Why is the genus (in terms of Euler characteristic) the maximum number of disj. closed curves that can be embedded in a surface without separating it? Let $S$ be a compact surface and $g$ its genus defined in terms of Euler characteristic.
Why is $g$ the maximum number of disjoint simple closed curves that can be embedded in the surface without disconnecting it?
I would be grateful if anyone could tell me how to prove it or give me a reference. Thanks.
 A: Welcome to MSE. I want to assume some familiarity about pants decompositions (see the references linked here). For simplicity, I'll assume that $S$ is orientable without boundary, but I think you can modify the argument to adapt to boundary components.

A pants decomposition for a connected genus $g$ surface $S$ consists of $2g - 2$ pairs of pants. We'll need to glue at least $(2g - 2) - 1 = 2g - 3$ of these pairs of pants together in order to form a connected surface.
At the outset, there are $3(2g - 2) = 6g - 6$ boundary cuffs which are all pairwise-glued to form $S$. In other words, we performed $(6g-6)/2 = 3g - 3$ gluings total. But we were forced to do at least $(2g - 3)$ of them, meaning that we did
$$(3g - 3) - (2g - 3) = g$$
gluings "just for fun." In other words, there were $g$ gluings that we didn't have to make in order to keep $S$ connected. These curves are the maximal set of disjoint scc's in this alternate characterization of genus.
A: Starting with the orientable case, consider a closed, connected, oriented surface $S$ of genus $g$. The genus is related to its Euler characterstic $\chi(S)$ by the formula $\chi(S)=2-2g$, and this number $g = 1 - \frac{\chi(S)}{2}$ is equal to that maximal number of pairwise disjoint circles in $S$ that do not separate $S$. Here's the proof.
When you cut along a circle, the Euler characteristic does not change. So if you cut $S$ along $k$ circles, and assuming those circles do not disconnect $S$, you will get a compact, connected, orientable surface-with-boundary that I'll denote $A_k$, that has $2k$ boundary circles, and that has Euler characterstic
$$\chi(A_k) = \chi(S)
$$
You can then glue a single disc to each of those $2k$ circles, to get a closed connected surface that I'll denote $B_k$. The same logic (gluing along circles does not change the Euler characteristic), together with the fact that each individual disc has Euler characterstic $1$, gives the result
$$\chi(B_k) = \chi(A_k) + 2k = \chi(S) + 2k
$$
It is known that the maximum Euler characteristic of a connected, closed, orientable surface is the Euler characteristic of the sphere which is equal to $2$, and therefore
$$\chi(S) + 2k \le 2
$$
Substituting $\chi(S)=2-2g$ one obtains $g \le k$.
To show that the maximum $k=g$ is obtained, start with a sphere whose Euler characteristic equals $2$, remove $2g$ discs which reduces the Euler characterstic to $2-2g$, and now glue those circles in pairs to give a closed oriented surface of the same Euler characterstic $2-2g$ as your surface $S$, together with a system of $g$ circles that do not separate the surface. Since the Euler characteristic is a complete topological invariant of closed, connected, oriented surfaces, and knowing that $\chi(S)=2-2g$, this surface is homeomorphic to the original surface $S$.

Now for the non-orientable case. Let $S$ be a closed, connected, nonorientable surface. The terminology here is less standard, so I'll be careful and refer to the nonorientable genus $g$ of $S$, which is related to Euler characterstic by the formula $\chi(S)=2-g$, and is again equal to the maximal number of pairwise disjoint circles in $S$ that do not separate $S$.
The proof is a bit trickier in the nonorientable case, due to the fact that there are two kinds of circles in $S$: two-sided and one-sided (in an orientable surface, every circle is two-sided). So, suppose that we have a pairwise disjoint collection of $k$ circles in $S$ that do not separate $S$. We may partition this collection into $k_1$ one-sided circles and $k_2$ two-sided circles, with $k = k_1 + k_2$. Again let $A$ be the connected surface-with-boundary obtained by cutting along these circles. Again $\chi(A)=\chi(S)$. But in this case the number of boundary circles of $A$ is equal to $k_1 + 2 k_2$. So, gluing in discs produces a closed surface $B$ with $\chi(B)=\chi(A)+k_1 + 2k_2 = \chi(S) + k_1 + 2 k_2$. We therefore have
$$2 - g + k = \chi(S) + k_1 + k_2 \le \chi(S) + k_1 + 2 k_2 = \chi(B) \le 2
$$
and so
$$k \le g
$$
To show that the maximum is achieved, one can get a hint by examining the above chain of inequalities. First, if $k_2 \ge 1$ then we get strict inequality $k < g$. Also, if $B$ is not the sphere then we again get strict inequality $k < g$. So, the only way that the maximum can be achieved is by a collection of $g$ one-sided circles which, when cut along, give a sphere with $g$ discs removed.
So, start from a sphere with $g$ discs removed, having $g$ boundary circles. Identify each boundary circle to itself by the antipodal map. The result is a non-orientable surface of Euler characteristic $2-g$. Since closed, connected, non-orientable surfaces are classified by their Euler characterstic, the resulting surface is homeomorphic to your original surface.

Here's one last comment relating the two concepts of genus. Given a closed, connected, orientable surface $S$, its genus is also equal to the number of toruses ($T^2 = S^1 \times S^1$) needed so that their connected sum is homeomorphic to $S$. And given a closed, connected, nonorientable surface $S$, its nonorientable genus equals the number of projective planes needed so that their connected sum is homeomorphic to $S$. Often these are the descriptions used to define genus and nonorientable genus.
