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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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The smooth inverse function theorem on manifolds

Statement

Let F:MN be a smooth map and let pM. If dFp:TpMTF(p)N is an isomorphism, then there are open neighbourhoods U of p and V of F(p) such that FU:UV is a diffeomorphism.

Facts & Assumptions

Given: A smooth map F:MN and a point pM with dFp an isomorphism.

[F1]

The differential of a smooth map is the induced linear map on tangent spaces (The differential of a smooth map).

[L1]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

[L2]

A Euclidean C1 map with invertible derivative at a point has a local C1 inverse (The Euclidean inverse function theorem).

[L3]

If the original Euclidean map is smooth, then its local inverse is smooth (A local inverse of a Ck regular map is Ck).

[L4]

Chart maps and their inverses are smooth diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1

Choose smooth charts (U0,φ) at p and (V0,ψ) at F(p) as in [L4], and let f:=ψFφ1. By [L4], the chart maps φ1 and ψ are diffeomorphisms, so their differentials are linear isomorphisms. Applying [L1] gives Df(φ(p))=dψF(p)dFpd(φ1)φ(p). Because dFp is an isomorphism by [F1], the Euclidean differential Df(φ(p)) is invertible.

F1L1L4given
2.1

By [L2], after shrinking to open neighbourhoods U~φ(U0) and V~ψ(V0) of φ(p) and ψ(F(p)), the restriction fU~:U~V~ is a bijection with C1 inverse. Since f is smooth, [L3] upgrades that inverse to a smooth one.

step 1.1L2L3
3.1

Put U:=φ1(U~) and V:=ψ1(V~). Then FU=ψ1(fU~)φ, so FU:UV is bijective. Its inverse is φ1(fU~)1ψ, which is smooth by [L4] and step 2.1.

step 2.1L4construct
4.1

Thus FU is a bijective smooth map with smooth inverse, hence a diffeomorphism of neighbourhoods.

step 3.1L4

Depends on

Used by

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Sources