Suppose $M$ is a smooth $m$-manifold and $Z$ a codimension-$d$ closed submanifold. Neither $M$ nor $Z$ has a boundary and neither is assumed to be compact. I believe that the inclusion $M \setminus Z \to M$ is $d-1$-connected, in that one can construct an argument based on the argument of Proposition 4.1 of https://people.math.osu.edu/anderson.2804/eilenberg/appA.pdf

The idea is that to show $i_*: \pi_n(M\setminus Z, m_0) \to \pi_n(M, m_0)$ is surjective when $n \le d-1$, you can take any class in the target and find a representative $f: S^n \to M$ that is smooth and meets $Z$ transversely. Dimension counting then shows that $f(S^n) \cap Z = \emptyset$, so the class is in the image of $i_*$.

To show it is injective when $n < d-1$, you argue in a similar way, but this time with smooth maps $F: D^{n+1} \to M$ that restrict to $f: S^n \to M \setminus Z$ on the boundary, where $f$ represents a class in the kernel of $i_*$. Again, you can choose a $D^{n+1}$ that is smooth and transverse to $Z$, and again conclude by counting dimensions.

Does anyone know of a reference for this fact in the (published) literature?

  • 3
    $\begingroup$ This is Theorem 1.1.4 in the notes ivv5hpp.uni-muenster.de/u/jeber_02/skripten/bordism-skript.pdf . It seems hard to track down a published reference. Many authors would just say ``By a standard transversality argument...". $\endgroup$ – Mark Grant Apr 1 at 12:16
  • $\begingroup$ Thanks! I'm glad it appears somewhere online. The treatment of transversality in Guillemin & Polalck's Differential Topology (ams.org/books/chel/370) is sufficiently strong to prove the result without too much extra work, since it handles manifolds with boundary. Other treatments I found restrict to manifolds without boundary for transversality, so aren't quite as convenient. $\endgroup$ – Ben Williams Apr 5 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.