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On Subdivisions of Complexes

Published online by Cambridge University Press:  24 October 2008

J. H. C. Whitehead
Affiliation:
Balliol College, Oxford

Extract

1. This paper is supplementary to an article by M. H. A. Newman on the superposition of manifolds. Guided by his main idea, I extend Newman's theorem that two equivalent manifolds have a common subdivision from equivalent manifolds to equivalent complexes of an arbitrary nature. The theorem is also sharpened by the possibility of leaving a certain subcomplex unaltered, and by substituting a partition for a general subdivision.

Type
Research Article
Copyright
Copyright © Cambridge Philosophical Society 1935

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References

Journal London Math. Soc. 2 (1926), 5664.Google Scholar

Proc. Royal Acad. Amsterdam, 29 (1926), 627Google Scholar, theorem 10.

§ Annals of Math. 31 (1930), 292320Google Scholar. Newman uses the term in a different sense.

Here a vertex means a point of Π which is not interior to any linear segment in Π. An n dimensional, convex polyhedral domain has at least n + 1 vertices, and two such domains coincide if they have the same vertices. For, assuming these properties for the (convex) boundary faces, they follow from induction on n.

That is to say the sequence does not contain a transformation of the form (A, a) if A belongs to L, or (A, a)−1 if a does.

This means that πA = A if A is any component of L, not merely that πL = L.

An n set is similar to a normal n set (Newman, Proc. Royal Acad. Amsterdam, loc. cit., p. 9) except that a k cell (k < n) need not be on the boundary of an n cell. In proving our theorem 3 we follow the main idea in the proof of Newman's theorem 10.

Alexander (loc. cit.), theorem 13·2. If we allow K 1K 2 to be “trivial on L”, we can adopt Newman's definition of starring and his simpler proof that an element can be starred (Journal London Math. Soc. 6 (1931), 186–93Google Scholar, theorem 10).

Journal London Math. Soc. 2 (1926), 5664, § 3.Google Scholar

Proc. Royal Acad. Amsterdam (loc. cit.), theorem 8a. See also Alexander (loc. cit.), theorem 14·3.

That is to say πK 0 has K 1 itself, not a subdivision of K 1, as a subcomplex.