Realizing homology classes on surfaces with boundary by simple closed curves Let $\Sigma$ be a compact oriented surface with boundary.  Assume that the genus of $\Sigma$ is positive.  We say that an element $h \in H_1(\Sigma)$ can be realized by a simple closed curve if there exists an oriented simple closed curve $\gamma$ on $\Sigma$ such that $[\gamma] = h$.
If $\Sigma$ has $0$ or $1$ boundary components, then $h \in H_1(\Sigma)$ can be realized by a simple closed curve if and only if $h$ is primitive, that is, if we cannot write $h = n \cdot h'$ for some $n \in \mathbb{Z}$ and $h' \in H_1(\Sigma)$ with $n > 1$.  This is a standard fact; for instance, it is contained in Farb and Margalit's Primer on Mapping Class Groups.
This brings me to my question : if $\Sigma$ has more than $1$ boundary component, then what elements of $H_1(\Sigma)$ can be realized by simple closed curves?
One might guess that the answer is still the primitive elements.  However, this guess is wrong.  Indeed, assume that $\Sigma$ has at least $2$ boundary components.  Let $\delta$ be an oriented simple closed nonseparating curve in the interior of $\Sigma$ and let $b$ be one of the boundary components of $\Sigma$.  Observe that $[b] \neq 0$, and hence that $2[\delta]+[b]$ is a primitive element of $H_1(\Sigma)$.  Assume that $\gamma$ is an oriented simple closed curve in $\Sigma$ such that $[\gamma] = 2[\delta]+[b]$.  Let $S$ be the surface obtained by gluing discs to all the boundary components of $\Sigma$.  There is then an inclusion map $i : \Sigma \hookrightarrow S$, and we have
$$[i(\gamma)] = 2[i(\delta)] + [i(b)] = 2[i(\delta)],$$
a contradiction.
 A: The case with boundary quickly reduces to the case without boundary.  The idea is to cap off your surface with discs to create a closed surface.  The map on homology is not injective.  But a homology class is realizable whenever it is realizable in the capped-off surface.  So you get a fairly simple answer:
A homology class in $\Sigma$ is realizable if and only if either 1) it is realizable in the capped-off surface or 2) it is a sum of boundary classes (coherently oriented) together with perhaps a realizable class in the capped-off surface. 
A: This is related to a nontrivial question, address in this paper of Chas and Krongold (there are other related papers of Moira Chas with Fabiana Krongold and Dennis Sullivan, which a google search will bring up).
The original question, however, is trivial, since if we take some curve (think of it as a hyperbolic geodesic) realizing the homology class, we can perform a surgery on each crossing, which removes it, and possibly disconnects the curve, so eventually we will have a multicurve realizing the homology class. Some components of this multicurve will be boundary-parallel. from this multicurve it is pretty easy to see when the class is realizable (unless I am confused, which is quite possible, you need to be realizable in the cupped-off surface, plus something that is not a multiple of a boundary component).
EDIT Firstly, the OP is apparently trying to win friends and influence people for downvoting my answer and Ryan's. Not cool at all.
Secondly, if you want a different answer, knock yourself out (notice that he is solving a more general, thus harder, problem):
MR2335737 (2008e:57015) Reviewed 
Granda, Larry M.(1-STL)
Representing homology classes of a surface by disjoint curves. (English summary) 
Houston J. Math. 33 (2007), no. 3, 807–813. 
57M50 (57M20 57N05) 
A more extensive discussion of the same problem solved in Granda's paper (without, however, a complete answer) is given by Allen Edmonds in:
Edmonds, Allan L.(1-IN)
Systems of curves on a closed orientable surface. 
Enseign. Math. (2) 42 (1996), no. 3-4, 311–339. 
Another edit
The best reference is W. Meeks and J. Patrusky, where Theorem 1 is:
enter link description here
For the link-challenged, it says that a class can be represented by simple closed curve if and only if it is primitive in the capped-off surface OR it is a sum of (some of the) boundary curves, which is what Ryan and I have been saying.
