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I know that every smooth manifold $M$ admits a triangulation. That means, there exists some simplicial complex $K$ homeomorphic to $M$.

Does this mean that it also admits a smooth $\Delta$-complex structure? In Hatcher, this is defined as

a collection of smooth maps $\sigma_\alpha :\Delta_\alpha^n\to M$ satisfying

  1. The restriction of each map to each face is injective, and each point of X is in the image of exactly one such restriction.
  2. Each restriction to a face is one of the maps $\sigma_\beta : \Delta^{nāˆ’1}\to M$.
  3. A set $U\subset M$ is open iff $\sigma^{-1}(A)$ is open in $\Delta^n$ for each $\sigma$.
David Roberts
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    A triangulation is a special case of a $\Delta$-complex. ā€“  Dec 13 '15 at 15:31
  • Yeah, but the formulations are different. Let me clarify my question. ā€“ David Roberts Dec 13 '15 at 15:32
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    First, note that the standard result is that a smooth manifold admits a smooth triangulation, not just a triangulation. That is to say, in the homeomorphism you write down, it's smooth on each simplex. But then my first comment applies again: the collection of maps from simplices you get from a triangulation gives you a $\Delta$-complex structure. The latter is a special case. ā€“  Dec 13 '15 at 15:41

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