Why do $2g+1$ distinct closed curves separate a compact orientable surface of genus $g$?

Question

The genus of a surface is the maximum number of pairwise disjoint simple closed curves that do not separate a surface. Why do $2g+1$ distinct closed curves separate a compact orientable surface of genus $g$?

Answer 1

Let me index the given collection of simple closed curves as $C_i$, $i=1,...,2g+1$. I'll give a solution under an additional hypothesis, namely that $C_i$ and $C_j$ intersect transversely if $i \ne j$; I believe there is an additional topological argument that would reduce the general case to this special case, but I'll ignore that in this answer.

CORRECTION: Some kind of additional assumption is actually necessary; see the counterexample by @MoisheKohan in the comments.

From transversality, it follows that there exists points $x_i \in C_i$ such that $x_i$ is not contained in any $C_j$ with $j \ne i$ (this is the only way that I will use the transversality hypothesis). Using this, we can construct a closed curve $\alpha_i$ that crosses $C_i$ once transversely at the point $x_i$, and does not cross any $C_j$ with $j \ne i$. First take a short path $\alpha'_i$ that starts at a point $y_i$ near $x_i$ on one side of $C_i$, then crosses $C_i$ transversely at $x_i$ to end at a point $z_i$ on the other side of $C_i$ still near $x_i$. But we know there is a path $\alpha''_i$ from $z_i$ to $y_i$ that misses $C_i$ and all of the other curves $C_j$. Now set $\alpha_i = \alpha'_i * \alpha''_i$.

For each $i$ we can can choose an orientation of $C_i$ and $\alpha_i$ so that we have an intersection number equation $$\langle C_i, \alpha_i \rangle = +1 $$ But $C_i$ and $\alpha_j$ are disjoint if $i \ne j$ and so it follows that $$\langle C_i,\alpha_j \rangle = 0 \quad \text{if $i \ne j$} $$ Now comes the algebraic topology: for each $i$, intersection number with $C_i$ defines a well-defined homology homomorphism $I_i : H_1(S,\mathbb R) \to \mathbb R$. From the two equations above it follows that the homomorphisms $I_1,...,I_{2g+1}$ are linearly independent elements of the dual vector space $(H_1(S,\mathbb R))^*$, and so the dimension of $(H_1(S,\mathbb R))^*$ is $\ge 2g+1$. It follows that the dimension of $H_1(S,\mathbb R)$ is $\ge 2g+1$, contradicting that the dimension actually equals $2g$.

Answer 2

OK, here is an example. Start with the cylinder $C=S^1\times [0,1]$. Pick two distinct points $p, q\in S^1$. These points yield points $x=(p,0), y=(q,0)\in S^1\times \{0\}\subset \partial C$. The points $x, y$ divide $S^1\times \{0\}$ in two arcs $\alpha_1, \alpha_2$ with the end-points $x, y$. Next, pick a simple arc $\alpha_3\subset C$ connecting the points $x=(p,0)$, $y'=(q,1)$ such that $\alpha_3\cap \partial C=\{x, y'\}$. The union $\partial C\cup \alpha_3$ has connected complement in $C$. The torus $T^2$ is obtained from $C$ by identifying boundary points of $C$ via the homeomorphism $(z,0)\mapsto (z,1)$, $z\in S^1$. Let $\pi: C\to T^2$ denote the quotient map. Set $a_i:= \pi(\alpha_i), i=1, 2, 3$. Then each $a_i$ is a simple arc in $T^2$. These three arcs share common end-points $\pi(x), \pi(y)$ and are disjoint otherwise. The union $$ G=a_1\cup a_2\cup a_3 $$ is the $\theta$-graph in $T^2$. Its complement in $T^2$ is homeomorphic to $C\setminus (\partial C\cup \alpha_3)$ and, hence, is connected. For $1\le i\ne j\le 3$ the union $a_i\cup a_j$ is a simple loop $c_{ij}$ in $T^2$. By the construction, each of these three loops is contained in the union of the other two. Thus, we obtained three distinct simple loops $c_{12}, c_{23}, c_{31}\subset T^2$ whose union has connected complement in $T^2$.