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I need to reference the following result. Do you know a good source?

The following conditions on an $n$-dimensional simplicial complex $S$ are equivalent: a) $S$ is an $n$ manifold; b) The link of every vertex of $S$ is homeomorphic to the $(n - 1)$-sphere; c) The link of every $k$-simplex is homeomorphic to the $(n - k - 1)$-sphere.

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I think this title is a little misleading as the term "simplicial manifold" is usually reserved for a simplicial object in the category of manifolds. – David Carchedi May 10 '10 at 11:49
Changed the title. – Kestutis Cesnavicius May 10 '10 at 15:01
To get a simple correct statement in the category implied by the tags, (a) should refer to S as a combinatorial n-manifold; and (b) and (c) should refer to the links in question as being PL homeomorphic to the appropriate sphere. – Allan Edmonds May 10 '10 at 17:49
up vote 5 down vote accepted

The usual term for objects like this is "combinatorial manifolds".

However, the result is not quite true as you have stated. Definitely b is true if and only if c is true, and b implies a. However, a does not imply b. There definitely exist simplicial complexes which do not satisfy b or c but which are topological manifolds. For example, the famous double suspension theorem of Cannon (weaker versions were proved by Edwards) says that if $X$ is a homology $n$-sphere, then the space $Y$ obtained by suspending $X$ twice is homeomorphic to the $(n+2)$-sphere. If neither $X$ nor the suspension of $X$ is an actual sphere (for instance, this holds if $X$ is the Poincare homology sphere), the vertices of $Y$ corresponding to the suspension points will then not satisfy b.

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Thanks! I saw an analogous result for homology manifolds and generalized homology spheres, and having heard that b) implies a) guessed that the same should hold. – Kestutis Cesnavicius May 10 '10 at 15:00
One needs to be a little more careful in this mixed category of triangulated topological manifolds. Consideration of the triple suspension of a triangulated homology 3-sphere shows that (b) does not imply (c). – Allan Edmonds May 10 '10 at 17:46
Whoops! Thanks Allan! – Andy Putman May 10 '10 at 20:28

I'm pretty sure you can find this kind of results in Rourke and Sanderson's "Introduction to Piecewise linear topology".

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It has parts of the claim as exercises on p. 22 and p. 24. – Kestutis Cesnavicius May 10 '10 at 10:03

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