Minimum number of closed simple curves that separates a surface I know from Jordan Curve Theorem that every simple closed curve separates the plane. For the torus, one curve isn't enough. Not even two of such curves necessarily separates this surface. What I would like to know is what (and why) are the minimum number of closed simple curves necessary to separate a torus, and more generally any other surface.
 A: Let $S$ be a surface. I will define two topological invariants of $S$. In both, $C=C(S)$ stands for the set of Jordan curves in $S$.
(a).
$$M(S)=\inf \{n: \forall \hbox{ distinct}\ c_1,.., c_k\in C, S\setminus \bigcup_i c_i \ \hbox{is disconnected}\}$$
(b). $$m(S)= \sup \{k: \exists \hbox{ distinct}\  c_1,..., c_n\in C,  S\setminus \bigcup_i c_i\ \hbox{is connected}\}$$
Clearly, $M(S)=m(S)+1$. You are interested in computing $M(S)$. I will be computing $m(S)$ instead, but only in the case $S=T^2$ since the general case looks like a difficult combinatorial problem. I will describe steps in computation but skip some of the more unpleasant details.

*

*Suppose that $S$ is a compact surface, $c_1,...,c_n\in C$ are such that $S\setminus \bigcup_i c_i$ is connected. Then the above union of curves $c_i$ is a finite graph embedded in $S$. This is the most unpleasant part of the proof, the key is to show that for all $i, j$, the intersection $c_i\cap c_j$ is a finite union of arcs and points in $c_i$.


*If $G$ is a finite graph in a compact surface $S$, then $S$ can be triangulated so that $G$ is a subgraph of the 1-dimensional skeleton of the triangulation. This step is essentially contained in Epstein's paper
Epstein, D. B. A., Curves von 2-manifolds and isotopies, Acta Math. 115, 83-107 (1966). ZBL0136.44605.


*In this step we assume that $S=T^2$, the 2-dimensional torus.  Suppose that $c_1,...,c_n\in C$ are distinct, such that $S\setminus \bigcup_i c_i$ is connected and $n>1$. Then $S\setminus \bigcup_i c_i$ is simply-connected, i.e. homeomorphic to the plane.

Indeed, since $S=T^2$, if $S\setminus \bigcup_i c_i$ is connected, then either $n=0$ or the above complement is homeomorphic to the (open) annulus and, hence, $n=1$.
Remark. Note that if $n=1$, then $n\ne m(S)$ since one can add to $c_1$ another Jordan curve $c_2$ which intersects $c_1$ in a single point. Then $c_1\cup c_2$ still does not disconnect $S$.


*Now, we are back to the case of a general compact surface $S$. Suppose that $c_1,...,c_n\in C$ are such that $S\setminus \bigcup_i c_i$ is simply-connected. Then $S$ can be obtained from a planar polygon $P$ by identifying boundary segments in pairs so that the projection of the boundary of $P$ to $S$ is the graph $G= \bigcup_i c_i$.


*Let $G\subset S$ be a graph as in (4). Without loss of generality we may assume that each vertex of $G$ has valence $\ne 2$ (otherwise we simply erase such a vertex). Moreover, since $c_1,...,c_n$ are Jordan curves, no vertex of $G$ can have valence $1$. Hence, each vertex has valence $\ge 3$.


*Suppose that $G\subset S$ is a graph where each vertex has valence $\ge 3$ and $S\setminus G$ is simply-connected. Let $V$ and $E$ denote the number of vertices and edges in $E$ respectively. We have:
$$
\chi(S)= V- E +1
$$
and, since each vertex has valence $\ge 3$,
$$
\frac{2}{3}E\ge V. 
$$


*Now, assume again that $S=T^2$. Then $\chi(S)=0$, $V=E-1$. By combining this equation and the inequality from (6), we obtain  $E\le 3$. Clearly, $2\le E$, thus, either $E=2$ or $E=3$. The former case is realized by the graph $G$ which is the projection of the boundary of the square to the torus under the standard identification. (Here $V=1$.) This graph contains exactly two distinct Jordan curves. However, the case $E=3$ is also realized! Specifically, take a regular planar hexagon $P$ and identify its opposite sides by translations. The resulting space is $T^2$ and the projection of the boundary of $P$ to $T^2$ is the $\theta$-graph, i.e. the graph with two vertices and three edges (the picture of the letter $\theta$). Observe that the $\theta$-graph  contains three distinct Jordan curves. Thus, we constructed distinct $c_1, c_2, c_3\in C$ such that
$C\setminus (c_1\cup c_2\cup c_3)$ is connected. Hence, $m(T^2)\ge 3$. At the same time, we also proved that $m(T^2)\le 3$.
Conclusion: $m(T^2)=3$, $M(T^2)=4$.
