The law of operation H×Sr(X;E)→Sr(X;E) is continuous.
In view of the definition of the topology of Sr(X;E) and General Topology, Chap. X, §3, no. 4, Th. 3, it suffices to prove that for any integers k≤r, the map f:H×X×Sk(X;E)→Jk(X;E) such that f(h,x,s)=jxk(hs) is continuous. For h∈H, denote by τh (resp. θh) the automorphism x→hx of X (resp. of E). Define maps
f1:H×X→Jk(X,X)
f2:H×E→Jk(E,E)
g:H×X×Sk(X;E)→Jk(X,E)
by f1(h,x)=jxk(τh),f2(h,v)=jvk(θh),g(h,x,s)=jhxk(s). We have
f(h,x,s)=f2(h,s(h−1x))∘g(h−1,x,s)∘f1(h−1,x)
and consequently it suffices, by Differentiable and Analytic Manifolds, Results, 12.3.6, to prove that f1,f2 and g are continuous.
Now g is the composite map
H×X×Sk(X;E)−−−−−−−⟶(m,Id)X×Sk(X;E)
−−−−−−−⟶(Id,jk)X×C(X;Jk(X,E))⟶εJk(X,E)
with ε(x,u)=u(x); the map ε being continuous (General Topology, Chap. X, §3, no. 4, Cor. 1 of Th. 3),g is continuous.
Let (h0,x0)∈H×X; we shall prove that f1 is continuous at (h0,x0). There exist charts (U,ψ,F) and (V,χ,F′) of X and an open subset Ω of H such that x0∈U,h0∈Ω and m(Ω×U)⊂V. By using the expression for Jk(X,X) in these charts, we are reduced to proving, for 1≤l≤k, the continuity at (h0,x0) of the map (h,x)→Δxl(τh) from Ω×U to Pl(F;F′), with Δxl(τh)(v)= l!1Dlτh(x).v for v∈F(Differentiable and Analytic Manifolds, Results, 12.2). But Dlτh(x) is simply the lth partial derivative of m(h,x) with respect to x, which is continuous by hypothesis; consequently, f1 is continuous. The proof that f2 is continuous is similar, hence the lemma.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.