Can homologous submanifolds be connected by an immersed manifold with boundary? Supposed I have an n-dimensional manifold M with a k-dimensional submanifold that is homologous to zero (or, equivalently, two homologous submanifolds). Can I always construct a k+1-dimensional manifold N and a smooth map $N\to M$ so that the boundary maps diffeomorphically to my submanifold? Can I just take abstract k+1-simplecies and glue them along boundaries to make N, and then somehow smooth it out? If not, is there some understandable obstruction?
I'm most interested in the smooth category, but if it makes more sense in some other category (or there is otherwise a better question I should've asked), do tell me. 
Update: As I first asked it, the question was a bit stupid because I forgot about cobordisms. However, in the case I care about, this does not seem to be a problem, since I want the boundary of N to be a union of two submanifolds which are diffeomorphic to each other. 
 A: Presumably, the answer is NO, since every manifold $K$ embeds in $\mathbb{S}^n$ of high dimension (where it is then null-homologous), but not every manifold is null-cobordant, in other words, there are $K$ such that there is no $N$ such that $\partial N = K.$
A: As Igor has noted, an obvious obstruction is that your embedded submanifold be null-cobordant.
It seems that you are really asking about the kernel of the realization map $MO_k(M)\to H_k(M;\mathbb{Z}_2)$ (assuming that you don't care about orientations). This appears as the edge homomorphism in the unoriented bordism spectral sequence, which collapses since every homology class is Steenrod realizable (as follows from the work of Thom). A basic reference for this fact is the book "Differentiable periodic maps" by Conner and Floyd. It follows that there is a module isomorphism $H_*(M; \mathbb{Z}_2)\otimes MO_*\cong MO_*(M)$. The bordism class of a map $f\colon A\to M$ is determined by its Stiefel-Whitney numbers (see Section 17 of Conner and Floyd). From this you should be able to piece together what you need.
More detail: Let $f\colon A^k\hookrightarrow M$ be the embedding of your submanifold. Every cohomology class $x\in H^\ell(M;\mathbb{Z}_2)$ and multi-index $(i_1,\ldots,i_r)$ with $i_1+\cdots + i_r=k-\ell$ gives a Stiefel-Whitney number of the map $f$, defined by 
$$\langle w_{i_1}(A)\cdots w_{i_r}(A)f^*(x),[A]\rangle\in\mathbb{Z}_2.$$
 Your map is null-bordant if and only if these are all zero. (Note when $x$ is the unit class we get the S-W numbers of $A$. Also the multi-index $(0)$ gives trivial numbers by your assumption that $f_*[A]=0$.)
A: First of all, it doesn't matter whether or not the map is smooth.  If you find any continuous map, then it will have a smooth approximation.
The other answers so far explain that the cobordism group gives you an obstruction to improving a singular-simplicial chain into a mapped-in manifold.  In fact, it is easy to see that this is basically the only obstruction, and that null corbodism directly gives you a way to improve the simplicial chain.  I'll work with integer coefficients rather than over $\mathbb{Z}/2$ so that things survive a little longer.  Say that you have this $(k+1)$-dimensional cobounding chain.  You can manifold-ize a $k$-dimensional face of the chain because various $k$-dimensional sheets meet the face with opposite sign and you can pair them.  This is basically using the fact that the reduced 0-corbodism group is trivial.  Then turn to the $(k-1)$-faces.  Because of what you did to the $k$-faces, the sheets meet the $(k-1)$-faces in a collection of circles.  But circles are null cobordant, so you can smooth them.  You can continue in this way using the fact that oriented surfaces and 3-manifold are all null-cobordant.  But when you try to improve a $(k-3)$-face, the link of the an incoming sheet can be a 4-manifold that is not null-cobordant, like $\mathbb{C}P^2$.  Then you're stuck.
The obstruction is fundamental because the original null-homologous $k$-cycle could have been an embedded $\mathbb{C}P^2$, and the original cobounding chain could have been a cone over it.
