Orientation induced on submanifolds

Question

Suppose you are given two oriented manifolds with boundary $M$ say $B, B'$ and $\partial B = M = \partial B'$. Identify the boundaries and form $C = B \sqcup_{Id: M \to M} B'$. I want to see why, with the orientation induced by being submanifolds of $C$, $B$ and $B'$ induce opposite orientations to their boundary $M$. I'm particularly interested in the way the fundamental classes of $C,B,B'$ and $M$ behave. Thanks a lot!

Answer 1

I'm assuming you're JuanOS from MO, and this question corresponds to the MO question: https://mathoverflow.net/questions/54278/orientation-of-a-glued-manifold

Here's how I interpret your question. You have an oriented manifold $C$ which is compact without boundary, and you've decomposed it into the union of two submanifolds $B$ and $B'$ with $B \cap B' = M$, $M$ a compact manifold. Let $n=dim(C)$, so $n-1=dim(M)$. The global orientation class for $C$ is a generator $\mu_C \in H_n C$. There are restriction maps $H_n C \to H_n(B,M)$ and $H_n C \to H_n(B',M)$ which give the corresponding global orientations $\mu_B$ for $B$ and $\mu_{B'}$ for $B'$ respectively. Then there is the pairs $M \to B \to (B,M)$ and $M \to B' \to (B',M)$ and you want to know how the two generators of $H_n(B,M)$ and $H_n(B',M)$ compare when mapped to elements of $H_{n-1}M$ via the two connecting maps for the above pairs, specifically you want to show that $\partial \mu_{B} + \partial \mu_{B'} = 0$. Moreover, you want an argument that's fairly generic, in particular not specific to triangulated smooth manifolds or anything like that.

The above "restriction maps" are formally induced by inclusion $C \to (C, C \setminus int(B') ) \leftarrow (B,M)$, one being an excision inclusion, the other just an inclusion.

I don't believe this is as complicated as Kuperberg makes out -- the complication comes when attempting to bridge the gap between the smooth or simplicial views with the strictly homological view, especially say in the singular homology setting. But the above formulation side-steps those complications as your question is phrased entirely in terms of Mayer-Vietoris type constructions.

Okay, so here's a cheap way to check that it's true. Since $M$ is a submanifold of $C$, given a point $p \in M$ you can find an orientation-preserving degree 1 map $f : C \to S^n$ such that $f(M) \subset S^{n-1} \times \{0\} \subset S^n$. Moreover, you can ensure $f$ when restricted to $M$ is an isomorphism on the top homology groups of $M$ and $S^{n-1}$ respectively, and that $f$ sends $B$ to the top hemi-sphere, and $B'$ to the bottom hemi-sphere. So by naturality, you've reduced your problem to the case $D^n \sqcup_{S^{n-1}} D^n = S^n$, i.e. $C= S^n$, and $B$ and $B'$ both discs, which one way or another boils down to a cellular homology computation (we are using singular homology but what I'm saying is this is singular homology of a CW-complex to effectively cellular homology). This is the equivalent step to using outward pointing normals and determinants for smooth manifolds.

Answer 2

I don't know if this answers what you meant by "why", but to my mind this is why the orientations are opposite: Given a point $P$ on the boundary, consider a neighbourhood $U$ of $P$ in $C$ which is homeomorphic to an open ball in $\mathbb{R}^n$, with the boundary mapped to a hyperplane bisecting the ball. An orientation of $C$ determines an orientation of this ball. For the ball, it is clear that its two halves induce opposite orientations on the hyperplane bisecting it. Then this has to be true also of the intersections of $M$, $B$ and $B'$ with $U$, and hence, since $P$ was arbitrary, of $M$, $B$ and $B'$ in general.

If this wasn't the sort of answer you were looking for, please elaborate.