1965/04/01 by D. B. A. Epstein · 3 citations
Mathematics · #Homotopy and Cohomology in Algebraic Topology #Advanced Topics in Algebra #Algebraic Geometry and Number Theory
paper · pdf · doi:10.1090/s0002-9939-1965-0208606-9
Let 717 be a compact differential manifold without boundary of dimension m.Theorem.Let M-x be embedded differentially in a differential manifold TV of dimension n, in such a way that the normal bundle of the embedding is fibre homotopically trivial.Then there is a map of degree one onto the smash product TV->Snm A 717. (The degree should be taken mod 2 if M or N is not orientable, and with respect to the compact cohomology of TV-Bd TV if TV is not compact or has boundary.)Proof.Let Dm be a closed disk in M such that xEDmCM.The normal (n -m)-dimensional disk bundle of 717-Int 7>m in 717 has as its total space a compact -dimensional manifold L with boundary.Let K = L/Bd L. L can be regarded as a submanifold of TV by the tubular neighbourhood theorem.The map N->K, which sends TV-L to a point and which sends L-K via the identification map, has degree one.We shall show that K is homotopy equivalent to Sn~m A 717.Let Li be the total space of the trivial (n -m) -dimensional disk bundle over M-lntDm.Let Ki = Li/BdLi.The fibre homotopy equivalence, referred to in the hypotheses, gives rise to a homotopy equivalence between K and Kx.Now Ki = M X D"-m/Dm X Dn~m KJ MX S"-m~l = 717 X Sn-m/Dm X Sn~m U MX* = MX S"-m/ * X Sn~m VJ MX* = 717 A Sn~m.This proves the theorem.Corollary 1.If 717 = 7(2, q), the three-dimensional lens space, then M -x cannot be differentially embedded in 54.Proof.If M-x is embedded in S4, then, by the theorem, there is a map of degree one S'-S1 AM.But Puppe [l, p. 416], shows that this is not true.(Puppe shows that