Representing relative homology classes orientable surfaces with boundary

Question

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?

Answer 1

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.