Let $(X;U,V)$ be an excisive triad and consider the corresponding part of the MayerVietoris sequence $H^{\bullet1}(U\cap V)\stackrel{\delta^*}{\to} H^\bullet(X)\to H^\bullet(U)\oplus H^\bullet(V)$. Now let $\alpha,\beta\in H^{\bullet1}(U\cap V)$. Can we say anything about $\delta^*(\alpha\smile\beta)$?

4$\begingroup$ I don't think you can in general, but if $\alpha$ is the restriction from $X$ of $\alpha'$, then $\delta^*(\alpha \cup \beta) = \alpha' \cup \beta$. You can check this by properly choosing a lift of $\alpha \cup \beta$ as a chain to $U$ ($\alpha'_U \cup \beta'$ for some lift $\beta'$). $\endgroup$– S. carmeliMay 3, 2018 at 17:00

$\begingroup$ Can you extend on our argument or do you have a reference for it? This formula would be great, but I only find the version for the relative cohomology sequence. $\endgroup$– FKranholdJul 8, 2018 at 12:49
1 Answer
Here are some comments.
The map $\delta^\ast$ is a composition of the form $\require{AMScd}$
\begin{CD} H^{\ast1}(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 EilenbergMac 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 $H^{k+\ell+1}(\Sigma K(\Bbb Z,k+\ell))$.

$\begingroup$ What is a reference for this description of the MayerVietoris map $\delta^\ast$? I haven't seen it in Hatcher, Dold, Spanier, etc. (albeit the standard references seem to only elaborate on the homological MV unless it's using de Rham cohomology). I take it "cones off" means pinches off $U(U\cap V)$ and $V(U\cap V)$ separately in $X$. I like this description because then it's easy to see that elements in $Im(\delta^\ast)$ have trivial (cup product) square. $\endgroup$ Dec 25, 2018 at 20:06

$\begingroup$ I don't know of a reference; it's probably folklore. It is a reinterpretation that uses the BarrattPuppe sequence. There's a cofiber sequence $X \to \Sigma (U \cap V) \to \Sigma U \vee \Sigma V$, where $\Sigma$ means unreduced suspension. The sequence wiill induce the MayerVietoris sequence on cohomology. $\endgroup$ Dec 26, 2018 at 4:05

$\begingroup$ "cones off" refers to the following: There is an inclusion $X = U \cup_{U\cap V} V \to CU \cup_{U\cap V} C V \simeq \Sigma (U\cap V)$, where $C$ means the unreduced cone. $\endgroup$ Dec 26, 2018 at 4:10