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)

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