Manuscript XV-E@1 · XV-E.tex · lemma · a3dc45f5

Lemma: β is surjective

Descending extensions in the punctured normal-jet module › Annihilator ideals · lines 615–626 · proof included or linked

XV-E@1#lem:surj

The map $σ : C^{\infty} (X)^{M_{r}} \to P_{r}^{jet}$ , $σ (g)_{α} (x, q) = g_{α} (x)$ , satisfies $β \circ σ = id$ . In particular $β$ is surjective.

Proof. If $a_{α}$ is independent of $q$ then $\partial_{q}^{α - γ} a_{α} = 0$ unless $α = γ$ , so (eq:beta) gives $β_{γ} (σ (g)) = g_{γ}$ .

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