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Let $A$ be a closed subspace $A$ of $[0,1]^4$---let's say, a subcomplex of some triangulation of the cube. I would like to show that the cup product $H^2(A)\times H^2(A)\to H^4(A)$ is trivial (or at least that the square map $x\mapsto x\smile x$ is trivial).

I tried analyzing simple examples of nontrivial cup product occurances and it seems to me that the basic "building blocks" are examples such as products of spaces (for example, $S^2\times S^2$) and/or attaching $4$-cells to something $2$-dimensional in a nontrivial way (for example, $\Bbb CP^2$) and none of these can be realized in a Euclidean $4$-spaces; however, I don't know how to prove the claim formally.

Thanks for possible hint.

Najib Idrissi
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Peter Franek
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  • I don't really have an idea, but just to show that this might not be very easy, think about the Milnor-Barrat higher dimensional analogues of the hawaiian Earring. These have non-trivial homology in arbitrary high dimension. I do not know the cohomology structure, but it might be non-trivial. Then you can take a product with an $S^1$ to obtain possibily something. These probably go in a different direction than you want though, as these spaces are not locally nice. – Thomas Rot Jan 05 '16 at 11:22
  • @ThomasRot Thank you, interesting. But at least in case of simplicial complexes, I hope the claim may be true. I'm even interested in a few generalziations (just wanted to keep it simple) and never know whether this is the right place to ask, or mathoverflow... – Peter Franek Jan 05 '16 at 12:39

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You may embed your complex $A$ into $S^4$. Then Alexander duality (see Hatcher, th. 3.44) gives you

$$ \tilde H^k(A)=\tilde H_{3-k}(S^4\setminus A), $$

so $H^4(A)$ are always $0$.

Andrey Ryabichev
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  • Doesn't Alexander duality require some additional hypothesis on $A$? The Wikipedia article requires local contractibility. – Qiaochu Yuan Jan 06 '16 at 16:35
  • @QiaochuYuan, if $A$ is a subcomplex of some triangulation of the sphere, all these requirements will be satisfied. – Andrey Ryabichev Jan 06 '16 at 16:38
  • Excelen answer. If it's easy, could you give me also a hint on the relative case $H^2([0,1]^4,A)\times H^2([0,1]^4,A)\to H^4([0,1]^4,A)$? Here the top cohomology may clearly be nontrivial but I'm convinced that the cup product is still trivial... – Peter Franek Jan 06 '16 at 19:50
  • @PeterFranek, actually, i don't know. the only one idea that i have is to use long exact sequence of pair, but i'm not sure how it respect a multiplication.. – Andrey Ryabichev Jan 06 '16 at 23:51
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    I never thought about it before, but $\tilde{H}_{-1}(\varnothing) = \mathbb{Z}$... (Of course here $A \subset \mathbb{R}^4 \subsetneq S^4$ so it doesn't matter.) – Najib Idrissi Jan 07 '16 at 08:23