Realizing a homology by a smooth immersion An alternative title is:  When can I homotope a continuous map to a smooth immersion?
I have a simple topology problem but it's outside my area of expertise and I worry may be rather subtle.  Any help would be appreciated.
The set-up is the following:
Let $M$ be some (closed say) $n$ dimensional manifold and suppose that $\Sigma_1$ and $\Sigma_2$ are two closed submanifolds of $M$ of dimension $k$.  Note that $\Sigma_1$ and $\Sigma_2$ are allowed to intersect (in my situation they are also embedded but I don't believe this is effects anything).  Suppose in addition that $\Sigma_1$ and $\Sigma_2$ are homologous.  If $k\leq n-2$ I would like a compact manifold with boundary $\Gamma$ with $\partial \Gamma=\gamma_1\cup \gamma_2$ and a smooth immersion $F:\Gamma\to M$ so that $F(\gamma_i)=\Sigma_i$.  In other, words the homology between $\Sigma_1$ and $\Sigma_2$ can be realized by a smooth immersion.  
I believe by approximation arguments one can always get a smooth such $F$ without restriction on $k$ but it need not be an immersion (especially if $k=n-1$).  My gut is that when you have $k\leq n-2$ since the dimension of the image of $F$ is codimension one you have enough room to perturb it to be an immersion.  That is that $F$ is homotopic rel boundary to our desired immersion.
Unfortunately, I don't know enough to formalize this and all my intuition comes from considering curves and domains in $\mathbb{R}^3$ so I'm afraid there may be obstructions in general.
References would be greatly appreciated.
Thanks!
Edit:
As suspected, the question is somewhat subtle .  To make it tractable lets assume that $M$ is a $C^\infty$ domain in $\mathbb{R}^3$ (so is a fairly simple three-manifold with boundary) and that the $\Sigma_i$ are curves.  This is where my intuition says that there should be such a smooth immersion.
 A: Reading between the lines, I suppose you have oriented manifolds $\Sigma_i$, so that each of them determines an element of the integral homology group $H_k(M)$. In fact the assumption that these two classes are equal does not even imply that there is a compact oriented $(k+1)$-manifold $\Gamma$ with boundary such that the boundary is $\Sigma_1\coprod \Sigma_2$. For example, $\Sigma_1$ could be $\mathbb CP^2$, $\Sigma_2$ could be $S^4$, and they could both be embedded in $S^n$ for $n>>0$. The boundary of a compact oriented $5$-manifold always has signature zero.
Maybe modify the question, so that insteading of assuming homologous you assume cobordant, meaning that you assume there is a compact oriented $\Gamma_0$ mapped into $M$ in such a way that $\partial \Gamma_0\to M$ is an immersion, and you ask whether there exists another thing $\Gamma\to M$ with $\partial\Gamma\to M$ isomorphic to $\partial \Gamma_0\to M$ but this time with $\Gamma\to M$ an immersion.
A: There is a general strategy for these kind of problems, which sometimes helps (the ''h-principle''): separate the homotopical and smooth aspects of the problem. Setup: $f:N \to M$ a map of smooth manifolds, $dim (N) < dim (M)$, $f|_{\partial N}$ is an immersion. 
Step 1: if your $f$ is going to be homotopic to an immersion $g$, then there should exist a bundle monomorphism $\phi:TN \to f^* TM$, extending the derivative of $f|_{\partial}$ to all of $N$. Otherwise, the hypothetical immersion won't exist, because there is no candidate for its derivative. Constructing this monomorphism is a purely bundle theoretic problem. Whether this can be successfully attacked depends very much on the concrete situation -- maybe it is trivial, maybe it is very hard.
In any case -- if there is no candidate for an immersion in sight -- it is be easier to construct a bundle map than an immersion.
Step 2: if you succeed in solving the homotopy problem, you are done! This is the content of the Smale-Hirsch-Theorem: once $\phi$ exists, $f$ will be homotopic to an immersion $g$ and the derivative is homotopic (as a bundle mono) to $\phi$. There is a relative statement as well; you have to extend the immersion $f_{\partial M}$ to an immersion of a collar of $\partial M$ in $M$. This is not trivial since the collar has one dimension 
more. If $dim(N)=dim(M)$, then the result still holds (in your situation, where $N$ is a connected cobordism with nonempty boundary), but you cannot fix $f$ on the boundary.
Your intuition in the low-dimensional case seems to be true: if $N$ is a surface and $M$ a domain in $R^3$, then both are parallelizable.
References: Eliashberg-Michachev: Introduction to the h-principle, Haefliger: Lectures on the theorem of Gromov, Ponaru: Homotopy theory and differentiable singularities. Adachi: Immersions and embeddings.
A: The answer is known in the special case you mentioned (curves in a 3-manifold)
Actually it is true more generally: Namely
if k = n-2 and your k-dimensional oriented submanifolds $\Sigma_1$ and $\Sigma_2$ represent the same INTEGER homology class in the ambient ORIENTED n-manifold M, and they are disjoint and embedded, 
then there is an embedded k+1 dimensional submanifold in the cylinder $M \times I$ with boundary  $\Sigma_1$ in $M\times {0}$ and $\Sigma_2$ in $M \times {1}.$
This a trivial consequence of the fact, that
$K(Z,2) = MSO(2).$
The homotopy classes [M, K(Z,2)] give the 2-dimensional cohomology group, hence also the k-th homology group of M.
The homotopy classes [M, MSO(2)] give the codimension 2 embedded submanifolds up to an embedded cobordism (by the Thom construction). QED.
In general the answer is negative, as it was mentioned by others.
