TITLE: Realizing a homology by a smooth immersion
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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.