If a topological set is triangulable, dose the triangulation map gives it the (regular) CW complex structure? From definitions, I see it seems to be, but I am not that sure, for may exist some strange examples.
Really, thanks for any help
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2$\begingroup$ A triangle is homeomorphic to a disk, a $3$-dimensional simplex is homeomorphic to a $3$-dim ball etc. $\endgroup$– Liviu NicolaescuCommented Oct 7, 2016 at 13:37
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$\begingroup$ Yes, in my understanding triangulation is very strong. So do you mean triangulation will induce a CW complex structure? $\endgroup$– lun zhangCommented Oct 7, 2016 at 13:47
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$\begingroup$ @lunzhang What is your definition of triangulation? If it is "homeomorphism with a simplicial complex", then the answer is yes (every simplicial complex has an obvious CW structure). $\endgroup$– Denis NardinCommented Oct 7, 2016 at 14:29
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$\begingroup$ Yes, that is the definition plus some like locally finite condition. So triangulation and regular CW complex are equivalent? $\endgroup$– lun zhangCommented Oct 7, 2016 at 15:34
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1$\begingroup$ They are not equivalent. A triangulation is a special case of a CW complex. $\endgroup$– Lee MosherCommented Oct 7, 2016 at 16:23
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Yes, you are right. You can see it as a consequence of Proposition 5.33 in Lee's book:
Introduction to Topological Manifolds, 2nd ed. Grad. Texts in Math., Springer, 2011.
I quote the proposition:
If $\mathcal{K}$ is a Euclidean simplicial complex, then the collection consisting of the interiors of the simplices of $\mathcal{K}$ is a regular CW decomposition of $|\mathcal{K}|$ (the polyhedron associated to $\mathcal{K}$).
