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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Transverse fibre products are embedded submanifolds

Statement

Let F:MN and G:PN be smooth and transverse. Then the fibre product

M×NP:={(p,q)M×P:F(p)=G(q)}

is an embedded submanifold of M×P.

Facts & Assumptions

Given: Smooth maps F:MN and G:PN with FG.

[F1]

Two smooth maps are transverse when their differential images span the target tangent space at every coincidence point (Transverse smooth maps).

[L1]

The diagonal ΔNN×N is an embedded submanifold (The diagonal is an embedded submanifold).

[L2]

Products of smooth maps are smooth, and the transverse preimage theorem applies to a map transverse to an embedded submanifold (Restrictions, corestrictions, and products of smooth maps are smooth, The transverse preimage theorem).

Proof

technique · direct
1.1

Define H:M×PN×N by H(p,q)=(F(p),G(q)). By [L2], H is smooth, and H1(ΔN)={(p,q):F(p)=G(q)}=M×NP.

L1L2given
2.1

At a point (p,q) with F(p)=G(q)=y, the tangent space to ΔN is {(u,u):uTyN}. Therefore transversality of H to ΔN means that every pair (a,b)TyN×TyN can be written as (a,b)=(dFpu,dGqv)+(w,w). This is equivalent to abdFp(TpM)+dGq(TqP), which is exactly [F1].

F1step 1.1algebra
3.1

Hence HΔN, so [L2] applied with [L1] shows that M×NP is an embedded submanifold of M×P.

L1L2step 2.1

Depends on

Used by

Dependency tree · two levels

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