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Early on in Hempel's book 3-Manifolds, he discusses two-sided submanifolds: if $N$ is a manifold of dimension $n$, and $M$ is a submanifold of dimension $(n-1)$, then $M$ is two-sided if there is an embedding of $M\times [1,-1]$ into $N$, with $M\times \{0\}$ the inclusion map.

He then proves that if M is any $(n-1)$-dimensional submanifold of $N$, and the inclusion map $i: M\rightarrow N$ induces the zero map on first homology ($i_\ast: H_1(M, \mathbb{Z}_2)\rightarrow H_1(N, \mathbb{Z}_2)$), then $M$ is two-sided. The proof is by contradiction: if $M$ is not two-sided, then it must not disconnect a tubular neighborhood. Thus there's a loop (call it $L$) in this neighborhood, that intersects $M$ in only one point. Since $L$ is homologous to $0$, this contradicts the homological invariance of intersection numbers (mod 2).

I have some questions about this proof that I just can't wrap my head around:

  1. Where does $L$ come from? I have a vague idea how to construct it, using the fact the tubular neighborhood of $M$ is non-trivial, but that isn't discussed in the book. I don't get how to construct the loop with just Hempel's definition of two-sided, together with $M$ not disconnecting its neighborhood. Is there an easy proof that $L$ must exist?
  2. Why is $L$ homologous to $0$? Is this because it is one-dimensional, and $i_\ast$ is zero on $M$'s first homology?
  3. Is there a resource that shows intersection numbers are "homologically invariant (mod 2)"? I tried searching around, but couldn't find anything.
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up vote 5 down vote accepted

To construct $L$, start with a short path $\alpha$ that cuts across $M$ transversely at a single point. The path is sufficiently short that its two endpoints lie in the given tubular neighborhood of $M$. Since $M$ does not disconnect this neighborhood, those two endpoints of $\alpha$ may be joined by another path $\beta$ which stays in the neighborhood and is disjiont from $M$. The concatenation of $\alpha$ and $\beta$ forms the loop $L$.

The reason $L$ is homologous to zero is because $L$ is homotopic to a loop in $M$ --- just project $L$ to $M$ using the projection map of the tubular neighborhood. Since $L$ is homologous to a loop in $M$, and since the inclusion induced map $H_1(M) \to H_1(N)$ is the zero map, it follows that $L$ is null-homologous.

For a reference, you might look up a basic book on differential topology, such as the one by Guilleman and Pollack.

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Nice answer, as always. If you know a better approach to (3) than in my answer, I'd like to hear it. – Mike Miller Mar 6 at 4:25
Your transversality argument is exactly how I learned to think about it. @MikeMiller – Lee Mosher Mar 6 at 14:26

As for why intersection numbers are homology invariant - this is easiest to see in the following case. Suppose I consider two submanifolds $L$ and $L'$ that are homologous, and this is evidenced by a map $f: C \to N$ where $C$ is a smooth manifold and $\partial C = L \sqcup \overline{L'}$; that is, $C$ is a cobordism from $L$ to $L'$. Then perturbing the manifold $M$ to be transverse to $C$, we see that $L \cap M$ is cobordant to $L' \cap M$, this cobordism realized by $C \cap M$. In the case that $L$ and $L'$ are one-dimensional, this is always possible.

In full generality, I would phrase it like this. Instead of being homologous submanifolds, there is a (let's say piecewise smooth) map from a simplicial complex $C \to N$ whose (simplicial) boundary is $L \sqcup \overline{L'}$, and such that $C$ is a manifold away from a codimension 3 subcomplex. (Hatcher very, very briefly sketches this idea somewhere in chapter 2.) Then we can still do our best to do transversality theory with this, and set it up so that $M$ only intersects $C$ away from this codimension 3 subcomplex. Then the above argument still goes through. It's just a bit harder to get intuition for what's going on here.

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To answer (3), in your specific case, the single point of intersection between $M$ and $L$ represents the dual to a non trivial cohomology class in $N$ (the cup product of the dual of $M$ with the dual of $L$); it's nontrivial because there's only one point. But if $L$ is homologous to zero, then this class should be zero, contradiction.

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