... Consequently £ is a congruence on Y. Following Notation 10.3, we easily verify that $a{* is a regular subsemigroup of C for every a £ Y. On each <£„{*, we define a relation 0„j* by a$a(*b if (7, a) p (y, b) for some 7 £ a. Let x = a ...
... Consequently, the corollary to Proposition 8 implies that the Conner-Floyd coindex of D(f) → A(f) must be P; 9. Finally Theorem 4 and the corollary to Theorem 2 of [13] imply that we may assume that D(f) is 5 connected and Ho (D(f)) ...
... Consequently, H is a direct sum of cyclic groups. See [2] for results relating p"G and G. To complete the proof, it ... Consequently, G|p]/S satisfies the Kulikov criterion in G/S. Using the purity of H, it can be shown that T(G[p]/ ...
... Consequently, there is an injective map: K/M x A* – X = G/K given by (kM,a) = (k, a) — ka. o. In addition, X'—the image of this map—is open and dense in X. PROOF FOR SL(n, R). Since M'/M = W is transitive on the Weyl chambers, it ...
... Consequently G,, contains the cyclic group generated by ( — 1)"-1S^. If K has characteristic 2 or if n is odd, we have ( — I)"-1 = 1 and hence Gv = K*. Thus ( — l)"-1x € Gv so that a solution of (4) with \X\ = x, Y\ = y exists in GL(n ...
... Consequently 5 satisfies J". Since 5 is in 0^^ ' , it must statisfy identity (4). Hence 5 € ^2, and thus A3 C ^42. Next let 5 £ /l2. Since 5 satisfies identity (4), we must have 5 £ 0&'& ' . Further, 5 satisfies ,/', which evidently ...
... Consequently ox" = — 1. But the fixed centrode of 25, namely + Z = + £x'. Therefore dź - , , , -, . is = ** + £ x = 0, which implies that the fixed centrode of 2 is a pair of diametrically opposite points and consequently, that 25 is a ...
... consequently is trivial. Since { is trivial, it may be extended to a subbundle o' of v(9*** : S"). The framing (ez", . . . , e.", Gi", . . . , Ga-n") may be extended to a framing of all t'. Then that framing and the exponential map may ...
... Consequently J3 is a closed set as well. Since M = m(p, q) leads to independent, identically distributed random variables, each such 'pure' measure belongs to %.. IffM(i+k,j + l) = fM(i,j)-fM(k,l) for all choices of the indices, the M ...
... Consequently, G(y)f(y) is strongly differentiable in the N-norm with derivative G' (y)f(y) + G(y)f'(y). It is easy to see that each term above is Bochner integrable. Consequently, using [5, Theorem 3.86], and choosing (c,d) to contain ...