Consider a triangulable space $X$ in the $n$-sphere $S^n$ whose complement $X^c=S^n\backslash X$ is also triangulable: there exists finite simplicial complexes $T_X$ and $T_{X^c}$ whose geometric realizations $|T_X|$ and $|T_{X^c}|$ (in $S^n$) are respectively homeomorphic to $X$ and $X^c$.
I'm interested in the relation between $|T_X|^c$ and $|T_{X^c}|$: do they have the same homology? (or even stronger, are they homotopy equivalent or homeomorphic?)
If $X$ is compact, I can show (using Alexander duality on $X$ and $|T_X|$), that $|T_X|^c$ and $|T_{X^c}|$ have the same (co)homology. However, I'm precisely interested in the case where $X$ is neither closed nor open.
Any idea?