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Extra info for A solution to a problem of Fermat, on two numbers of which the sum is a square and the sum of their squares is a biquadrate, inspired by the Illustrious La Grange

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Given a decomposition G for which HG is a countable collection of locally shrinkable sets, one might expect to be able to shrink them one at a time, in the limit effecting a total shrinking. Bing showed how to make this work in case NG forms a Ga-subset. Justification of the need for some hypotheses on HG will be given after the proof has been completed. 43 7. Countable Decompositions and Shrinkability Theorem 4. Suppose G is a use decomposition of a complete metric space (S, p) such that HG is a countable collection of locally shrinkable sets and NG is a Gs-subset of S.

39 6 . Cellular Sets f(B) FIG. 6-2 Now consider the slightly more complicated situation where C1 and C2 denote disjoint bicollared (n - 1)-spheres in S". Let (El, &) denote that component of S" - (CI u &) having both these spheres in its frontier, [El, &) = C1 u ( X I , C2)and [El, &] = C1 u & u (El, Cz). The conjecture that [ X I , CZ]is topologically S"-' x [- 1 , 11 is a famous statement known as the annulus conjecture. It is now known in all dimensions, having been proved primarily by Kirby [ 11 for n 1 5 and more recently by Quinn [2] for n = 4.

If w is a map of S" onto itself with exactly two inverse sets A and B, then each of A and B is cellular. Proof. We will show that B is cellular. Let Q be an n-cell in S" containing A u B in its interior. Define a map f of Q into S" as the identity on w-'(V) and as w-'Oty on Q - A . Then f(1nt Q) is open and B is the only inverse set off. By Proposition 3, B is cellular. 38 11. The Shrinkability Criterion Proposition 5 . I f f is a map of S" onto itself with only afinite number of inverse sets, then each is cellular.

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