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In many situations in topology, (like the poincare duality) they put a distinction between the space being a manifold or just a cell or simplicial complex. I want to know why this is important, in other words are there manifolds without the latter structure?

7779052
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A homeomorphism from a manifold to a simplicial complex is called a triangularization. The $E_8$ manifold (a four-dimensional manifold) does not have a triangularization. I don't know the history of this result but I suspect it's from the late 80s as the proof uses Casson's invariant.

More recently -- just over a year ago! -- Ciprian Manolescu proved that there are manifolds of any dimension greater than 4 which do not admit an triangularization. He uses Pin(2)-equivariant Seiberg-Witten Floer homology. Here is the paper.

Adam Saltz
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    Smooth manifolds are triangularizable, though. The question does not make it too clear whether it is using manifold to mean smooth or topological manifolds. – Mariano Suárez-Álvarez May 02 '14 at 03:24
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    The proof does not use Casson's invariant, the first proof of non-smoothability used Donaldson's invariants (1983); one can now prove it easier using Seiberg-Witten invariants. Of course, existence of E8 manifolds requires Freedman's work (1982). – Moishe Kohan May 02 '14 at 16:28
  • @studiosus Thank you for the correction. – Adam Saltz May 04 '14 at 14:14
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    Is it not possible to make a topological manifold smooth? so that a triangulization for a smooth manifold gives one for a topological manifold? – 7779052 May 04 '14 at 16:21
  • @7779052: Not every topological manifold is smoothable. See http://math.stackexchange.com/questions/408221 for a detailed discussion. – Moishe Kohan May 04 '14 at 23:37