Mayer-Vietoris Sequence with Pairs Applied to (Oriented) Connected Sum of Manifolds

Question

I was trying to figure out why the (oriented) connect sum of two oriented manifolds should be oriented. Here the oriented connect sum $M_1\# M_2$ of two $n$-manifolds $M_1$ and $M_2$ is taken to be the manifold gotten by gluing the two manifolds with interior of closed disks $B_1^0$ and $B_2^0$ removed via an orientation reversing homeomorphism. An $n$-manifold $M$ is orientable if its top homology group relative to its boundary $H_n(M, \partial M)$ is non-trivial, in fact, isomorphic to $\mathbb{Z}$. An orientation is a choice of the generators of this group (c.f. Knots and Links by Rolfsen).

Let's take $M_j$ to be manifolds without boundary to make it easier. I approached this problem by applying Mayer-Vietoris sequence with pairs. On P152 of Hatcher's Algebraic Topology, it says "If one has a pair of spaces $(X, Y ) = (A \cup B, C \cup D)$ with $C \subset A$ and $D \subset B$ , such that $X$ is the union of the interiors of $A$ and $B$, and $Y$ is the union of the interiors of $C$ and $D$, then there is a relative Mayer–Vietoris sequence" \begin{equation} \cdots \rightarrow H_n(A\cap B, C\cap D) \xrightarrow{\Phi} H_n(A, C) \oplus H_n(B, D)\xrightarrow{\Psi} H_n(X, Y) \xrightarrow{\partial}\cdots \end{equation} I applied this with $A = M_1\setminus B_1, B = M_2\setminus B_2, C = \partial B_1, D = \partial B_2$. Note that both $M_i$ should be enlarged a bit such that the interior is a cover of $M_1\# M_2$. I got \begin{equation} 0\rightarrow H_n(S^{n-1}\times I)\xrightarrow{\Phi_n} H_n(M_1\setminus B_1, \partial B_1)\oplus H_n(M_2\setminus B_2, \partial B_2)\xrightarrow{\Psi_n} H_n(M_1\# M_2)\xrightarrow{\partial_n} H_{n-1}(S^{n-1}\times I)\xrightarrow{\Phi_{n-1}} H_{n-1}(M_1\setminus B_1, \partial B_1)\oplus H_{n-1}(M_2\setminus B_2, \partial B_2)\rightarrow\cdots, \end{equation} by the fact that $(M_1\setminus B_1)\cap(M_2\setminus B_2) \cong S^{n-1}\times I$, $\partial B_1 \cap \partial B_2 = \emptyset$ and interior of $\partial B_1 \cup \partial B_2$ is also empty.

Now consider $\Phi_{n-1}$. This maps the generator of $H_{n-1}(S^{n-1}\times I)$ to the generators of $H_{n-1}(\partial B_j)$, so $\Phi_{n-1} = 0$. Therefore by exactness, $\partial_n$ is surjective. Since $H_{n-1}(S^{n-1}\times I) \cong \mathbb{Z}$, $H_n(M_1\# M_2)\not= 0$, and in fact, $H_n(M_1\# M_2) \cong \mathbb{Z}$. Since we do subtraction for the map $\Psi_n$ if $\Phi_n$ is taken to be positive embeddings in both components, the gluing homeomorphism should be orientation reversing.

Up to here everything was good, until I observed that I have turned the above sequence to be \begin{equation} 0\rightarrow 0 \rightarrow \mathbb{Z}\oplus \mathbb{Z}\xrightarrow{\Psi_n} \mathbb{Z}\xrightarrow{\cong} \mathbb{Z}\xrightarrow{0} \cdots. \end{equation} Then, $\Phi_n$ will be an injective homomorphism by exactness, which is impossible.

My first question is, what's the problem above? Did I apply Mayer-Vietoris with pairs correctly?

If we use homology instead of relative homology for the Mayer-Vietoris, the sequence should be

\begin{equation} 0\rightarrow 0 \rightarrow 0 \oplus 0\xrightarrow{\Psi_n} \mathbb{Z}\xrightarrow{\cong} \mathbb{Z}\xrightarrow{0} \cdots. \end{equation}

Now everything makes sense. However, I am wondering if there is a rigorous but elementary way to see $H_n(M_j\setminus B_j) = 0$. On one hand, this is intuitively true by triangulation, but it's hard to make it rigorous. On the other hand, using Lefschetz Duality seems to be an overkill.

Any help would be greatly appreciated!