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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?

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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 $(\Delta^n,\partial\Delta^n)$).

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

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