Todays’ random exercise will be the pretty famous exercise 11-C one can find in Milnor’s Book. Let me recall the statement:
Let and be compact oriented manifolds with smooth embedding . Let . Show that the Poincare duality isomorphism maps the cohomology class dual to to the homology class . Assume moreover that the normal bundle is oriented so that is orientation preserving isomorphic to .
Hint: consider the following commutative diagram:
where is a tubular neighbourhood of in .
Before going all in with the proof, let me recall two things:
Theorem [Corollary 11.2 page 117 in Milnor’s Characteristic Classes] If is embedded as a closed subset of , the cohomology ring associated with the normal bundle in is canonically isomorphic to the cohomology ring . Here can be any coefficient ring.
Definition: Let be oriented, let be its Thom Class (denoted occasionally also ) Define to be the image of the Thom Class under the above iso.
Theorem [Thm 11.3 page 119]: If is embedded as a closed subset of , then the composition of the two restriction homomorphisms
with coefficients, maps the fundamental class to the top Stifle-Whitney class of the normal bundle. Similarly, if is oriented, then the corresponding composition with integer coefficients maps the integral fundamental class to the Euler Class .
Definition: The image of is called the dual cohomology class to the submanifold of codimension . The terminology will be explained in the exercise
Proof of exercise: Denote with the fundamental class of . Start picking in the upper left angle the element . We need to chase this element a little bit. The left “path” ( and then ) is straightforward. We end up with the element .
For the “right” path ( and then and then ) we can use the so called coherency of the fundamental class and Corollary 11.2 to prove that is sent to which is the generator of ( See Bredon’s Topology and Geometry Theorem 7.8 page 344). Following the next arrow, which is an iso, we end up in with . For the curious reader (yes please!) interested in this kind of stuff, I worked with compactly supported sections of the orientation bundle of , which we know being trivial since 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 ) 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 to (up to a sign and using ). The conclusion follows at once.
(inspired by my answer here in M.Se)