X
\end{CD}
where the first map is given by the suspension isomorphism and the second is induced by the map $X\to \Sigma(U\cap V)$ which cones off $U$ and $V$ in $X$ (here $\Sigma$ means suspension). This second displayed map is cup product preserving so it is sufficient to investigate the first map.
Setting $Z = U\cap V$,  represent your cohomology classes as maps
$a: X \to K(\Bbb Z,k)$, $b: X \to K(\Bbb Z,\ell)$ (where we the Hopf classification theorem to identify $H^k(X)$ with  $[X,K(\Bbb Z,k)]$, where $K(\Bbb Z,k)$ is the Eilenberg-Mac Lane space). Then $a\cup b$ is given by
$\require{AMScd}$
\begin{CD}
X @>\text{diag} >> X \wedge X @>a\wedge b>> K(\Bbb Z,k)\wedge K(\Bbb Z,\ell) @> m >> K(\Bbb Z,k+\ell)\, .
\end{CD}
where $m$ represents the evident cohomology class in degree $k+\ell$ of the smash product (using the Künneth formula).
The class $\sigma(a\cup b)$ is given by
\begin{CD}
\Sigma X @>\Sigma \text{diag} >> \Sigma X \wedge X @>\Sigma a\wedge b>> \Sigma K(\Bbb Z,k))\wedge K(\Bbb Z,\ell) @>\Sigma m>> \Sigma K(\Bbb Z,k+\ell)\to K(\Bbb Z,k+\ell+1)\, ,
\end{CD}
where the last map represents the generator of 
$H^{k+\ell+1}(\Sigma K(\Bbb Z,k+\ell))$.
