I am reading the John Lee's Introduction to Riemannian manifolds, proof of the Theorem 9.7 and stuck at understanding some calculation.
Theorem 9.7 ( The Gauss-Bonnet Theorem ). If $(M,g)$ is a smoothly triangulated compact Riemannian 2-manifold, then $ \int_{M}KdA =2\pi \chi(M) $, where $K$ is the Gaussian curvature of $g$ and $dA$ is its Riemannian density.
( C.f. As the definition of a smooth triangulation of $M$ and curved triangle-more generally a curved polygon-, refer to his book p.276, p. 271 : 
)
Proof. We may as well assume that $M$ is connected, because if not we can prove the theorem for each connected component and add up the results ( Can anyone explain this more in detail ? ).
First consider the case in which $M$ is orientable. In this case, we can choose an orientation for $M$, and then $\int_MKdA$ gives the same result whether we interpret $dA$ as the Riemannian density or as the Riemannian volume form, so we will use the latter interpretation for the proof. Let $\{\Omega_i : i= 1, \dots, F \}$ denote the faces of the triangulation, and for each $i$, let $\{\gamma_{ij} : j= 1,2,3 \}$ be the edges of $\Omega_i$ and $\{\theta_{ij} : j=1,2,3 \}$ its interior angles. Since each exterior angle is $\pi$ minus the corresponding interior angle, applying the Gauss-Bonnet formula to each triangle and summing over $i$ gives
$$ \Sigma_{i=1}^{F} \int_{\Omega_i}KdA + \Sigma_{i=1}^{F}\Sigma_{j=1}^{3}\int_{\gamma_{ij}}\kappa_Nds + \Sigma_{i=1}^{F}\Sigma_{j=1}^{3} (\pi- \theta_{ij}) = \Sigma_{i=1}^{F}2\pi$$
Note that each edge integral appears exactly twice in the above sum, with opposite orientation, so the integrals of $\kappa_N$ all cancel out. Thus the above formula becomes
$$ \int_{M}KdA + 3\pi F - \Sigma_{i=1}^{F}\Sigma_{j=1}^{3}\theta_{ij} =2\pi F$$
Note also that .. (omitted).
Q. My question is, I don't know why $\Sigma_{i=1}^{F} \int_{\Omega_i}KdA = \int_{M} KdA$ is true, altough intuitively it seems to be true. I'm trying to use next theorem ( John Lee's smooth manifolds, Proposition 16.8 ) and can't make formal proof until now.
Proposition 16.8 (Integration Over Parametrizations). Let $M$ be an oriented smooth $n$-manifold with or without boundary, and let $w$ be a compactly supported $n$-form on $M$. Suppose $D_1, \dots, D_k$ are open domains of integration in $\mathbb{R}^{n}$, and for $i=1, \dots , k$, we are given smooth maps $F_i : \overline{D_i} \to M$ satisfying (i) $F_i$ restricts to an orientation-preserving diffeomorphism form $D_i$ onto an open subset $W_i \subseteq M$ ; (ii) $W_i \cap W_j = \varnothing$ when $i \neq j$; (iii) $\operatorname{supp} w \subseteq \bar{W_i} \cup \cdots \cup \bar{W_k}$. Then $\int_{M} w = \Sigma_{i=1}^{k} \int_{D_i}F_i^{*}w$ .
First trial of usage of the Proposition 16.8. : Since each $\Omega_i$ is face of a curved triangle, there is an oriented smooth coordinate disk $(U_i, \phi_i )$ containing $\overline{\Omega_i}$ under whose image the edges form a curved triangle in the plane ( c.f. Fig 9.12, above image ). Let $w := KdA$. Define for each $i$ that $D_i := \phi_i(\Omega_i)$. Then each $D_i$ is a open domain of integration in $\mathbb{R}^2$ ; (c.f. Lee's smooth manifolds, p.402 ) i.e., a bounded subset whose boundary has measure zero. Let $\iota_i : \overline{\Omega_i} \hookrightarrow M$. Note that $\phi_i(\overline{\Omega_i})= \overline{\phi_i(\Omega_i)} = \overline{D_i}$ since $\phi_i$ is homeomorphism. Define $F_i : \overline{D_i} \to M$ by $F_i := (\iota_i ) \circ ( \phi_i |_{\overline{\Omega_i}}^{\overline{\phi_i(\Omega_i)}} )^{-1}$. Then note that
(i) $F_i$ restricts to an orientation-preserving diffeomorphism from $D_i$ to an open subset $W_i:=\Omega_i \subseteq M$.
(ii) $W_i \cap W_j = \Omega_i \cap \Omega_j = \varnothing $ when $i \neq j$.
(iii) $\operatorname{supp}w \subseteq \overline{W_1} \cup \cdots \cup \overline{W_F} = \overline{\Omega_1} \cup \cdots \cup \overline{\Omega_F} = M$.
So, by the above Proposition 16.8. , we have
$$ \int_{M}w = \Sigma_{i=1}^{F} \int_{D_i}F_i^{*}w $$
So if we can show that $\int_{D_i}F_i^{*}w = \int_{\Omega_i} w$, then we are done and this is a point that I stuck now.
Here, I think that for $\Sigma_{i=1}^{F} \int_{D_i}F_i^{*}w $, more correct notation is $\Sigma_{i=1}^{F} \int_{D_i} (F_i|_{D_i})^{*}w $ since the domain of $F_i$ is $\overline{D_i}$, not $D_i$. Note that $F_i|_{D_i} = j_i \circ (\phi_{i} |_{\Omega_i}^{\phi(\Omega_i)})^{-1} : D_i := \phi_i(\Omega_i) \to \Omega_i \subseteq M$ where $j_i : \Omega_i \hookrightarrow M$ is the inclusion. ( True? )
Note that since $\operatorname{supp}(w|_{\Omega_i}) \subseteq \overline{\Omega_i} \subseteq (U_i , \phi_i)$, by the definition of integral over manifold ( c.f. Lee's smooth manifold book p.404 )
$$\int_{\Omega_i}w = \int_{\phi_i(U_i)} (\phi_i^{-1})^{*}w $$
On the other hand,
$$ \int_{D_i} (F_i |_{D_i})^{*}w = \int_{\phi_i(\Omega_i)}(j_i \circ (\phi_{i} |_{\Omega_i}^{\phi(\Omega_i)})^{-1})^{*}w = \int_{\phi_i(\Omega_i)}((\phi_{i} |_{\Omega_i}^{\phi(\Omega_i)})^{-1})^{*} j_i^{*} w = \int_{\phi_i(\Omega_i)}((\phi_{i} |_{\Omega_i}^{\phi(\Omega_i)})^{-1})^{*} w|_{\Omega_i} $$
Q. Are these two really the same? ( Or not? ) Why? Perhaps, is there a point that I made mistake? Could there be something that I've overlooked?
Or..what theorem can be used to prove that? Can anyone helps?