Let $S$ be compact oriented surface without boundary. Then it is a classical result that a primitive class $\gamma \in H\_1(S; \mathbb{Z})$ is always represented by a simple closed curve. It implies that any class $\beta \in H\_1(S; \mathbb{Z})$ is represented by a disjoint union of simple closed curves (take $\beta = k \gamma$ with $\gamma$ primitive and consider $k$ parallel simple closed curves representing $\gamma$).


Let now $\Sigma$ be a compact oriented surface with non-empty boundary:


Is it true that I can always represent any element $\gamma \in H\_1(\Sigma, \partial \Sigma; \mathbb{Z})$ by a disjoint union of simple closed curves and properly embedded arcs?



Representing relative homology classes orientable surfaces with boundary
Yes, this can be done. You can do this directly for surfaces but it's as much a "codimension one" as a "dimension one" phenomenon and so useful to see the general argument.


For any compact oriented n-manifold with boundary, duality says that $H\_{n-1}(M,\partial M) \cong H^1(M)$. From simple obstruction theory, $H^1(M)$ is identified with homotopy classes of maps to a circle. Assuming your manifold was smooth, take a smooth map $f$ representing your cohomology class and a regular value $p\in S^1$. (I guess I mean regular value for both $f$ and the restriction of $f$ to the boundary.) Then $f^{-1}(p)$ is a codimension one submanifold in the given homology class.


For surfaces, the submanifold is a union of properly embedded arcs and simple closed curves. If you want to do so, you can assume that every component is an arc.


