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14 Sep 2023: If χ: A B is an algebra homomorphismthen there is b B such that χ(x) = bxb1 for all x A. ... Evidently such a pair is also an element of AutA(M/X/G). (b) Given a group homomorphism σ : G AutA(M/X/G) such that pr1 σ =1G, the restriction

29 Nov 2023: 10. )〉. So. R2 =〈(. 10. )〉〈(. 01. )〉is the only such decomposition in this case. ... Ws such that W = sj=1Wj. Thus. V =ri=1. Ui sj=1.

31 Jan 2024: G = H1 BH2 BH3 BH4 B BHn = {e}. such that each quotient Hi1/Hi is simple. ... Since they have the same cycle type, there is a σ Sn such that (a b c) = σ(1 2 3)σ1.

25 Oct 2023: Then there is a point y / E such. that B(y,r) E 6= whenever r > 0. ... Since Uj1 isopen we can find a δj1 > 0 with δj1 δj/4 such that B(xj1,δj1) Uj1.

25 Oct 2023: Theorem 4.2. If f : [a,b] [a,b] is a continuous function, then we canfind a w [a,b] such that f(w) = w. ... Supposethat we can find a a0 < a1 < < an b such that, writing σ = ‖fP‖we have either.

20 Dec 2023: Proposition 2.1.1. ([25, Exercise 2F]) For every ring homomorphism φ: A Band element b B such that bφ(a)φ(a)b = φ(δ(a)) for all a A, ... there is a uniquering homomorphism ψ : A[T; δ] B such that ψ|A = φ and ψ(T) = b.

31 Jan 2024: Theorem. There is no group homomorphism F : {0} Z such that the composition. ... such that. f g ' IdY f ′ g′ ' IdZ g f ' IdX g′ f ′ ' IdY.

24 Oct 2023: Vn such that V ' V1 Vn. Theorem 4.12. Suppose that G is finite and |G| 6= 0 in F. ... central character, such that z v = λ(z) v forall z Z(G) and v V.

29 Jan 2024: 4Perhaps the Revelations are not flagged as such? Perhaps checking that something isa revelation involves establishing its independence from earlier revelation? ... Ramsey is such a theorem; Van der Waerdenisn’t! vdW does not prove itself in that way;

24 Nov 2023: Let a,b K such that the polynomials f = Xm a, g = Xm b are irreducible. ... Use asuitable Lagrange resolvent to find a Q(ζ3) such that K = Q(ζ3, 3.

15 Apr 2024: b) Consider a polygonal decomposition of Σ such that at least three edges meet at each vertex andeach face has at least three edges. ... Deduce. that there exists δ > 0 such that in any hyperbolic triangle the union of the δ-neighbourhoods oftwo of

29 May 2024: Then there is abijection h: A B. Proof. We seek partitions A = P Q of A and B = R S of B such thatf(P) = R and g(S) = Q. ... then h is a bijection. To see that such partitions exists, let X = P(A) and define H : X X byH(P) = A g(B f(P)).

27 Nov 2023: X̃1, that is, β̃1 = (X̃T1 X̃1). 1X̃T1 Ỹ. (c) Find a condition on X0 and X1 such that β̃0 = β̂0. ... b) The model is to be reparametrised in such a way that it can be specialised to (i) two parallellines for the two groups, (ii) two lines with

25 Nov 2023: Let θ : A B be a -homomorphism from a Banach -algebra A to a C-algebraB. ... H) such thatz(T) = T, where z(λ) = λ for all λ K.

5 Oct 2023: Theleast such n is called the order of g.). (b) Show from first principles that there is a positive integer N such that gN = e for all g G. ... 9. Let H and G be groups and let X G such that 〈X〉 = G.

5 Jun 2024: I)for some small ϵ > 0, but the choice of such is contractible. ... 2.3.1(ii)]. We will write W dp p for such a manifold.

5 Jun 2024: Such a functor F : J Top is said to be continuous if theevaluation map. ... For any closed subset 0 B, let ˜BunMap0 (ξ,π)• denote the semi-simplicialset whose n-simplices consist of bundle maps G : n ξ τn π := (τn B′) (n π) such