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If a 3-dim'l cell complex $X$ is embedded in $\mathbb{R}^3$, no subcomplex of $X$ is a real projective plane. Intuitively, this suggests that all cup-squares vanish in $H^*(X;\mathbb{Z}_2)$. Is this statement true? If so, can you provide a reference?

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It is true. Let's first assume that $X$ is a finite complex. Then by Alexander duality, $H^2(X;\mathbb{Z})=H_0(S^3\setminus X,\mathbb{Z})$ must be torsion-free. Now the squaring map $H^1(X;\mathbb{Z}/2)\to H^2(X;\mathbb{Z}/2)$ coincides with the Bockstein map. This means that if the squaring map is nonzero, then $H_1(X;\mathbb{Z})$ must have $\mathbb{Z}/2$ as a direct summand. But then $H^2(X;\mathbb{Z})$ would have $\mathbb{Z}/2$ as a summand, and in particular cannot be torsion-free.

Now let $X$ be arbitrary, and write $X=\bigcup X_n$ as a union of finite complexes; from now on we only use $\mathbb{Z}/2$ coefficients. Then $H^1(X_n)$ is finite for all $n$, so $\lim^1 H^1(X_n)=0$. It follows that the map $H^2(X)\to\lim H^2(X_n)$ is an isomorphism. Now for any $x\in H^1(X)$, the image of $x^2$ in $H^2(X_n)$ must be 0 for all $n$ by the previous paragraph. Thus $x^2=0$.

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