Killing a cohomology class in a manifold by taking out a submanifold representing its Poincaré-dual

Question

Let $M$ be an oriented $n$-manifold, $\alpha \in H^k(M;R)$ a cohomology class so that its Poincaré-dual $a\in H_{n-k}(M;R)$ is represented by an embedded $(n-k)$-manifold $F$.

Question: Is the restriction of $\alpha$ to $M-F$ always trivial?

In particular, I am interested in the case where $n=4$, $k=2$, $R=\mathbb{Z}/2$ and $\alpha$ is the second Stiefel-Whitney class of $M$. This case is used in the book 4-manifolds and Kirby Calculus by Gompf and Stipsicz (p.187). If the Statement in my Question is false, how would I try to prove this special case?

Answer 1

Yes, let us fix a cohomology class $\beta \in H^i (M)$ and denote its dual homology class in $H_{n-i}(M)$ by $b$. Poincare-Lefschetz duality allows us to identify $H^*(M - F)$ with $\bar{H}_{n-*}(M/F)$ (the topological quotient). There is an interaction between the Poincare duality of $M$ and the Poincare-Lefschetz duality of $(M-F)$, namely the restriction $\iota^* (\beta)$ is Poincare dual to $q_*(b)$, the pushforward of $b$ under the quotient map. If the class $\beta$ is Poincare dual to a homology class, this has a very geometric proof via intersection theory for noncompact manifolds. If $\beta$ is a general cohomology class, the easiest proof of this is likely via the Thom isomorphism theorem.

Specializing to the case $\beta =\alpha$, we see that the Poincare-Lefshcetz dual of $\iota^* (\alpha)$ is the image of $[F]$ under the quotient map collapsing $F$ to a point, i.e. is $0$. Since Poincare-Lefschetz duality is an isomorphism, this implies that the restriction of $\alpha$ to $M-F$ is 0.

Answer 2

As Connor pointed out, the answer to my question is positive. I just wanted to write down an easy proof with precise references.

Let $\nu(F)$ be a tubular neighborhood of $F$. We apply Poincaré-Lefschetz duality in the form of [Bredon, VI.8.4] to $M$ and $M-\nu(F)$ to obtain a commutative diagram of the following form.

\begin{array} 0& & H^k(M;R) &\xrightarrow{i^*}& H^k(M\setminus \nu(F);R) \\ &&\downarrow&&\downarrow\\ H_{n-k}(\nu(F);R) & \xrightarrow{j_*} & H_{n-k}(M;R) & \xrightarrow{u} & {H_{n-k}(M,\nu(F);R)} \end{array}

The vertical maps are isomorphisms induced by Poincaré duality and I will denote them by PD. A class $\alpha\in H^k(M)$ which is Poincaré-dual to $j_*[F]$ evidently maps to zero under $i^*$ as the diagram commutes and $u(j_*[F])$ is trivial.