Poincare duality isomorphism problem in the book "characteristic classes"

Question

This is the problem from the book, "characteristic classes" written by J.W. Milnor.

[Problem 11-C] Let $M = M^n$ and $A = A^p$ be compact oriented manifolds with smooth embedding $i : M \rightarrow A$. Let $k = p-n$. Show that the Poincare duality isomorphism $\cap \mu_A : H^k(A) \rightarrow H_n(A)$ maps the cohomology class $u^{'}|A$ dual to $M$ to the homology class $(-1)^{nk} i_{*} (\mu_M)$. [We assume that the normal bundle $v^k$ is oriented so that $\tau_M \oplus v^k$ is orientation preserving isomorphic to $\tau_A|M$. The proof makes use of the commutative diagram where $N$ is a tubular neighborhood of $M$.

I think that it suffices to show that the right vertical map is sending the fundamental cohomology class of $H^k(N,N-M)$ to the fundamental homology class of $H_n(N) \cong H_n(M)$ because the right vertical map is an isomorphism. However, I can not catch the reason why the right vertical map is an isomorphism.

Can anybody give me a hint?

Thank you.

Answer 1

So this is one quite interesting exercise in Milnor-Stasheff. The exercise is not too hard, so I will give you a hint but also motivation to solve it. Here we go:

What the exercise means: So the exercise means that whenever you have a submanifold in a compact manifold, you can look at the fundamentalclass of the submanifold and the euler class of its normal bundle included/restricted to the (co)homology of the supermanifold and will see Poincaré duals! One very fundamental Theorem which is well-known (but somehow seldomly proved) can be deduced by this exercise together with naturality of characteristic classes:

Corollary: Every 1-codimensional homologyclass can be realised as a submanifold.

Note for this Corollary the canonical bijection $H^1(A) \cong [A,S^1]$ (boundary might be empty), that every class in the latter set can be realized by a smooth map which has regular values and that points admit only trivial bundles.

On how to solve the exercise: The plan is to start with an interesting element in the upper left corner. Then chase it the outer left and outer right way through the diagram. You will end up at $H_n(A)$ one time with the Poincaré dual of the restricted normal bundle (from left) and with the included fundamental class (coming from the right). Commutativity gives you equality of those and hence the important result of the exercise.

If you need further assistance with the proof of the exercise or with the proof of the corollary I will be happy to help. Good luck and have fun.

Answer 2

Start picking in the upper left angle the element $u' \otimes \mu_A \in H^k(A,A\setminus M)\otimes H_p(A)$ (using Milnor's notation $u'$ is the dual cohomology class of $M$), then you have to chase this element a little bit. The left "path" $( \downarrow$ and then $\rightarrow)$ is straightforward. You end up with $u'|A\cap \mu_A$.

For the "right" path $(\rightarrow$ and then $\downarrow$ and then $\leftarrow )$ use the so called coherency of the fundamental class $\mu_A$ and Corollary 11.2 in Milnor's Characteristic Classes to prove that you end up in $H^k(N,N\setminus M)\otimes H_p(N,N\setminus M)$ with $th(\nu)\otimes \mu_{N,N\setminus M}$ (work with compactly supported sections of the orientation bundle of $A$, which we know being trivial since $A$ is orientable, for more details see Bredon's Topology and Geometry chapter VI.7).

Since the vertical right arrow (after choosing the Thom class of the normal bundle as the cohomology class in $H^k(N,N\setminus M)\cong H^k(E(\nu),E(\nu)_0)$) is the so called homological Thom isomorphism (tom Dieck's Algebraic Topology Theorem 18.1.2 page 439), it's an isomorphism so it maps $\mu_{N,N\setminus M}$ to $\mu_M$ (up to a sign and using $H_n(N)\cong H_n(M)$). The conclusion follows at once.