Map transverse to a submanifold - Imported Knowlpedia Content

Definition

Let $f:M\to N$ be a smooth map between finite-dimensional smooth manifolds without boundary, and let $S\subseteq N$ be an embedded submanifold. The map $f$ is transverse to $S$ at $x\in M$ , written $f\pitchfork_xS$ , if either $f(x)\notin S$ , or

df_x(T_xM)+T_{f(x)}S=T_{f(x)}N.

It is transverse to $S$ , written $f\pitchfork S$ , if it is transverse at every $x\in M$ . Equivalently, $f$ and the inclusion $S\hookrightarrow N$ are transverse smooth maps.

Normal-space formulation

At $x\in f^{-1}(S)$ , compose the differential $df_x:T_xM\to T_{f(x)}N$ with the quotient map to $T_{f(x)}N/T_{f(x)}S$ . Transversality is equivalent to surjectivity of this composite. This says that $f$ supplies every direction normal to $S$ , even though $df_x$ need not be surjective onto all of $T_{f(x)}N$ .

Preimage theorem

If $f\pitchfork S$ , then $f^{-1}(S)$ is an embedded submanifold of $M$ , with

T_xf^{-1}(S)=(df_x)^{-1}(T_{f(x)}S)

and codimension equal to $\operatorname{codim}_N S$ . This is the transverse-preimage theorem Hirsch, Chapter 3.

Examples and non-examples

Every submersion is transverse to every embedded submanifold of its target. A constant map whose value lies in a positive-codimension submanifold is not transverse, because its differential contributes no normal directions. A map whose image misses $S$ is transverse by the vacuous clause.

Conventions and scope

Some authors state the condition only at points of $f^{-1}(S)$ ; declaring it automatic elsewhere yields the same global notion. Boundary and corner versions require additional compatibility conditions not included here.

References

M. W. Hirsch, Differential Topology, Springer, 1976. Springer DOI record. Relevant: Chapter 3, transversality and the inverse-image theorem.
J. M. Lee, Introduction to Smooth Manifolds, 2nd ed., Springer, 2012. Springer DOI record. Relevant: Chapter 6, transversality.