The canonical Seifert surface argument

One of the earliest "proper" obstruction-theory arguments in topology goes back to Bruschlinsky, Math. Ann. 109 (1934), stating that the homotopy-classes of maps from a space to the circle are in one-to-one correspondence with the 1-dimensional cohomology of the space. 

[X,S1] = H1(X)

As a grad student I think I learned this as being one of Serre's theorems, as it was he and Thom that used this theorem heavily to resolve various Steenrod realization problems.  Much later it was pointed out (Milgram SLN 685) that this machinery gives a very slick formalism for proving the existence of Seifert surfaces for knots -- probably it was slow to be noticed due to it being a relative argument.  I'll provide a sketch below. 

Theorem: Let M be an (n-2)-dimensional compact, connected, oriented, boundaryless submanifold of Sn. Provided n>2, there is an orientable submanifold W of Sn with dW=M.  Here d stands for boundary. 

Of course, it does not hold for d=2, as a point is not a boundary.  

The proof involves two key observations. 

1) The Bruschlinsky theorem is functorial.  Meaning if A is a subspace of B, then the restriction map [B,S1] → [A,S1] is identified with the cohomological restriction H1(B) → H1(A). 

2) There is the inclusion M → Sn, with the corresponding pair (Sn, M), which is homologically equivalent to the pair (CM, dCM), where CM is the complement of a small open tubular neighbourhood of M in Sn.  This is via excision.   Moreover, dCM is a fibre bundle over M with fibre a circle.  

Although the fibre bundle dCM → M turns out to be trivial, I find it isn't best to try attacking this problem head-on.  Focus your attention on the map:

[CM, S1] → [dCM, S1]. 

equivalently, 

H1(CM) → H1(dCM)

This map fits into the cohomological long exact sequence of the pair (dCM, CM). 

... →H1(CM,dCM) → H1(CM) →H1(dCM) → H2(CM, dCM) →...

Via the excision identification Hk(CM, dCM) ~ Hk(Sn, M) and a little chase with the cohomology long exact sequence of the pair (Sn, M) we see that we have an extension:

0 → H1(CM) →H1(dCM→ H1(M) → 0

with the map H1(dCM→ H1(M) being a section of induced map on H1 corresponding to the bundle map dCM → M. 

This gives us a direct-sum decomposition

H1(dCM) = H1(CM H1(M)

A quick Poincare Duality (+LES+excision...) argument gives us that H1(CM) = Hn-1(CM, dCM) = Hn-1(Sn, M) = Hn-2(M) = Z. 

This gives us that the restriction map

H1(CM) →H1(dCM)

can be identified with the inclusion

Z → Z  H1(M),

given by mapping t → (t,0).

We're in a good position to deal with the bundle dCM → M now.  The above argument shows that the map dCM → M induces (on 1-dimensional cohomology) precisely what one would expect provided you knew the bundle was trivial, i.e. dCM ~ M x S1.  But it turns out that circle bundles are very special, and it takes relatively little for them to be trivial!

First off, we know dCM is an orientable manifold, because it is the boundary of an orientable manifold. Since M is also orientable, this means the fibres admit an orientation.  So the bundle dCM → M is trivial over the 1-skeleton of M.   Think through a cellular homology computation of H1dCM.  The only way we can have the splitting H1(dCM) = H1(M)  H1(S1) is if the bundle is trivial over the 2-skeleton.  

Perhaps surprisingly, this finishes the job.  This is due to the fact that the group Diff(S1) has the homotopy-type of O2, whose only non-trivial homotopy groups are in dimension zero and one.  So once you know a bundle with fibre a circle is trivial over the 2-skeleton of a space, you know it is trivial over the entire space. 

So where are we? We've proven that the generator of [CM, S1] restricts to a very special map in [dCM, S1]. dC= M x S1, so there is a very special map dC→ S1, retraction onto the fibre.  The generator of  [CM, S1] restricts to this!

Technically, these are all homotopy-classes of continuous maps, but with a little thought you can ensure all the maps are smooth, and the above arguments ensures a smooth representative of the generator restricts to a retraction onto the fibre.  Let f : CM → Sbe such a map.  The pre-image of a regular value of this map is a Seifert surface for (a manifold isotopic to) M. 

That's the basic idea of the argument. 


I should note, there are more sophisticated ways to prove the bundle dCM→M is trivial.  In Bredon's text Topology and Geometry he formulates "triviality over the 2-skeleton" in terms of the Euler Class of the bundle, a concept I've avoided above.  He argues that the Euler class of M, which is an element of H2(M) is the pull-back of the Thom Class of M in Sn, which is itself an element of H2(Sn), but this latter group is trivial. Ultimately, Bredon's argument is equivalent to the argument above.  In the above argument, more obstruction theory appears "out" of the black box of slick machinery, in Bredon's argument there is more "packaging".  These arguments appear in VI.12.1, VII.14.6, VII.14.7 of Bredon's book.