Nontrivial cup product realized in $\Bbb R^4$

Question

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.

Answer 1

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 $0$.