Let $M$ be a set. We call a set $\mathcal{A} = \{(U_i, \phi_i)\mid i \in I\}$ an atlas of dimension $m$ on $M$ if the following conditions are satisfied:
(1) $U_i \subseteq M$
(2) $\phi_i: U_i \to \mathbb{R}^m$ is injective.
(3) $\phi(U_i)$ is open.
(4) $\bigcup_{i \in I} U_i = M$
(5) If $i,j \in I, U_i \cap U_j \ne \emptyset$, then $\phi_i(U_i \cap U_j)$ is open.
(6) If $i,j \in I, U_i \cap U_j \ne \emptyset$. Then $\phi_j \circ \phi_i^{-1}: \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$ is smooth (of class $C^\infty$).
Prove that the following is an equivalence relation on the set of all atlases of dimension $m$ on $M$:
$\mathcal{A} \sim \mathcal{A'} \iff \mathcal{A} \cup \mathcal{A'}$ is an atlas of dimension $m$ on $M$.
My attempt:
Reflexivity and symmetry are immediate. Suppose $\mathcal{A} = \{(\phi_i,U_i)\}, \mathcal{B} = \{(\psi_j, V_j)\}, \mathcal{C} = \{(\varphi_k, W_k)\}$ are atlases such that $\mathcal{A} \cup \mathcal{B}$ and $\mathcal{B} \cup \mathcal{C}$ are atlases.
We show that $\mathcal{A} \cup \mathcal{C}$ is an atlas. (1)-(2)-(3)-(4) are evident.
(5): Suppose $i \in I, k \in K$ with $U_i \cap W_k \neq \emptyset$. Then
$$\phi_i(U_i \cap W_k) = \phi_i\left((U_i \cap W_k) \cap \bigcup_{j \in J}V_j\right)$$
$$=\bigcup_{j \in J}\phi\left(U_i \cap V_j \cap W_k\right)$$
and $$\phi_i(U_i \cap V_j \cap W_k) = (\phi_i \circ \psi_j^{-1})(\psi_j((U_i\cap V_j) \cap (V_j \cap W_k))$$
$$= (\phi_i \circ \psi_j^{-1})(\psi_j(U_i \cap V_j)\cap \psi(V_j \cap W_k))$$
and the argument of this last line is open, as intersection of open sets. So the image is open, as it is an open mapping. Hence, $(\phi_i \circ \psi_j^{-1})$ is open, and it follows that that $\phi_i(U_i \cap W_k)$ is open, because it is the union of open sets.
(6) I'm sure my proof of this is correct, so I don't bother posting it.
Is this correct, especially (5)?