Poincare duality and orientation

Question

I can't see why we need $M$ to be orientable in the proof of Hatcher, i.e. where do we need the orientation assigns each $x\in M$ a generator of $H_n(M|x;R)$? I wonder if $M$ is not orientable where will go wrong in the proof?

Answer 1

An orientation of $M$ is used to define the map $D_M$ to begin with; you can't prove that a map is an isomorphism without defining what the map is! The way $D_M$ is defined in terms of an orientation is used repeatedly in the proof of Lemma 3.36 and the proof you have shown. First, you need to know that $D_M$ forms commutative diagrams with the maps $D_U$ for each open set $U$ (this uses the fact that $D_U$ is defined by using the same orientation as $D_M$, just restricted to $U$). Second, in step (1) you need to know that in the case $M=\mathbb{R}^n$, the definition of $D_M$ is such that you can explicitly compute that it is an isomorphism (this uses the fact that $D_M$ is defined in terms of capping with fundamental classes, and you can explicitly compute the fundamental class and what capping with it does for $(\Delta^n,\partial\Delta^n)$).