$R$-Orientability of a Topological Manifold Let $M$ be a $n$-manifold and $R$ a PID ring. $M$ is then called $R$- orientable if for each $x \in M$ there exists a family of generators $[M]_x \in H_n(M, M - \{x\}) \cong H_n(\mathbb{R}^n, \mathbb{R}^n- \{x\}) \cong \widetilde{H}_{n-1}(S^{n-1}) \cong R$ of $R$ such that for every $x \in M$ there exist an open $U \subset M$ with $x \in U$ and an $\alpha \in H_n(M, M- U)$ with following property:
For every $ z \in U $we have that
$\alpha \mapsto [M]_z$ for the canonical map $H_n(M, M-U) \to H_n(M, M - \{z\}) $.
Therefore locally $\alpha$ is mapped to the choosen generators $[M]_z$.
Now my question:
Let $B \subset M$ a closed $n$- disc on $M$ (so $B \cong \overline{B_r(0)}$).
I want to show that $M$ is $R$-orientable $\Leftrightarrow$ $M\backslash B$ is $R$-orientable.
"$\Rightarrow$" is trivial. Now the problem is "$\Leftarrow$"
My idea was following: If $x \in M \backslash B$ that's ok since local problem. So wlog $x \in B$. If I can show that there exist a smaller $n$ disc $B_{r'}(0) \cong B' \subset B$ with $x \in B \backslash B'$ such that the $R$ orientation of $M \backslash B$ is extendable to $M \backslash B'$ then I win since $x$ was arbitrary but I don't see a way to get it. 
Or does there exist another way to prove the statement?
 A: Let us prove
Theorem 1. Let $M$ be a manifold and  $M_1, M_2$ be open subsets (hence they are also manifolds). Assume that $M_1, M_2$ are orientable and $M_1 \cap M_2$ is connected. Then $M$ is orientable.
This doesn't apply exactly to your question concerning  the complement of a closed disk in $M$, but if you read the comments, you will see that this is probably the best what can be done ($M_2$ is an open disk and $M_1 \cap M_2$ an annulus). Anyway it suffices to treat connected sums.
Let us now introduce the concept of an orientation.
Let $\omega = (\omega_x)_{x \in M}$ be a a family of generators $\omega_x$ of $H_n(M,M \setminus x) \approx R$. Then for any two $x, z \in M$ we get a unique isomorphism $[\omega]^x_z : H_n(M,M \setminus x) \to H_n(M,M \setminus z)$ such that $[\omega]^x_z (\omega_x) = \omega_z$. Note we have $[\omega]^x_x = id$ and $[\omega]^z_y \circ [\omega]^x_z = [\omega]^x_y$.
We say that an open $U \subset M$ is $\omega$-trivializing if there exists an $\alpha \in H_n(M, M \setminus U)$ with the following property:
For every $z \in U$, $(i_U^z)_*(\alpha) = \omega_z$ (with the inclusion of pairs $i_U^z : (M, M \setminus U) \to (M, M \setminus z)$).
If each $x \in M$ has an open neighborhood $U$ which is $\omega$-trivializing, we call $\omega$ an orientation of $M$.
Clearly, a manifold is orientable if and only it has an orientation. However, it may have more than one.
Let us say that an open $U \subset M$ is isomorphically trivializing if $(i_U^z)_*$ is an isomorphism for all $z \in U$. Then for any two $x,z \in U$ we obtain an isomorphism $[U]^x_z = (i_U^z)_* \circ ((i_U^x)_*)^{-1} : H_n(M, M \setminus x) \to H_n(M, M \setminus z)$. This isomorphism is completely independent from any family of generators $\omega = (\omega_x)_{x \in M}$.
Lemma 1. For each $x \in M$ and each open neighborhood $V$ of $x$ in $M$ there exists an open neighborhood $U \subset V$ of $x$ which is isomorphically trivializing.
Proof. Using a chart around $x$ we find a homeomorphism $h : B_{2r}(0) \to V' \subset V$ such that $V'$ is open in $M$ and $h(0) = x$. Define $U = h(B_r(0))$. It is easy to see that the inclusion $j^z_U : M \setminus U \to  M \setminus z$ is a homotopy equivalence for all $z \in U$ (in fact, $M \setminus U$ is a strong deformation retract of $M \setminus z$). We have $i^z_U = (id_M,j^z_U) : (M, M \setminus U ) \to (M, M \setminus x)$. Now $i^z_U$ induces homomorphisms connecting the long exact sequences of the pairs $(M, M \setminus U)$ and $(M, M \setminus z)$. But all $(id_M)_*$ and all $(j^z_U)_*$ are isomorphisms (on all $H_k$), thus the five lemma proves that all $(i_U^x)_*$ are isomorphisms.
Lemma 2. Let $U \subset M$ be isomorphically trivializing. Then $U$ is $\omega$-trivializing if and only if $[\omega]^x_z = [U]^x_z$ for all $x, z \in U$.
