The following is written just before the boundary theorem in Guillemin & Pollack :

enter image description here

But I see that if Z is not transversal to X this is not true,why the book did not consider this case? why the book build the boundary theorem on the transversality case only? could anyone explain this for me please?

Boundary Theorem: suppose that $X$ is the boundary of some manifold $W$ and $g: X \to Y$ is a smooth map. If $g$ may be extended to all of $W$ then $I_2 (g,Z)=0$ for any closed submanifold $Z$ in $Y$ of complementary dimension.

  • 3
    $\begingroup$ Aren't these things only defined in the transverse case, because otherwise there's too many pathologies? Then there's some theorem that says you can homotope a non-transverse case into a transverse one or something. $\endgroup$
    – Randall
    Dec 9, 2018 at 2:57
  • $\begingroup$ yes you are right Randall @Randall $\endgroup$
    – Secretly
    Dec 9, 2018 at 2:59

1 Answer 1


They first build a map which is an extension of the map g on the W and then by using The Transversality Homotopy Theorem they ensure the transversality and Showed that the restriction of those two homotopic maps to the boundary are also homotopic, and so the theorem follows.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .