According to the above definition of a global orientation, don't we have to show that both $\tilde\mu_{x}, \tilde\mu_{y}$ in $H_n(\tilde M, \tilde M \backslash \mu_x )$ and $H_n(\tilde M, \tilde M \backslash \mu_y )$ respectively come from the same generator in $H_n(\tilde M , \tilde M \backslash U(\mu_B))$? I can't justify this fact rigorously.
I tried to prove that the isomorphism between $H_n(\tilde M , \tilde M \backslash U(\mu_B)$ and $H_n(M,M \backslash B)$ induced by 8.7.1 above is the same isomorphism for all x, but I wasn't sucessful.
I would be grateful for any help. Thanks.
Let us go back to the roots and discuss the concept of local orientation and the local consistency condition for orientations. Let $M$ be an arbitrary manifold.
$H_n(M \mid x) = H_n(M, M \setminus \{x\})$ is an infinite cyclic group. To see this, choose any open neighborhood $U$ of $x$ which is homeomorphic to $\mathbb R^n$ or, if you prefer, to an open Euclidean ball $B_R(0) \subset \mathbb R^n$ of finite radius $R > 0$ (note that for $R = \infty$ we obtain $B_R(0) = \mathbb R^n$). Then by excision $H_n(M, M \setminus \{x\}) \approx H_n(U, U \setminus \{x\}) \approx \mathbb Z$. A local orientation of $M$ at $x$ is a generator of $H_n(M \mid x)$. There are exactly two of them.
An open $B \subset M$ is called a small ball if there exists an open $U \supset B$ (called "surrounding set") and a homeomorphism $h : (U,B) \to (B_R(0),B_r(0))$ with $0 < r < R \le \infty$. The small balls clearly form a basis for the topology of $M$. Note that $B \approx B_r(0)$ does not imply the existence of a surrounding set $U$ (as an example take $B = S^n \setminus \{\xi\}$ for $\xi \in S^n$). For later use also observe that each small ball admits a surrounding set $U$ which is itself a small ball.
Hatcher has shown the following: For each small ball $B$ one has
$H_n(M \mid B) = H_n(M, M \setminus B)$ is an infinite cyclic group. This follows from the excision isomorphism $H_n(M, M \setminus B) \approx H_n(U, U \setminus B)$ and the the fact that the inclusion $U \setminus \ h^{-1}(\{0\}) \to U \setminus B$ is a homotopy equivalence.
Note that $U$ does not occur in the group $H_n(M \mid B)$, but its existence is essential in the proof.
For each $x \in B$ the inclusion-induced $j^B_{x} : H_n(M \mid B) \to H_n(M \mid x)$ is an isomorphism.
A family of local orientations $\mu_x$ at the points $x \in M$ is said to be consistent on $B$ if all $(j^B_x)^{-1}(\mu_x)$, $x \in B$, have the same value $\mu_B$. This $\mu_B$ is of course a generator of $H_n(M \mid B)$. Equivalently we can therefore require $\mu_x = j^B_x(\mu_B)$. We can also avoid using $\mu_B$. For any two $x,y \in B$ we get the isomorphism $j^B_{x,y} = j^B_y \circ (j^B_x)^{-1} : H_n(M \mid x) \to H_n(M \mid y)$. Then $(\mu_x)$ is consistent on $B$ iff $\mu_y = j^B_{x,y}(\mu_x)$ for all $x,y \in B$.
An orientation function on $M$ assigns to each $x \in M$ a local orientation $\mu_x$ of $M$ at $x$. An orientation of $M$ is an orientation function on $M$ satisfying the local consistency condition which means that each point of $M$ has a small ball neighborhood $B$ such that the $\mu_x$ are consistent on $B$.
Let us now come to $\tilde M$. I would prefer to write $$\tilde M = \{(x,\mu) \mid x \in M, \mu \in \Omega(x) \} = \bigcup_{x \in M} \{x\} \times \Omega(x)$$ where $\Omega(x)$ denotes the two-element set of local orientations of $M$ at $x$. Hatcher's $\mu_x$-notation seems to suggest that we already have an orientation function $x \mapsto \mu_x$ which is not the case. But that is just matter of taste.
