Bourbaki

Exercise 14, § 5

Algebra, IV, p. 97

With the notation as in Ex. 13, suppose that a∈S(E,F)a \in S(E, F) is of order ≥1\geq 1. a) Show that for each p∈Np \in \mathbf{N}, σ(p)\sigma^{(p)} is of order ≥p\geq p. Deduce that the family (σ(p))p∈N(\sigma^{(p)})_{p \in \mathbf{N}} is summable. Writing A(σ)=∑pσ(p)A(\sigma) = \sum_p \sigma^{(p)}, verify that for every n∈Nn \in \mathbf{N},

A(σ)(Γn(E))⊂Γ0(F)+⋯+Γn(F). A(\sigma)(\Gamma_n(E)) \subset \Gamma_0(F) + \cdots + \Gamma_n(F) .
b) Show that if Ker(a)⊃Γm(E)\mathrm{Ker}(a) \supset \Gamma_m(E) for all m≠1m \neq 1, then A(σ)=Γ(a∘γ1)A(\sigma) = \Gamma(a \circ \gamma_1). c) Show that A(σ)A(\sigma) is a homomorphism of the cogebra Γ(E)\Gamma(E) into the cogebra Γ(F)\Gamma(F), compatible with the counits (cf. Ex. 13, e)). d) Show that if a,τ∈S(E,F)a, \tau \in S(E, F) are two series of order ≥1\geq 1, then A(σ+τ)=A(σ)A(τ)A(\sigma + \tau) = A(\sigma) A(\tau), the product A(σ)A(τ)A(\sigma) A(\tau) being defined by the product in Γ(E)\Gamma(E) (cf. Ex. 13, c)).

Solution

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