Proof. Let $U$ be $\omega$-trivializing. Then there exists $\alpha \in H_n(M, M \setminus U)$ such that $(i_U^z)_*(\alpha) = \omega_z$ for all $z \in U$, i.e. $\alpha = ((i_U^z)_*)^{-1}(\omega_z)$ for all $z \in U$. But now $[U]^x_z(\omega_x) = (i_U^z)_*(((i_U^x)_*)^{-1}(\omega_x)) = (i_U^z)_*(\alpha) = \omega_z$, hence $[U]^x_z = [\omega]^x_z$. Conversely, let $[\omega]^x_z = [U]^x_z$ for all $x, z \in U$. For $x \in U$ define $\alpha_x = ((i_U^x)_*)^{-1}(\omega_x)$. Then $\omega_z = [\omega]^x_z(\omega_x) = [U]^x_z(\omega_x) = (i_U^z)_*(((i_U^x)_*)^{-1}(\omega_x)) = (i_U^z)_*(\alpha_x)$, i.e. $U$ is $\omega$-trivializing.
Let us say that an open $U \subset M$ is strongly $\omega$-trivializing if it is both $\omega$-trivializing and isomorphically trivializing. If $U$ is strongly $\omega$-trivializing, then for all $x, z \in U$ we have $[\omega]^x_z = [U]^x_z$ which shows that on $U$ the isomorphisms $[\omega]^x_z$ are independent of $\omega$.
Lemma 3. Let $V \subset M$ be $\omega$-trivializing. Then each isomorphically trivializing $U \subset V$ is strongly $\omega$-trivializing.
Proof. Let $\beta \in H_n(M, M \setminus V)$ such that for all $z \in V$, $(i_V^z)_*(\beta) = \omega_z$. Define $\alpha = j_*(\beta)$, where $j : M \setminus V \to M \setminus U$ denotes inclusion. Then for all $z \in U$, $(i_U^z)_*(\alpha) = \omega_z$.
Corollary 1. $\omega$ is a orientation if and only each $x \in M$ has an open neighborhood $U$ which is strongly $\omega$-trivializing.
Lemma 4. Let $\omega^1, \omega^2$ be orientations of $M$. Then the sets $E(\omega^1,\omega^2) = \{ x \in M \mid \omega^1_x = \omega^2_x \}$ and $N(\omega^1,\omega^2) = \{ x \in M \mid \omega^1_x \ne \omega^2_x \}$ are open in $M$.
Proof. For $x \in M$ choose $\omega^i$-trivializing open neighborhood $U_i$. $U_1 \cap U_2$ contains an isomorphically trivializing neighborhood $U$ of $x$. If $x \in E(\omega^1,\omega^2)$, then for $z \in U$ we get $\omega^1_z = [\omega^1]^x_z(\omega_x) = [U]^x_z(\omega^1_x) = [U]^x_z(\omega^2_x) = [\omega^2]^x_z(\omega^2_x) = \omega^2_z$, i.e. $U \subset E(\omega^1,\omega^2)$. The same argument applies for $x \in N(\omega^1,\omega^2)$.
Corollary 2. Two orientations $\omega^1, \omega^2$ of a connected manifold are equal if and only if $\omega^1_\xi = \omega^2_\xi$ for a single $\xi \in M$.
Now let $M$ be a manifold, $\xi \in M$ and $\omega$ be a family of generators. For each $g  \in G_\xi$ = set of generators of $H_n(M,M \setminus \xi)$ define a family of generators $\omega^{\xi,g} = (\omega^{\xi,g}_x)$ by $\omega^{\xi,g}_x = [\omega]^\xi_x(g)$. For $g = \omega_\xi$ we have $\omega^{\xi,g} = \omega$. Moreover, $\omega^{\xi,g}_\xi = g$ and $[\omega]^x_z(\omega^{\xi,g}_x) = [\omega]^x_z([\omega]^\xi_x(g)) = [\omega]^\xi_z(g) = \omega^{\xi,g}_z$ which implies $[\omega]^x_z = [\omega^{\xi,g}]^x_z$.
Lemma 5. Each strongly $\omega$-trivializing open $U \subset M$ is strongly $\omega^{\xi,g}$-trivializing. Thus, if $\omega$ is an orientation, then so is $\omega^{\xi,g}$.
Proof. For $x, z \in U$, $[(\omega^{\xi,g})]^x_z =  [(\omega)]^x_z = [U]^x_z$. Now Lemma 2 applies.
Corollary 3. Let $M$ be a connected manifold, $\xi \in M$ and $\omega$ be an orientation of $M$. Then the set of all orientations of $M$ is given as $\{ \omega^{\xi,g} \mid g \in G_\xi \}$. In other words, for each $g \in G_\xi$ there exists a unique orientation $\omega^{\xi,g}$ of $M$ such that $\omega^{\xi,g}_\xi = g$.
To prove Theorem 1 it clearly suffices to prove
Theorem 2. Let $M$ be an orientable manifold and $N \subset M$ be open and connected. Then each orientation of $M$ induces an orientation of $N$ (this is well-known) and each orientation of $N$ extends to an orientation of $M$.
Proof. It suffices to extend any orientation $\omega^N$ of $N$ to the component$M'$ of $M$ which contains $N$. All other components of $M$ can be oriented independently. Let $\xi \in N$. If $k : (N,N \setminus \xi) \to (M',M' \setminus \xi)$ denotes inclusion, we see by excision that $g = k_*(\omega^N_\xi)$ is a generator of $H_n(M',M' \setminus \xi)$. Since $M'$ is orientable, there exists a unique orientation $\omega^{\xi,g}$ of $M'$ such that $\omega^{\xi,g}_\xi = g$. Restriction induces an orientation $\omega^{\xi,g} \mid_N$ on $N$. But by construction $(\omega^{\xi,g} \mid_N)_\xi = \omega^N_\xi$ so that $\omega^{\xi,g} \mid_N = \omega^N$.  
