Can you explain a step in a proof about 2-sided surfaces in 3-manifolds?

Following is an argument given by Hempel where I am unable to understand his comment about choosing a loop close enough to a surface. Can somebody please elucidate this:

Lemma: If $F$ is a compact connected surface properly embedded in a $3$-manifold $M$ and if $image(i_*:H_1(F;Z/2Z)\rightarrow{H_1(M;Z/2Z)})=0$, then $F$ is 2-sided in $M$.

Proof: By regular neighborhood theory, it suffices to show that $F$ separates some connected neighborhood of $F$. If this is not the case then there is a loop $J\subset{M}$ such that $J\cap{F}$ is a single point, with transverse intersection. We may choose $J$ close enough to $F$ so that $J$ is homologous to zero (mod 2) in $M$. This contradicts homological invariance (mod 2) of intersection numbers. QED.

Doubt: Everything else is clear to me except the bold part in the proof. I don't think we will be able to bring $J$ close to $F$ unless it already bounds a disc. So I can see the proof only in the simply connected $M$ case where after suitably perturbing the loop and using the loop theorem to bound, I can apply an ambient isotopy to bring the disc itself closer to $F$. But how do we see this for the non-simply connected $M$? A complicated loop in $M$ might go all around it.

Reference: Lemma 2.1, Chapter 2 (Heegard Splittings), 3-manifolds(book) by Hempel.

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2 Answers

We want to prove that in the case F is not one-sided, we may replace J by a curve J' that is contained in a small neighborhood of F and interesects F in the same way as J. By assumtion F is one sided. Consider the boundary B of a small neighborhood N of $F$. Since F is one-sided, B is connected. Now, conisder the intersection of J with B. There are even number of intersections, since B is the boundary. So you can throw the part of J that does not belong to the neighbohood N an close it to a connected curve J' by segments in B (we assumed that B is connected). This explanes the words written in bold.

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Thanks Dmitri. I think this works! – Maharana Dec 24 '09 at 21:23

We have to prove that $F$ separates some connected neighborhood. Choose a neighbourhood $U$ that can be contracted onto $F$ (e.g. a tubular one) and suppose $F$ does not separate $U$. Then there is a curve $J$ in $U$ that intersects $F$ transversally at one point.

Every homological class in $U$ becomes zero mod 2 in $M$ (since the image of the homology mod 2 of $U$ = the image of the homology mod 2 of $F$ is zero).

In other words, for tubular neighborhoods "close enough" is not necessary, any curve would do.

Sorry, got the notation wrong the first time around.

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thanks algori, both yours and Dmitri's idea is same but his is somewhat more explicit so I am accepting it! You get my up vote! cheers! – Maharana Dec 25 '09 at 9:50