Orientation of a "glued"-manifold Im wondering if there's a short way to prove that when two manifolds with diffeomorphic boundaries are glued together along the boundaries the orientations of these must be inverse to each other. That is to say, suppose you have $M$ and $N$ oriented $n$-dimensional manifolds with $\partial M \cong \partial N$ under a diffeomorphism $\phi: \partial M \to \partial N$, you form $ C = M \cup_\phi N$, in order to do this you need that the orientation of $\partial M$ be opposite to that of $\partial N$, why is that? By homological means...
I understand the reason via the orientation of the tangent spaces and the outward-first orientation of the boundaries, but how can i prove it with fundamental classes?
I know the homology of the pair $(C,\partial M) \cong (M,\partial M) \oplus (N,\partial N)$ (relative Mayer-Vietoris) and the inclusion $j: (C, \emptyset) \to (C,\partial M)$ induces a monomorphism in the top homology because of the exact sequence of the pair $(C,\partial M)$, and I came up with a "proof" using this, but it is way to lengthy, maybe there is a "quick way" to do this?
Thanks
 A: I agree that the difficulty in the question is that you are relying on the homological definition of an orientation of a manifold.  As Ryan implies in the comments, the solution is undergraduate-level mathematics if you define the orientation of a smooth manifold as an equivalence class of bases of its tangent spaces.  But actually, in the question itself you said that you already know this.
What's left, if you press the point, is to prove that this definition of an orientation in differentiable topology is equivalent to the homological definition.  This is not an easy theorem!  You either have to work with singular homology, or if you want geometric simplicial homology you need the theorem that you can triangulate manifolds.  After that, I would use the de Rham theorem, that de Rham cohomology is isomorphic to simplicial or singular cohomology.  It's easy to see that the orientation class in de Rham cohomology is equivalent to a class of tangent orientations.
You can argue similarly in the PL category, using a triangulation of a PL manifold.
You can ask the question again for topological manifolds, and then...maybe it is the most in the spirit of your question.  I think that indeed, the exact sequence solution that you sketch is the best one.  In a sense, even that is the easy end of the question.  It is not easy to distinguish the boundary of a manifold from the interior, and it is not so easy to prove Poincaré duality for topological manifolds either.
