Let $M$ be a connected discrete surface. From the last post we know

$\textrm{dim Im }d_1=\#F-1$

and we already knew

$\textrm{dim Ker }d_0=1$.

From

$\textrm{Im }d_0 \subset \textrm{Ker }d_1$

we see that the first Betti number

$\beta_1(M) := \textrm{dim Ker }d_1 -\textrm{dim Im }d_0$

is non-negative. Let us apply the dimension formula for linear maps to $d_0$ and $d_1$:

$\beta_1=\#\tilde{E}-\textrm{dim Im }d_1-(\#V-\textrm{dim Ker }d_0)=2+\#\tilde{E}-\#F-\#V$.

Thus the so-called Euler characteristic

$\chi(M):=\#V – \#\tilde{E} + \#F$

satisfies

$\chi(M) \leq 2$.

Theorem: $\chi(M)$ is an even number.

Proof: Since the permutation $\rho$ of $E$ consists of $\#\tilde{E}$ transpositions, its parity is

$\textrm{sgn } \rho =(-1)^{\#\tilde{E}}$.

The parity of a cyclic permutation

$\sigma: \{1, \ldots ,n\} \rightarrow \{1, \ldots ,n\}$

$\sigma(k) = k+1 \quad \textrm{mod }n$

is given by

$\textrm{sgn } \sigma =(-1)^{n-1}$.

From this we see

$\textrm{sgn } s=(-1)^{\#E+\#F}=(-1)^{\#F}$.

Similarly,

$(-1)^{\#F}(-1)^{\#\tilde{E}}= (\textrm{sgn } s)(\textrm{sgn } \rho)=\textrm{sgn } s\circ \rho=(-1)^{\#V}$.

$\square$

Thus also $\beta_1(M)$ is even and we can write

$\beta_1(M) = 2g$

for a natural number $g$ called the genus of $M$. The genus is related to the Euler characteristic by

$\chi= 2-2g$.