Surface filling simple closed curves

Question

At least how many simple closed curves do we need to fill a surface?

Definition: Let $A$ be set of simple closed curves. $A$ is filling set if all other curves on the surface that are not parallel to boundary components or does not bound a disc must intersect the curves in A that fill the surface.

Answer 1

The answer is that you can fill any surface $S$ with two simple closed curves (assuming $S$ is compact and connected).

For the proof, take two CW decompositions of $S$ each having a single 2-cell, but having disjoint 0-skeleta, and whose $1$-skeleta $X_1,X_2$ have the property that $$X_1 \cap X_2 - \partial S$$ is a set of transverse intersections points in the interior of $S$. Choose $\chi_i : D^2 \to S$ to be a characteristic map for the unique 2-cell of the $i^{\text{th}}$ CW decomposition, so $\chi_i(\partial D^2) \subset X_i$ and $\chi_i(\text{int}(D^2)) = S - X_i$.

Let $N_1,N_2$ be very small regular neighborhoods of $X_1,X_2$ respectively. The boundaries of $N_1,N_2$ are simple closed curves $C_1,C_2$: we may take $C_i$ to be equal to the image under $\chi_i$ of a circle in $D^2$ which is concentric to and very close to the boundary circle $\partial D^2$.

The circles $C_1,C_2$ fill the surface. To see why, consider a simple closed curve $\gamma$ in $S$ which is disjoint from $C_1$ and $C_2$. The curve $\gamma$ cannot be contained in the open disc $S - N_i$ because then $\gamma$ itself bounds an open disc. Thus $\gamma \subset N_1 \cap N_2$. Also $\gamma$ cannot be contained in a component of $N_1 \cap N_2$ which contains an interior vertex of either $X_1$ or $X_2$, because the interior vertices of $X_1$ are contained in $S-N_2$ and vice versa. Thus, $\gamma$ must be contained in a component of $N_1 \cap N_2$ which contains a boundary component and no interior vertex. Any such component of $N_1 \cap N_2$ is an annulus sharing a boundary circle with a component of $\partial S$, and so $\gamma$ is either parallel to that component of $\partial S$ or $\gamma$ bounds a disc.

I will add, as a comment, that the answer would be almost the same if one intended to ask a stronger question about how many simple closed curves it would take to fill a surface assuming none of those curves bounds a disc nor is parallel to a boundary component. But the proof would be different and somewhat more difficult, and anyway that wasn't the question.

Answer 2

You can fill with any finite number of touching closed loops or arbitrary number of polygons with/without of geodesic boundaries. As a minimum bifurcation we can take two parts with a single common continuous boundary apart from the originally given surface boundary. Topologically when going around the neighboring loops, two loops should have same curvatures, each loop should also satisfy the Gauß-Bonnet theorem involving geodesic and integral curvatures.