Coboundary of a cup-product

Question

Let $A\subset X$ be CW-complexes (or even manifolds). In cohomology with coefficients in a commutative ring $R$, we have a long exact sequence $$\cdots \rightarrow H^p(X,A)\rightarrow H^p(X)\rightarrow H^p(A)\stackrel{\partial_p }{\rightarrow }H^{p+1}(X,A)\rightarrow \cdots$$ Let $\alpha $ in $H^p(A)$, $\beta$ in $H^q(A)$. Is there a formula for $\partial_{p+q} (\alpha \smile\beta )$?

Answer 1

I will assume $(X,A)$ is a "good pair", i.e. that $A$ is a CW-subcomplex of $X$. Cup product works on mixed relative cohomologies, $H^p(X,A_1)\times H^q(X,A_2)\to H^{p+q}(X,A_1\cup A_2)$, so we can take $A_1=A_2=A$ or $A_1=A_2=\varnothing$ or $A_1=A$ and $A_2=\varnothing$ and vice versa. The coboundary map $H^n(A)\to H^{n+1}(X,A)$ is obtained by taking co-chains on $A$ and viewing them as co-chains on $X$ which vanish on $X-A$ and then pre-composing with the differential $C_{n+1}(X)\to C_n(X)$, i.e. it is obtained directly from the co-differential. If you work out the formula for the co-differential of the cup product of co-chains (which is the Leibniz rule), this should respect the values of the relative co-chains... but now I'm stuck, and the best I can say is the following:

If $\beta=i^\ast\eta$ where $i:A\hookrightarrow X$ and $\eta\in H^\ast(X)$, then the desired formula is $$\partial_{p+q}(\alpha\smile i^\ast\eta)=\partial_p\alpha\smile\eta$$

This is a "stability" result found in chapter VII section 8 of Dold's Lectures on Algebraic Topology. Exercise #3 of that section asks for a generalization of this result, which mimics the corresponding result for cross products (given in section 7 and section 2 of the same chapter). But while the cross product makes sense for general pairs $(X,A)$ and $(Y,B)$, the cup product needs $X=Y$ so that we may apply the diagonal map $\Delta:X\to X\times X$ (and appropriate relative versions). So I think Dold's desired "generalization of stability" only considers larger spaces such as $(X\times Y,A\times Y)$.

I originally wrote down a "Leibniz rule", but it's not defined (see the comments). Though for the cross product it is the case that $\partial_{p+q}(\alpha\times\beta)=\partial_p\alpha\times\beta=(-1)^p\alpha\times\partial_q\beta$.

Answer 2

Let me give an answer for $R=\mathbf Z$, the ring of integers, and let us translate to sheaf cohomology. Your long exact sequence comes from the short exact sequence $$ 0 \rightarrow j_! \mathbf Z \rightarrow \mathbf Z \rightarrow i_\star \mathbf Z \rightarrow 0 $$ of sheaves of abelian groups on $X$, where $i$ is the inclusion of the closed subset $A$ in $X$, and $j$ is the inclusion of its complement. In the derived category $\mathcal D^+$ of sheaves of abelian groups on $X$, it gives rise to a triangle $$ j_! \mathbf Z \rightarrow \mathbf Z \rightarrow i_\star \mathbf Z \rightarrow j_! \mathbf Z[1]. $$ The last morphism $$ \delta\colon i_\star \mathbf Z \rightarrow j_! \mathbf Z[1] $$ is responsable for the coboundary map in the long exact sequence of cohomology: $$ \partial(\alpha)=\delta[p]\circ\alpha, $$ where $\alpha$ is alternatively interpreted as an element of $\mathrm H^p(A)$ and $\mathrm{Hom}(\mathbf Z,i_\star \mathbf Z[p])$. Here $\mathrm{Hom}$ means morphisms in the derived category $\mathcal D^+$. If one continues by identifying both groups with yet another, $\mathrm{Hom}( i_\star \mathbf Z, i_\star \mathbf Z[p])$, the formula $$ \partial(\alpha\cup\beta)=\delta[p+q]\circ\alpha[q]\circ\beta $$ makes sense and is valid for $\beta\in\mathrm{Hom}(\mathbf Z, i_\star \mathbf Z[q])=\mathrm H^q(A)$, since composition in the derived category coïncides with cup product.

If we go on and interpret $\delta$ as a cohomology class in the local cohomology group with support in $A$ and coefficients in $j_! \mathbf Z$ $$ \mathrm H_A^1 (X,j_! \mathbf Z)=\mathrm{Hom}(i_\star \mathbf Z, j_! \mathbf Z[1]), $$ one has $$ \partial(\alpha)=\delta[p]\circ\alpha=\delta\cup\alpha, $$ for the extra-ordinary cup product $$ \cup\colon \mathrm H_A^1(X,j_! \mathbf Z) \times \mathrm H^p(A) \rightarrow \mathrm H_A^{p+1}(X,j_! \mathbf Z). $$ The only formula for $\partial (\alpha\cup\beta)$ that I can make out of this is $$ \partial(\alpha\cup\beta)=\delta\cup(\alpha\cup\beta)=(\delta\cup\alpha)\cup\beta, $$ formula where among the $5$ cups, $2$ are ordinary, and $3$ are extra-ordinary.