A submersion is a smooth map between manifolds for which
is surjective at every .
The local submersion theorem says that around every there are coordinates centered at and centered at in which
To prove it, surjectivity lets us choose source coordinates such that are independent. Complete them by source coordinates and define
The derivative of is invertible at , so the inverse function theorem makes a local coordinate system; in these coordinates is the displayed projection.
For , the fiber is locally given by
These are slice coordinates, so is an embedded submanifold of dimension .
Choose a Riemannian metric on , using a partition of unity if necessary. For each , let
Since is a submersion, is an isomorphism. Its inverse depends smoothly on , so
defines a smooth vector field on satisfying . The formula also gives whenever .
Choose a coordinate ball around and a smaller convex coordinate ball whose closure lies in . For , let be the constant coordinate vector from to . Choose a smooth bump function supported in and equal to one on , and define in the chart
extending it by zero outside . This is a compactly supported smooth vector field. The segment stays in , where , so uniqueness for ordinary differential equations gives
In particular, .
Fix and choose as in part c. For , choose the compactly supported moving to , and lift it by part b to . The support of is contained in
which is compact because is a proper map. Hence is compactly supported and complete. If and are the flows of and , then
Uniqueness of integral curves gives
Therefore the diffeomorphism maps onto . Every equivalence class is open. Its complement, being a union of the other open classes, is also open; thus each class is clopen. If is connected, there is only one class. This proves the fiber-diffeomorphism conclusion of the Ehresmann fibration theorem.
Properness is essential. The projection
is a submersion but is not proper. Its fiber over is diffeomorphic to , whereas its fiber over zero is and has two connected components.
Merely requiring every fiber to be a submanifold is also insufficient. The surjective map
has every fiber finite and therefore a zero-dimensional embedded submanifold. Some regular values have three preimages and others have one, so the fibers are not all diffeomorphic. The map fails to be a submersion at .

Articles by others on the same topic (0)

There are currently no matching articles.