TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)

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The transverse preimage theorem

Statement

Let $F : M^{m} \to N^{n}$ be smooth and let $Z \subseteq N$ be an embedded submanifold of codimension $c$ , and suppose $F ⋔ Z$ . If $c \leq m$ , then $F^{- 1} (Z)$ is an embedded submanifold of $M$ of codimension $c$ (possibly empty). If $c > m$ , then $F^{- 1} (Z) = \emptyset$ . In the first case, for each $p \in F^{- 1} (Z)$ ,

$T_{p} F^{- 1} (Z) = {v \in T_{p} M : d F_{p} (v) \in T_{F (p)} Z} .$

Facts & Assumptions

Given: A smooth map $F : M^{m} \to N^{n}$ transverse to an embedded submanifold $Z \subseteq N$ .

[L1]

Embedded submanifolds admit local defining submersions (Embedded submanifolds admit local defining submersions).

[L2]

Transversality to $Z$ is equivalent to surjectivity on the normal quotient (Transversality is equivalent to surjectivity on the normal quotient).

[L3]

A regular level set is an embedded submanifold, and its tangent space is the kernel of the defining submersion differential (A regular level set is an embedded submanifold, The tangent space of a regular level set is the kernel).

[F3]

An embedded $k$ -submanifold has $0 \leq k \leq m$ and is characterized by its slice-chart condition (Embedded submanifolds and slice charts).

Proof

technique · direct

1.1L1L2givencasescontradictionchoose

First suppose $c > m$ . If $p \in F^{- 1} (Z)$ , [L2] says the composite $T_{p} M \to T_{F (p)} N / T_{F (p)} Z$ is surjective. Its target has dimension $c > m = \dim T_{p} M$ , which is impossible, so the preimage is empty. Now suppose $c \leq m$ . If the preimage is empty, the embedded-submanifold conclusion in this dimension range is vacuous; otherwise fix $p \in F^{- 1} (Z)$ and put $y = F (p)$ . By [L1], choose a neighbourhood $V$ of $y$ and a submersion $h : V \to R^{c}$ such that $Z \cap V = h^{- 1} (0)$ .

2.1L2step 1.1algebra

At $p$ , the differential $d h_{y}$ kills exactly $T_{y} Z$ . Therefore [L2] implies that $d h_{y} \circ d F_{p} = d (h \circ F)_{p}$ is surjective. So $0$ is a regular value of $h \circ F$ near $p$ .

3.1L3step 2.1algebra

Since $(h \circ F)^{- 1} (0) = F^{- 1} (Z) \cap F^{- 1} (V)$ , [L3] shows that this set is an embedded codimension- $c$ submanifold near $p$ , with tangent space $\ker d (h \circ F)_{p} = {v : d F_{p} (v) \in T_{y} Z} .$

4.1F3step 1.1step 3.1cases∎

The local models in step 3.1 cover every point of $F^{- 1} (Z)$ , so they give its embedded-submanifold structure and the stated tangent-space formula. If $c \leq m$ and the preimage is empty, it is also an embedded submanifold of dimension $m - c$ by [F3], whose pointwise slice-chart condition is vacuous. Together with step 1.1 this proves both dimension cases.

The transverse preimage theorem

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources