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A $\textbf{local orientation}$ of a manifold $M$ at a point $x$ is a choice of generator $\mu_x$ of the infinite cyclic group $H_n(M, M- \{x\} )$.

For example, in the case of $M= \mathbb{R}^n$, $H_n(\mathbb{R}^n, \mathbb{R}^n - \{x \}) \cong H_n(S^{n-1})$ in which case $\mu_x$ would refer to a generator of the homology group of $S^{n-1}$ centered at $x$.

Hatcher defines $\textbf{orientation}$ as a function $x \to \mu_x$ assigning to each $x \in M$ a local orientation $\mu_x$. Such a function must satisfy a 'local consistency' property.

I am having trouble grasping this 'local consistency' property which is stated as follows:

For each $x \in M$ there exists a neighborhood $\mathbb{R}^n \subset M$.( I am assuming that this neighborhood is the one homeomorphic to $\mathbb{R}^n$ as in the definition of manifold) containing an open ball $B$ of finite radius about $x$ such that all local orientations $\mu_y$ for $y \in B$ are the images of one generator $\mu_b$ of $H_n(M, M-B) \cong H_n(\mathbb{R}^n, \mathbb{R}^n - B)$ under the natural map $H_n(M, M-B) \to H_n( M, M - \{y\})$.

  1. What exactly is the meaning of $\textbf{natural map}$ $H_n(M, M-B) \to H_n(M, M-\{y\})$? I know this is a map between local homology groups but in what way is this 'natural'?
Yuugi
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1 Answers1

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"Natural" here just means that it is the map on $H_n$ induced by the map of pairs $(M,M-B)\to (M,M-\{y\})$ (given by the identity $M\to M$), as opposed to some random other map $H_n(M,M-B)\to H_n(M,M-\{y\})$.

Eric Wofsey
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  • How do we define maps between pairs $(M, M-B) \to (M, M - {y})$ ? I can't recall or locate a definition of this. – Yuugi Jan 22 '16 at 02:29
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    It's just the identity map on $M$, which maps $M-B$ to (a subset of) $M-{y}$ since $y\in B$. – Eric Wofsey Jan 22 '16 at 02:36
  • Oh ok. Sure. Thank you! – Yuugi Jan 22 '16 at 02:37
  • @EricWofsey Is it necessary here to show the identity map induce a chain hotomopy to the identity map between the relative singular chains $S_\bullet(M, M-B)$ and $S_\bullet(M, M - {y})$? If $M$ is not a manifold and $B$ is not an open ball, does the map induced by the identity map still exist? – onRiv Feb 24 '22 at 16:03