Transition maps between surface patches Before I formulate my question I want to give our definitions:
Let $M \subseteq \mathbb{R}^3$ be connected. Then $M$ is called a smooth surface, if for every $p \in M$ there exists an open subset $V \subseteq \mathbb{R}^3$ such that $p \in V$ and a Diffeomorphism (smooth) $\phi \colon V \to W$ to another open subset of $\mathbb{R}^3$ such that $\phi(V \cap M)=W \cap \{x_3=0\}$.
Let M be a smooth surface, a smooth map $\sigma \colon U \to \mathbb{R}^3$ is called a surface patch for M, if $U \subseteq \mathbb{R}^2$ open, it is injective, $D\sigma$ is injective and $\sigma(U) \subseteq M$.
Let $\sigma \colon U \to \mathbb{R}^3$, $\tau \colon V \to \mathbb{R}^3$ be two surface patches for M with nonempty $\Omega=\sigma(U) \cap \tau(V)$. The map:
$\sigma^{-1} \circ \tau \colon \tau^{-1}(\Omega) \to \sigma^{-1}(\Omega)$ is called transition map.
Now the lecture notes state that transition maps are diffeomorphisms. The proof says it follows from the definition of a surface. I have many questions:

*

*Is $\tau^{-1}(\Omega)$ an open set of $\mathbb{R}^2$? I dont see why $\Omega$ should be an open set in the image of $\sigma$

*Why would it be a differentiable function?

 A: For each surface patch $\sigma$ define $\bar \sigma : U \stackrel{\sigma}{\to} M$. Moreover let $\bar{\mathbb R}^2 = \{(x_1,x_2,x_3 \in \mathbb R^3 \mid x_3 = 0 \}$ the $x_1$-$x_2$-plane, $p : \mathbb R^3 \to \mathbb R^2, p(x_1,x_2,x_3) = (x_1,x_2)$ the projection and $i : \mathbb R^2 \to \mathbb R^3, i(x_1,x_2) = (x_1,x_2,0)$ be the embedding of $\mathbb R^2$ as the $x_1$-$x_2$-plane. The maps $p, i$ are smooth.

*

*$\bar \sigma$ maps $U$ homeomorphically onto an open subset $U' \subset M$. This proves your first bullet point.

Let $x \in U$. There exist open $V_x,W_x \subset \mathbb R^3$ with $\sigma(x) \in V_x$ and a diffeomorphism $\phi : V_x \to W_x$ such that $\phi(V_x \cap M) = W_x \cap \bar{\mathbb R}^2$. The set $U_x = \sigma^{-1}(V_x)$ is an open neigbhorhood of $x$ in $U$. Consider
$$\gamma : U_x \stackrel{\sigma}{\to} V_x \stackrel{\phi}{\to} W_x \stackrel{p}{\to} \mathbb R^2 .$$
$\gamma$ is smooth with $D_x \gamma$ injective. To check injectivity, note that
$$\gamma^* : U_x \stackrel{\gamma}{\to} \mathbb R^2 \stackrel{i}{\to} \mathbb R^3$$
has image in $W_x$, thus we can consider
$$\bar \gamma :  U_x \stackrel{\gamma^*}{\to}  W_x \stackrel{\phi^{-1}}{\to} V_x .$$
This map agrees with $\sigma \mid_{U_x}$, thus $D_x \gamma^*$ is injective (recall that $\phi$ is a diffeomorphism). This implies that also $D_x \gamma$ must be injective since $D_x \gamma^* = D_{\gamma(x)}i \circ D_x \gamma$. We conclude that $D_x \gamma$ is an isomorphism.
By the inverse function theorem $\gamma$ maps an open neighborhood $U'_x \subset U_x$ of $x$ diffeomorphically onto an open neighborhood $N_x$ of $\gamma(x)$ in $\mathbb R^2$. The set $i(N_x)$ is open in $\bar{\mathbb R}^2$, thus $\phi^{-1}( i(N_x))$ is open in $M$. By construction $\bar \sigma(U'_x) = \phi^{-1}( i(N_x))$ and $\bar \sigma$ establishes a homeomorphism between $U'_x$ and $\bar \sigma(U'_x)$. Note that at the moment it does not make sense to say that $\bar \sigma$ establishes a diffeomorphism  between $U'_x$ and $\bar \sigma(U'_x)$ because that conecpt has not yet been introduced for maps into surfaces.
This implies that $\bar \sigma$ is an open map (because it is locally open). Since $\bar \sigma : U \to \sigma(U)$ is a bijection, we have proved 1.


*For each $y \in \sigma(U)$ there exists an open neighborhood $V_y \subset \mathbb R^3$ and a smooth map $\psi : V_y \to U$ such that $\psi \circ \sigma \mid_{\sigma^{-1}(V_y)} = id$.

For $x = \sigma^{-1}(y)$ consider the map $\gamma$ introduced above. Now define $V_y = (p \circ \phi)^{-1}(N_x)$  and
$$\psi : V_y \stackrel{p \circ \phi}{\to} N_x \stackrel{\gamma^{-1}}{\to} U'_x .$$
By definition for $z \in \sigma^{-1}(V_y)$
$$\psi(\sigma(z)) = \gamma^{-1}(p(\phi(\sigma(z)))) = \gamma^{-1}(\gamma(z)) = z .$$
Your second bullet point follows from 2. because differentiability is a local property. In fact for each $\xi \in \Omega$ define $y = \tau (\xi)$ and choose $\psi$ as in 2. Then on $\tau^{-1}(V_y)$ we have $\sigma^{-1} \circ \tau = \psi \circ \tau$.