The canonical map $(x,\mu) \mapsto x$ is a two-to-one surjection $p : \tilde M \to M$ and we can endow $\tilde M$ with a topology making $p$ a two-sheeted covering space projection. This is done by taking all pairs $(B,\mu_B)$ where $B$ is a small ball and $\mu_B$ is a generator of $H_n(M \mid B)$. Let $$U(B,\mu_B) = \{ (x, j^B_x(\mu_B)) \in \tilde M \mid x \in B \} .$$ As Hatcher says, it is easy to check that these sets form a basis for a topology on $\tilde M$, and that the projection $p : \tilde M \to M$ is a covering space with two sheets. The sheets over $B$ are the two sets $U(B,\mu^\pm_B)$, where $\mu^\pm_B$ are the two generators of $H_n(M \mid B)$.
Let us check that $p$ induces for each $(x,\mu) \in \tilde M$ a canonical isomorphism $$p^{(x,\mu)}_* : H_n(\tilde M \mid (x,\mu)) \ \to H_n(M \mid x) .$$ In fact, choose $(B,\mu_B)$ such that $(x,\mu) \in U(B,\mu_B)$ and use excision to define $$p^{(x,\mu)}_* : H_n(\tilde M \mid (x,\mu)) \to H_n(U(B,\mu_B) \mid (x,\mu)) \xrightarrow{p_*} H_n((B \mid x) \to H_n(M \mid x) .$$ Note that $p$ maps $U(B,\mu_B)$ homeomorphically onto $B$, thus the map $p_*$ in the middle is an isomorphism.
The map $p^{(x,\mu)}_* = p^{(x,\mu)}_*[B,\mu_B]$ does not depend on the choice of $(B,\mu_B)$: If $(B',\mu_{B'})$ is another choice, then we may choose a small ball $B''$ with $x \in B'' \subset B \cap B'$. There exists a generator $\mu_{B''}$ of $H_n(M \mid B'')$ such that $(x,\mu) \in U(B'',\mu_{B''})$. Clearly $U(B'',\mu_{B''}) \subset U(B,\mu_{B}) \cap U(B',\mu_{B'})$, thus we can use once more excision that the maps $p^{(x,\mu)}_*[B,\mu_B]$ and $p^{(x,\mu)}_*[B',\mu_{B'}]$ both agree with $p^{(x,\mu)}_*[B'',\mu_{B''}]$.
This shows that $H_n(\tilde M \mid (x,\mu))$ has $\tilde \mu_x = (p^{(x,\mu)}_*)^{-1}(\mu)$ as its canonical generator.
Thus $\tilde M$ has a canonical orientation function $$(x,\mu) \to \tilde \mu_x .$$
It remains to verify the local consistency condition. I think Hatcher is a bit short here.
Consider a point $\xi_0 = (x_0,\mu_0) \in \tilde M$. Take a small ball neighborhood $B$ of $x_0$ in $M$ as above and let $\tilde B$ denote the sheet over $B$ containing $\xi_0$.
Let $B' \supset B$ be a surrounding set which is itself a small ball. Let $\tilde B'$ denote the sheet over $B'$ containing $\xi_0$. Then $\tilde B' \supset \tilde B$. Since $p : \tilde B' \to B'$ is a homeomorphism, we get an excision-induced isomorphism $$p_*^{B} : H_n(\tilde M \mid \tilde B) \to H_n(\tilde B' \mid \tilde B) \xrightarrow{p_*} H_n(B' \mid B) \to H_n(M \mid B) .$$ This means that $H_n(\tilde M \mid \tilde B)$ is infinite cyclic with a generator $\tilde \mu_{\tilde B}$ such that $p_*^{B}(\tilde \mu_{\tilde B}) =\mu_B$. By construction the followimg diagram commutes:
$\require{AMScd}$ \begin{CD} H_n(\tilde M \mid \tilde B) @>>> H_n(\tilde B' \mid \tilde B) @>{p_*}>> H_n(B' \mid B) @>>> H_n(M \mid B) \\ @V{j}VV @V{j}VV @V{j}VV @V{j}VV \\ H_n(\tilde M \mid (x,\mu)) @>>> H_n(\tilde B' \mid (x,\mu)) @>{p_*}>> H_n(B' \mid x) @>>> H_n(M \mid x) \end{CD} for all $(x,\mu) \in \tilde B$. That is, $\tilde \mu_x = j^{\tilde B}_{(x,\mu)}(\tilde \mu_{\tilde B})$ which shows consistency.