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 $\delta^*(\alpha\smile\beta)$?

  • 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. carmeli
    May 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. $\endgroup$
    – FKranhold
    Jul 8, 2018 at 12:49

1 Answer 1


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 $H^{k+\ell+1}(\Sigma K(\Bbb Z,k+\ell))$.

  • $\begingroup$ What is a reference for this description of the Mayer-Vietoris 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 Barratt-Puppe 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 Mayer-Vietoris sequence on cohomology. $\endgroup$
    – John Klein
    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. $\endgroup$
    – John Klein
    Dec 26, 2018 at 4:10

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