Let $X$ and $Y$ be two triangulable CW complexes which are homeomorphic.
Is it true that there exists a triangulation of $X$ and a triangulation $Y$ which have a common subdivision?
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5$\begingroup$ Look up the Hauptvermutung from geometric topology and Milnor's work on the topic. $\endgroup$– skdCommented Sep 1, 2019 at 12:40
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1$\begingroup$ For a concrete example of two homeomorphic simplicial complexes with no common subdivision, let $X$ be the double suspension of any triangulation of a homology 3-sphere that is not the 3-sphere itself, and let $Y$ be the boundary complex of a 6-dimensional simplex. $\endgroup$– Richard StanleyCommented Sep 1, 2019 at 23:12
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1$\begingroup$ The key to showing that two simplicial complexes cannot be refinements of each other to consider the link of a simplex. The link of a subdivision is a subdivision of a link, thus homeomorphic to the original. In @RichardStanley's example, the link of a 1-simplex is a sphere or a non-simply connected, so not homeomorphic. But showing that the two complexes are homeomorphic is the very difficult Cannon-Edwards theorem. $\endgroup$– Ben WielandCommented Sep 6, 2019 at 15:54
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$\begingroup$ Whereas Milnor's example is a 3d lens space times $D^4$ with the boundary coned off. So some links are 3d lens spaces. These are distinguished up to homeo by Reidemeister torsion, which is a lot more difficult than fundamental group, but found in some intro textbooks. For homeo, the two spaces are the 1-point compactification of the lens space cross $\mathbb R^4$, so it suffices to show the homeo of the open manifolds. First, $L_1\hookrightarrow L_2\times\mathbb R^4$ by Whitney and its tube nbd is $L_1\times\mathbb R^4$. Then apply Mazur's argument $\endgroup$– Ben WielandCommented Sep 6, 2019 at 15:55
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