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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

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    $\begingroup$ You don't need them to be opposite. To get an orientation on the union you only need to ensure that the induced orientations on the boundary are consistent. So when you compare $M$'s induced orientation on its boundary to $N$'s, they're either the same or opposite. But you need that "sameness" or "oppositeness" to be constant from boundary component to boundary component. There is a slight preference to say "opposite" since that's what happens when the orientation on $M$ and $N$ (as submanifolds of the union) are induced by a global orientation on the union. $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2011 at 2:09
  • $\begingroup$ Thank you Ryan, you're correct and that is precisely what I want to know, I guess I wrote everything wrongly. How can you see that the orientation of $C = M \cup_\phi N$ induces opposite orientations on $M$ and $N$? $\endgroup$
    – Juan OS
    Commented Feb 4, 2011 at 2:24
  • $\begingroup$ No, an orientation of $C$ does not induce "opposite" orientations of $M$ and $N$ -- those orientations are not comparable. But it does induce opposite orientations on $M \cap N$ (the common boundary of $M$ and $N$). This is due to the orientation conventions chosen for boundaries of manifolds (the "inside" direction for $M$ is the "outside" for $N$). $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2011 at 3:00
  • $\begingroup$ In the above, $M \cap N$ can be oriented in two ways, as components of $\partial M$ or $\partial N$. This is the "opposite". $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2011 at 3:01
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    $\begingroup$ IMO your question would be more appropriate for math.stackexchange.com, rather than here. These are issues confronted in multi-variable calculus classes and intro manifolds classes, so that's really the more appropriate forum. You can encounter them as well when studying Poincare duality in an intro algebraic topology course but I think that's not as common. Anyhow, I'd be happy to write up a long answer there. $\endgroup$
    – Ryan Budney
    Commented Feb 4, 2011 at 3:03

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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.

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