Preïmage theorem

Let $𝑓 : 𝑋 \to 𝑌$ be a $𝐶 \infty$ differentiable map between $𝐶 \infty$ differentiable manifolds $𝑋, 𝑌$ of dimensions $𝑛, 𝑚$ respectively. Let $𝑦 \in 𝑌$ be a regular value of $𝑓$ . Then the fibre $𝑆 = 𝑓 - 1 {𝑦}$ is a $𝐶 \infty$ submanifold of $𝑋$ of dimension $𝑛 - 𝑚$ . #m/thm/geo/diff

Proof

Since $𝑦$ is a regular value of $𝑓$ , $𝑓$ is submersive at every $𝑥 \in 𝑆 = 𝑓 - 1 {𝑦}$ , so by the local submersion theorem we may define charts such that the following diagram commutes in $𝖬 𝖺 𝗇 \infty$

with $𝜑 : 𝑥 \mapsto 𝑣$ and $˜ 𝜑 : 𝑦 \mapsto ˜ 𝑣$ . Now

𝑉 \cap 𝑗 - 1 {˜ 𝑣} = 𝑉 \cap ({˜ 𝑣} \times ℝ 𝑛 - 𝑚)

Therewithal

𝑓 - 1 {𝑦} = 𝜑 - 1 𝑗 - 1 ˜ 𝜑 {𝑦} = 𝜑 - 1 𝑗 - 1 {˜ 𝑣}

so $𝜑 (𝑈 \cap 𝑆) = 𝑉 \cap ({˜ 𝑣} \times ℝ 𝑛 - 𝑚)$ which is diffeomorphic to an open subset of $ℝ 𝑛 - 𝑚$ . Thus $𝑓 - 1 {𝑦}$ is an $(𝑛 - 𝑚)$ -dimensional $𝐶 \infty$ differentiable manifold.

Direct proof

Note that this is essentially the same as the above proof, just with the content of the proof of the local submersion theorem absorbed.

Let $𝑥 \in 𝑓 - 1 {𝑦}$ . Since $𝑦$ is a regular value, the tangent map $𝑇 𝑥 𝑓 : 𝑇 𝑥 𝑋 ↠ 𝑇 𝑦 𝑌$ is a linear epimorphism (i.e. has full rank). We define $𝐾 = k e r 𝑇 𝑥 𝑓$ where $d i m 𝐾 = 𝑛 - 𝑚$ by the Rank-nullity theorem, and let $𝑃 : ℝ 𝑁 ↠ 𝐾$ be a projection operator onto $𝐾$ (note $𝐾 \leq 𝑇 𝑥 𝑋 \leq ℝ 𝑁$ ). We can then define

𝐹 : 𝑋 \to 𝑌 \times 𝐾 𝜉 \mapsto (𝑓 (𝜉), 𝑃 (𝜉))

which has the tangent map

𝑇 𝑥 𝐹 : 𝑇 𝑥 𝑋 \to 𝑇 𝑥 𝑌 \times 𝐾 ⃗ 𝐚 \mapsto (𝑇 𝑥 𝑓 ⃗ 𝐚, 𝑃 ⃗ 𝐚)

which is clearly a Linear isomorphism, so by the inverse function theorem $𝐹$ is locally a diffeomorphism at $𝑥$ , i.e. maps some open neighbourhood $𝑈$ of $𝑥$ diffeomorphically onto a neighbourhood $˜ 𝑈$ of $(𝑦, 𝑃 (𝑥))$ , Thus $𝐹$ maps $𝑓 - 1 {𝑦} \cap 𝑈$ diffeomorphically onto $({𝑦} \times 𝐾) \cap ˜ 𝑈$ which is diffeomorphic to an open subset of $ℝ 𝑛 - 𝑚$ . Thus $𝑓 - 1 {𝑦}$ is an $(𝑛 - 𝑚)$ dimensional $𝐶 \infty$ differentiable manifold.

Further properties

$𝑇 𝑥 𝑆 = k e r 𝑇 𝑥 𝑓$

#state/tidy | #lang/en | #SemBr