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By E. M. Friedlander, M. R. Stein

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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.

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