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.

heartof the question is which $b$ can occur, and Ryan's argument says nothing about this. $\endgroup$ – Roger Jan 2 '14 at 20:25