The Jordan-Brouwer separation theorem tells us that the $(n-1)$-sphere separates $\mathbb{R}^n$ into two connected components. I want to know when a simplicial complex embedded in $\mathbb{R}^n$ separates $\mathbb{R}^n$ into two connected components.
For a 2-complex embedded in $\mathbb{R}^3$ my intuition tells me that a manifold of any genus should separate the space, but so can other types of complexes that are not manifolds such as the pinched torus. What is the most general type of complex that separates $\mathbb{R}^n$?
The answer to my question comes from the Alexander duality theorem, which related the $q$th reduced homology group of a subcomplex $K$ of $S^n$ with the $(n - q - 1)$th reduced cohomology group of $K$ by the isomorphism $\tilde{H}_q(S^n \setminus K) \cong \tilde{H}^{n - q - 1}(K)$. Assume we are working with (co)homology over a finite field, so the homology and cohomology groups coincide. Setting $n=2$ and $q=0$ we get $\tilde{H}_0(S^2 \setminus K) \cong \tilde{H}_2(K)$. The dimension of $H_0$ counts the number of connected components, so when $\dim H_2(K) = k$ we see that $K$ separates $\mathbb{R^3}$ into $k + 1$ components.
To answer my question, I will say that a 2-complex separates $\mathbb{R}^3$ when its second homology group is nontrivial.