TITLE: Homologically trivial submanifolds
QUESTION [15 upvotes]: Unuseful prequel
Let $M$ be a (compact, oriented, differentiable) manifold. Before knowing anything about homology theory a naif but clever mathematician may want to measure the holes in $M$ by the following procedure. One lets $B_k(M)$ be the abelian group generated by embedded submanifolds of $M$ of dimension $k$ with border. There is an obvious border operator $\delta \colon B_{k}(M) \to B_{k-1}(M)$ and trivially $\delta^2 = 0$ since the border of a manifold with border is a closed manifold. So one may define $H^K_{naif}(M)$ by taking the quotient of cycles modulo boundaries.
Ok, this does not really work, for at least two reasons. The first there is no way to make this functorial without considering more degenerate objects (singular simplices or currents being two possible choices). The second is that even if one adjusts the definition to get, for instance, the bordism groups of $M$, these will be highly nontrivial for differential geometric reasons, rather than for the topology of $M$ (for instance they will be nontrivial for a point).
Anyway, one finally chooses a working (co)homology theory, for instance singular, and then can use the fundamental class of submanifolds to link this theory to the naif one. Two questions naturally arise:

Are fundamental classes of manifolds enough to generate the cohomology? This is nicely answered here.
Are submanifolds with border enough to generate homology relations? This is the purpose of the present question.

The actual question
Let $M$ be a (compact, oriented, differentiable) manifold and let $N \subset M$ be a (closed) submanifold. The problem is to test whether $N$ is the boundary of submanifold with boundary of $M$. There are two obvious obstructions:

$N$ should be bordant as an abstract variety
The class $[N] \in H_{*}(M, \mathbb{Z})$ should be $0$.


Assume 1 and 2 hold. Are there any conditions on $M$, $N$ or the embedding $N \to M$ which guarantee that $N$ is the boundary of submanifold with boundary of $M$?

There are a few classical cases which come to mind:

If $N$ and $M$ are spheres, 1 and 2 are vacuous and the thesis is true by the (generalized) Jordan curve theorem.
If $N = S^1$ and $M = \mathbb{R}^3$, the thesis is still true by the existence of Seifert surfaces.
If $\dim N = 1$ and $\dim M = 2$ the result seems true to me by the classification of surfaces as quotients of polygons and the usual Jordan curve theorem (but I did not check the details).

On the other hand I'm sure there are plenty of negative example even in $3$-dimensional topology. Is there anything nontrivial which can be said about this question, whether on the positive or negative side?

REPLY [4 votes]: (A similar question has since been asked here.)
This is a supplement to the answers of Jeff and Oscar, who note that first you must ask if the embedding $f\colon M \to N$ is the boundary of a smooth map $F\colon W\to N$, then ask if $F$ can be made an embedding. The bordism problem can be attacked using characteristic numbers of the map $f$. This is described in section 17 of the book "Differentiable Periodic Maps" by Conner and Floyd. In particular, their Theorem 17.5 states that

If the torsion of $H_*(N;\mathbb{Z})$ consists only of elements of order two, then maps $f_i\colon M_i\to N$, $i=1,2$ of oriented closed manifolds represent the same oriented bordism class if and only if all their Stiefel-Whitney and Pontrjagin numbers agree.

So, under these homological hypotheses, a necessary and sufficient condition for $f\colon M\to N$ to be null-bordant is that all the numbers $\langle w_I\cup f^*(x),[M]_2\rangle$ and $\langle p_J\cup f^*(x),[M]\rangle$ are zero, where $x$ ranges over the cohomology of $N$ with appropriate coefficients, and the $w_I$ and $p_J$ are monomials in the Stiefel-Whitney and Pontrjagin classes of $M$.