# Random Exercise #2: Problem 11-C in Milnor’s Characteristic Classes

Todays’ random exercise will be the pretty famous exercise 11-C  one can find in Milnor’s Book. Let me recall the statement:

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  $\frown \mu_A : H^k(A) \rightarrow H_n(A)$ maps the cohomology class $u^{'}|A$ dual to $M$ to the homology class $(-1)^{nk} i_{*} (\mu_M)$. Assume moreover that the normal bundle $\nu^k$ is oriented so that $\tau_M \oplus \nu^k$ is orientation preserving isomorphic to $\tau_A|M$.

Hint: consider the following commutative diagram:

where $N$ is a tubular neighbourhood of $M$ in $A$.

Before going all in with the proof, let me recall two things:

Theorem [Corollary 11.2 page 117  in Milnor’s Characteristic Classes] If $M$ is embedded as  a closed subset of $A$, the cohomology ring $H^*(E,E_0; \Lambda)$ associated with the normal bundle $\nu^k\colon E \to M$ in $A$ is canonically isomorphic to the cohomology ring $H^*(A,A\setminus M; \Lambda)$. Here $\Lambda$ can be any coefficient ring.

Definition: Let $\nu^k$ be oriented, let $u \in H^k(E,E_0\setminus M; Z)$ be its Thom Class (denoted  occasionally also $Th(\nu^k)$)  Define $u' \in H^k(A,A\setminus M; Z)$ to be the image of the Thom Class under the above iso.

Theorem [Thm 11.3 page 119]: If $M$ is embedded as a closed subset of $A$, then the composition of the two restriction homomorphisms

$H^k(A,A\setminus M) \to H^k(A)\to H^k(M)$

with $\mathbb{Z}_2$ coefficients, maps the fundamental class $u'$ to the top Stifle-Whitney class $w_k(\nu^k)$ of the normal bundle. Similarly, if $\nu^k$ is oriented, then the corresponding composition with integer coefficients maps the integral fundamental class $u'$ to the Euler Class $e(\nu^k)$.

Definition: The image of $u' \in H^k(A)$ is called the dual cohomology class to the submanifold $M$ of codimension $k$. The terminology will be explained in the exercise

Proof of exercise:  Denote with $\mu_A \in H_p(A)$ the fundamental class of $A$. Start picking in the upper left angle the element $u' \otimes \mu_A \in H^k(A,A\setminus M)\otimes H_p(A)$. We need to chase this element a little bit. The left “path” ($\downarrow$ and then $\rightarrow$) is straightforward. We end up with the element $u'|A\frown \mu_A$.

For the “right” path ($\rightarrow$ and then $\downarrow$ and then $\leftarrow$) we can use the so called coherency of the fundamental class $\mu_A$ and Corollary 11.2 to prove that $\mu_A$ is sent to $\mu_{A,A\setminus M}$ which is the generator of $H_p(A, A\setminus M)$ ( See Bredon’s Topology and Geometry Theorem 7.8 page 344). Following the next arrow, which is an iso, we end up in $H^k(N,N\setminus M)\otimes H_p(N,N\setminus M)$ with $Th(\nu)\otimes \mu_{N,N\setminus M}$ . For the curious reader (yes please!) interested in this kind of stuff, I worked 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 $H^k(N,N\setminus M)\cong H^k(E,E_0)$) 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.

(inspired by my answer here in M.Se)