By Nigel Ray, Grant Walker

J. Frank Adams had a profound effect on algebraic topology, and his paintings maintains to form its improvement. The overseas Symposium on Algebraic Topology held in Manchester in the course of July 1990 was once devoted to his reminiscence, and almost all the world's prime specialists took half. This quantity paintings constitutes the lawsuits of the symposium; the articles contained the following diversity from overviews to experiences of labor nonetheless in growth, in addition to a survey and entire bibliography of Adam's personal paintings. those complaints shape a huge compendium of present learn in algebraic topology, and person who demonstrates the intensity of Adams' many contributions to the topic. This moment quantity is orientated in the direction of homotopy idea, the Steenrod algebra and the Adams spectral series. within the first quantity the subject is especially volatile homotopy thought, homological and express.

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**Sample text**

Thus, as jvj is always even, we see that if K(n)°"(X) = 0 then (E(n)/II)odd (X) = 0 and the homomorphisms (E(n)/In+1)*(X) --- (E(n)/In)*(X) are epimorphisms for all k. As E(n)*(X) for a finite complex X is also given by lim_k(E(n)/In)*(X), we have: Lemma 13. If K(n)odd(X) = 0 then the map E(n)*(X) K(n)*(X) is epimorphic. This completes the proof of Theorem 11. Proof of Theorem 12. It suffices to prove the result for X connected. Up to a multiple by a power of v,,, an element x E K(n)2r(X) can be considered as a map X -) K(n)'2r.

However, the corresponding statements are false for p = 2, even when KO(2) is used in place of K(2). 1 above. 13]. The above theorem opens the way to an Adams spectral sequence with only three nontrivial rows. 3 Theorem. ZT(£ *RTX K*RTY) where E2't(X,Y) = 0 for s > 3 and E3't(X,Y) = E1 (X, Y). " In particular, it reduces to the KO*-Adams spectral sequence when X = S° or, more generally, when KO*X is a free i*KO-module, and reduces to the K*-Adams spectral sequence when the generator rt E ir1KO - Z/2 annihilates KO*X or KO*Y.

If we let Qq. and Qp denote the induced maps on indecomposables, then (3) and (4) show that ker Qq* C ker Qp since p acts by tensoring with the coefficients and hence certainly kills any element [vI] o V or [vI] o o el with some iq, q > n, non-zero. The commutative diagram above gives the opposite inclusion, and so these kernals are equal. Then ker q* = ker p since H*(BP(n)',,) is a free algebra. Hence TBP(n) is isomorphic for all r sufficiently small. The result for TE(n) now follows by passing to direct limits.