# 37. Orientability and the cap product

An $n$-dimensional topological manifold $M$ is *orientable*** **if for each compact set $K \subset M$, there exists a homology class $ \langle o_K \rangle \in H_n(M, M \setminus K)$ with the property that for any $x \in K$, the natural map $H_n(M, M \setminus K) \to H_n(M , M \setminus x)$ induced by inclusion maps $ \langle o_K \rangle$ to a generator of $H_n(M , M \setminus x) \cong \mathbb{Z}$. This is equivalent to the fibre bundle $ \mathcal{O}(M)$ from last lecture being a trivial bundle.

We being this lecture we begin by proving the main result from last time: that the two functors $H_n(M, \square ;A)$ and $ \Gamma_c(M \setminus \square ; A)$ are naturally isomorphic, and then use this to define orientability as above.

The lecture concludes with the introduction of (yet another) product, called the *cap product*. This works for arbitrary topological space and is a "mixed" product:

$$ H^{ \bullet}(X) \otimes H_{ \bullet}(X) \to H_{ \bullet}(X), \qquad \langle \alpha \rangle \otimes \langle c \rangle \mapsto \langle \alpha \rangle \frown \langle c \rangle.$$

Although useful in its own right, for us the most important property of the cap product is for *Poincaré Duality*, which states the following: *if $M$ is a closed oriented topological manifold then $ \square \mapsto \square \frown \langle o_M \rangle$ is an isomorphism. *We will prove this in Lecture 39.

Comments and questions?