Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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The composite of holomorphic maps is holomorphic and its complex Jacobian is the product

Statement

Let m,n,p≥1, let U⊆Cm and V⊆Cn be open, let F:U→Cn have F(U)⊆V and be holomorphic at a∈U, and let G:V→Cp be holomorphic at F(a). Then G∘F is holomorphic at a with

D(G∘F)(a)=DG(F(a))∘DF(a),JC(G∘F)(a)=JCG(F(a)) JCF(a).

Facts & Assumptions

Given: Open sets U⊆Cm and V⊆Cn, a map F:U→V holomorphic at a, and G:V→Cp holomorphic at F(a); the spaces are read through Complex m-space and its real coordinate dictionary.

[L1]

F is holomorphic at a when there is a C-linear L with F(a+h)=F(a)+L(h)+r(h) and ∥r(h)∥/∥h∥→0; L is unique, written DF(a), and JCF(a) is its matrix in the standard bases (Holomorphic maps Cm→Cn and the complex Jacobian matrix, Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases, Finite rectangular matrices over a commutative ring, their entries, rows and columns).

[L2]

A map into Cn is holomorphic exactly when each component is, with DF(a)h the tuple of the component differentials (A map into Cn is holomorphic exactly when each of its components is).

[L3]

A holomorphic function of several variables is continuous (A holomorphic function of several variables is continuous and separately holomorphic).

[L4]

[S∘T]BD=[S]CD[T]BC for linear T:U′→V′ and S:V′→W′ with ordered bases B,C,D ([S∘T]BD=[S]CD[T]BC).

[L5]

For every linear L:Rm′→Rn′ there is K≥0 with ∥Lh∥2≤K∥h∥2 (Every Euclidean linear map has a unique matrix and satisfies ∥Lh∥2≤K∥h∥2 for some K≥0), the notion of linear map being that of A linear map L:Rm→Rn in Euclidean coordinates.

[L6]

If f is totally differentiable at a and g at f(a), then g∘f is totally differentiable at a with D(g∘f)(a)=Dg(f(a))∘Df(a) (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Proof

technique · direct
1.1givenL1L5

Write F(a+h)=F(a)+DF(a)h+rF(h) and, for κ∈Cn small, G(F(a)+κ)=G(F(a))+DG(F(a))κ+rG(κ) as in [L1], with ∥rF(h)∥=o(∥h∥), ∥rG(κ)∥=o(∥κ∥) and rG(0)=0. By [L5], read through the dictionary, there are K,K′≥0 with ∥DF(a)h∥≤K∥h∥ and ∥DG(F(a))κ∥≤K′∥κ∥.

2.1step 1.1L8L9

Put κ(h)=DF(a)h+rF(h), so F(a+h)=F(a)+κ(h); by step 1.1 and [L8] there is δ>0 with ∥κ(h)∥≤(K+1)∥h∥ whenever ∥h∥<δ and a+h∈U, and F(a+h)∈V because F(U)⊆V.

3.1step 1.1step 2.1L8

Substituting, G(F(a+h))=G(F(a))+DG(F(a))DF(a)h+ϱ(h) with ϱ(h)=DG(F(a))rF(h)+rG(κ(h)). By step 1.1 the first summand has norm at most K′∥rF(h)∥=o(∥h∥); by step 2.1 the second has norm o(∥κ(h)∥) with ∥κ(h)∥≤(K+1)∥h∥, hence o(∥h∥), the value at h with κ(h)=0 being 0. So ∥ϱ(h)∥=o(∥h∥) by [L8].

4.1step 3.1L1L6

The composite DG(F(a))∘DF(a) is C-linear, being a composite of C-linear maps, so step 3.1 and [L1] make G∘F holomorphic at a with D(G∘F)(a)=DG(F(a))∘DF(a); this agrees with the real chain rule of [L6] read through the dictionary, by the uniqueness in [L1].

5.1step 4.1L1L2L3L4L7∎

Taking matrices in the standard bases, [L4] turns step 4.1 into JC(G∘F)(a)=JCG(F(a))JCF(a), the entries being read off at the basis vectors by [L2], [L3] and [L7].

Depends on

Used by

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Sources