Published online by Cambridge University Press: 24 October 2008
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.
† Journal London Math. Soc. 2 (1926), 56–64.Google Scholar
‡ Proc. Royal Acad. Amsterdam, 29 (1926), 627Google Scholar, theorem 10.
§ Annals of Math. 31 (1930), 292–320Google 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 1 → K 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), 56–64, § 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.