# Poincare duality and orientation

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$is not orientable where will go wrong in the proof? ## 1 Answer An orientation of$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)\$).

• Thank you, do you know any motivation of orientation? And how do we find fundamental class, can we require it to be a generator? – 6666 Nov 7 '16 at 16:35
• Moreover, why do we require the coefficient to be a ring not just a group? – 6666 Nov 7 '16 at 16:36
• I would suggest asking those as separate questions, as they're a bit involved to answer in comments. – Eric Wofsey Nov 7 '16 at 20:14