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 $\partial_{p+q} (\alpha \smile\beta )$?
[Math] Coboundary of a cup-product
at.algebraic-topology
Related Solutions
One of the standard proofs of Poincaré duality, at least for those manifolds that have handle decompositions, provides a reason for some of these naturality properties. Every piecewise linear manifold, or every smooth manifold, has a handle decomposition, and many but not all topological manifolds also do. (Amazingly enough, the only exceptions are in 4 dimensions.) A handle decomposition gives rise to two different CW cellulations on the manifold, one using cores and the other using co-cores. Then this proof of Poincaré duality posits that the CW chain complex of one cellulation is identical to the CW cochain complex of the other cellulation.
You can extend this coincidence of chain complexes to both of your examples, the Mayer-Vietoris sequence and the exact sequence of a pair. Obtaining identical chain complexes also gives you other information, for instance that the Bockstein maps are the same.
$\DeclareMathOperator{\Sq}{Sq}\newcommand{\Z}{\mathbb{Z}}$The short version is that every cohomology operation can be interpreted as a Bockstein operator for an "exact sequence" (read: fiber sequence) of grouplike $E_\infty$ spaces.
Any cohomology operation $\delta:H^*(-;A)\Rightarrow H^{*+k}(-;B)$ such as $\Sq^i$ gives a morphism of Eilenberg-MacLane spectra $f_\delta:HA\to \Sigma^k HB$. You can take the fiber of $f_\delta$ to obtain a spectrum $F$ which defines a generalized cohomology theory $E_F^*:X\mapsto \pi_{-*} F^X$, and there is a resulting long exact sequence $$ \dots\to E_F^i(X)\to H^i(X;A)\xrightarrow{\delta}H^{i+k}(X;B)\to E_F^{i+1}(X)\to\dots $$ If $\delta = \Sq^1$, the fiber is again an Eilenberg-MacLane spectrum (namely $F\simeq H\Z/4$). If the degree of $\delta$ is bigger than $1$, it will have two non-zero homotopy groups, namely $A$ in degree $0$ and $B$ in degree $k-1$.
The connection to the crossed modules you mention is that applying the functor $\Omega^{\infty-1}$ to the fiber of $\Sq^2$ gives rise to a $2$-group, i.e. a homotopy type $X$ whose homotopy groups vanish outside degrees $1$ and $2$. As you mention, these are classified by the two groups $A = \pi_1(X),B = \pi_2(X)$, the action of the former on the latter, and a $k$-invariant in $H^3(A;B)$, which together can be packaged into the datum of a crossed module. However, these deloop once if and only if the action is trivial and the $k$-invariant vanishes. It is still possible to find reasonably easy algebraic models for spaces whose homotopy groups vanish outside degrees $k$ and $k+1$ (given by braided ($k=2$) and symmetric ($k\ge 3$) monoidal Picard (every object has a tensor inverse) groupoids), although I do not know a definition of $\Sq^2$ in this language.
Questions in the comments
- You already discuss the resulting cohomology operation in the case that the action of $A$ on $D$ is trivial. For the general case, one first construts a natural transformation from $H^1(-;A)$ to local systems: $H^1(-;A)$ are isomorphism classes of $A$-principal bundles, and this natural transformation sends a principal bundle $P$ to $P\times_A D$. A cohomology class in $H^k(A;D)$ then gives a natural transformation from $H^1(-;A)$ to $H^k$ of this local system, by the same construction as in the trivial case.
- Yes, for a $(2,3)$-type $X$ the $2$-group $\Omega X$ is split, i.e. the action of $\pi_1(\Omega X)\cong \pi_2(X)$ on $\pi_2(\Omega X)\cong \pi_3(X)$ is trivial (this can be shown by the Eckmann-Hilton argument) and the $k$-invariant vanishes.
- For degree $2$ cohomology operations, there is in fact a complete classification (compare (Co)homology of the Eilenberg-MacLane spaces K(G,n) and the cited references): operations $H^2(-;A)\to H^4(-;B)$ are given by quadratic functions $q:A\to B$ (i.e. such that $q(x+y) - q(x) - q(y)$ is bilinear and $q(kx) = k^2q(x)$), and the resulting cohomology operation is given by a suitable version of the Pontryagin square. For $k\ge 3$, operations are in bijection with linear maps from $A\otimes \Z/2$ to $B$ (observe that such a linear map is also quadratic by the "freshman's dream"), and the resulting cohomology operation is the composition
$$ H^*(-;A)\to H^*(-;A\otimes Z/2)\xrightarrow{\Sq^2} H^{*+2}(-;A\otimes\Z/2)\to H^{*+2}(-;B) $$
The relation to Picard groupoids is a consequence of the Homotopy hypothesis, and given a braided Picard groupoid $C$, you can associate to it its abelian group $\pi_0 C$ of isomorphism classes and the (abelian!) group $\pi_1 C$ of automorphisms of the unit $1$, together with the map $q: \pi_0 C\to \pi_1 C$ which sends $x$ to the composition
$$ 1\cong x\otimes x^{-1}\xrightarrow{\beta_{x,x^{-1}}} x^{-1}\otimes x\cong 1 $$
It's a fun exercise to show that $q$ is quadratic in the above sense. It is not straightforward to give an inverse to this construction, i.e. construct the braided Picard groupoid from the quadratic map; for a reference in the symmetric setting, see Cegarra, A. M.; Khmaladze, E., Homotopy classification of graded Picard categories, Adv. Math. 213 (2007).
A chain level representative of $\Sq^2$ (and higher Steenrod squares) can be found in Ralph M. Kaufmann, Anibal M. Medina-Mardones. Cochain level May-Steenrod operations.
Best Answer
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 $\partial_{p+q}(\alpha\times\beta)=\partial_p\alpha\times\beta=(-1)^p\alpha\times\partial_q\beta$.