Euler Charateristic of a surface of genus $g$ Using the definition of Euler charateristic from the theory of intersection numbers that is done in Hirsch's Differential Topology , I am trying to see that $\chi(G)=2-2g$, where $G$ is a closed surface of genus
$g$. Now my idea for this was to go by induction on $g$, and the case where $g=0$ it's true since we have that $\chi(S^2)=2$. Now I am having some trouble it the induction hypothesis. I am trying to use the following result

Suppose a compact $n$-manifold can be expressed by $A\cup B$ where $A$ and $B$ are compact $n-$dimensional submanfiolds and $A\cap B$ is a $(n-1)$-dimensional submanifold, then $\chi(A\cup B)=\chi(A)+\chi(B)-\chi(A\cap B)$.

I have tried using this but I can't seem to find what to use for the sets $A$ and $B$. Any help with this is aprecciated. Thanks in advance.
 A: We will first recall facts Hirch proves in the chapter.

$$ \chi\left(S^{n}\right)=1+(-1)^{n}=\left\{\begin{array}{ll} 2 &
\text { if }  n \text { is even } \\ 0 & \text { if }  n \text {
is odd. } \end{array}\right. $$
Let $M$ and $N$ be compact oriented manifolds without boundaries.
Then $\chi(M \times N)=\chi(M) \chi(N)$

So we can compute
$$\chi (S^1)=0$$
$$\chi(S^2)=2$$
$$\chi(T)=\chi(S^1 \times S^1)=0$$
We also need to compute $\chi(D)$. We have that $S^2=D_S \cup D_N$ where $D_N$ is the northern hemisphere and $D_S$ is the southern hemisphere. Thus by your formula we have
$$2=\chi(S^2)=\chi(D_S)+\chi(D_N)-\chi(S^1)=2\chi(D),$$
hence $\chi(D)=1$.
Let $S_g$ be the surface of genus $g$. We have $S_g=D\cup S_g\backslash \mathring{D}$ so $$\chi(S_g)=\chi(D)+\chi(S_g\backslash \mathring{D})+\chi(S^1).$$
The equation above implies that $\chi(S_g\backslash \mathring{D})=\chi(S_g)-1$.
Those computations set the stage for an induction arguments. We write $S_g=S_{g-1}\backslash \mathring{D}\cup T \backslash \mathring{D}$, plugging this to the formula we get
$$\chi(S_g)=\chi(S_{g-1}\backslash \mathring{D})+\chi(T \backslash \mathring{D})-\chi(S^1).$$
Useing the induction hypothisis we get
$$\chi(S_g)=(2-2(g-1)-1)+(-1)-0=2-2g+2-2=2-2g$$.
Note that we have allready shown the base case $T=S_1$ so this completes the proof.
