An equivalence relation for atlas on an abstract set X Let $X$ be an abstract set. A chart for a set $X$ is a bijection $\phi$ from a subset $U$ of $X$ to an open subset of $\mathbb{R}^n.$ Denote $(U, \phi).$ Two charts $(U, \phi)$ and $(U', \phi')$ will be called compatible if:
1) $\phi(U\cap U')$ and $\phi'(U\cap U')$ open in $\mathbb{R}^n;$
2) $\phi \circ \phi’^{-1}$ (respectively $\phi’ \circ \phi^{-1}$) from $\phi’(U\cap U’)$ on $\phi(U\cap U’)$ (respectively from $\phi(U\cap U’)$ on $\phi’(U\cap U’)$) is a homeomorphism.
An atlas for a set $X$ is a collection $\{(U_{\alpha}, \phi_{\alpha})\}$ of pairwise compatible charts on $X$ such that $\bigcup U_{\alpha}=X.$
Two atlases $\mathcal{A}$ and $\mathcal{B}$ will be called equivalent if their charts are compatible of pairwise. 
In some books about differential topology, we have the statement: this is an equivalence relation. 
If $X$ is a topological space and $\phi$ are homeomorphisms, it's very easy to prove transitivity. But I don't know how to prove that for an abstract set $X.$ Help!
 A: Henno Brandsma, let $X$ be a topological space. 
Let $\mathcal{A}=\{(U_{\alpha}, \phi_{\alpha})\}, \mathcal{B}=\{(V_{\beta}, \psi_{\beta})\}, \mathcal{C}=\{(W_{\gamma}, \theta_{\gamma})\}$ are atlases on $X$ and $\mathcal{A} \sim \mathcal{B}, \mathcal{B}\sim \mathcal{C}.$ Now, $"\sim"$ means that charts of $\mathcal{A}$ and $\mathcal{B}$ (respectively $\mathcal{B}$ and $\mathcal{C}$) are compatible of pairwise. We show that $\mathcal{A}\sim \mathcal{C}.$ 
1) We fix arbitrary charts $(U_{\alpha}, \phi_{\alpha})$ and $(W_{\gamma}, \theta_{\gamma}).$ Denote $(U,\phi)$ and $(W, \theta).$ Let $U \cap W \ne \varnothing.$ Since $U, W$ are open in $X$, their intersection $U\cap W$ is also open. So $\phi(U\cap W)$ and $\theta(U\cap W)$ open in $\mathbb{R}^n$ because $\varphi, \theta$ are homeomorphisms. 
2) $\phi \circ \psi^{-1}$ and $\psi \circ \phi^{-1}$ are homeomorphisms from $\psi(U\cap W)$ on $\phi(U\cap W)$ (respectively from $\phi(U\cap W)$ on $\psi(U\cap W)$).
If $X$ is an abstract set we can not show that $\phi(U\cap W)$ open in $\mathbb{R}^n$ as easily as in the case of a topological space.
A: If I understand correctly, David Nejaty said (my interpretation):
$\theta(U\cap W)=\bigcup \theta(U\cap W \cap V_{\beta})$ and $\phi (U \cap W)=\bigcup \phi(U\cap W\cap V_{\beta})$ on $(V_{\beta}, \psi_{\beta}) \in \mathcal{B}.$
For each open subsets of $\theta(U\cap W)$ is homeomorphic to an open subset of $\phi(U\cap W)$, we have the result that 
$\phi \circ \theta$ is a homeomorphism.
A: I proved this! Let $X$ be an abstract set.
1) We fix arbitrary charts $(U_{\alpha}, \phi_{\alpha})$ and $(W_{\gamma}, \theta_{\gamma}).$ Denote $(U,\phi)$ and $(W, \theta).$ Let $O=U \cap W \ne \varnothing.$ 
The set $\{V_{\beta} \colon \bigcup V_{\beta}\cap O=O\}$ is a cover for $O.$ We have that $\psi_{\beta}(V_{\beta}\cap U)$ and $\psi_{\beta}(V_{\beta}\cap W)$ are open in $\mathbb{R}^n.$ Than $\psi_{\beta}(V_{\beta}\cap U)\cap \psi_{\beta}(V_{\beta}\cap W)$ is open in $\mathbb{R}^n.$ 
And $\psi_{\beta}(V_{\beta}\cap U)\cap \psi_{\beta}(V_{\beta}\cap W)=\psi_{\beta}(V_{\beta}\cap U\cap W)$ because $\psi_{\beta}$ is a bijection. 
So $\psi_{\beta}(V_{\beta}\cap U\cap W)$ is open in $\mathbb{R}^n.$ Finally, $\phi(U\cap W)$ and $\theta(U\cap W)$ are open in $\mathbb{R}^n$ because $\psi_{\beta} \colon (U\cap V_{\beta})\to \psi_{\beta}(U\cap V_{\beta})$ and $\psi_{\beta} \colon (W\cap V_{\beta})\to \psi_{\beta}(W\cap V_{\beta})$ are homeomorphisms and it doesn't exist $\beta$ that $\phi(U\cap W\cap V_{\beta})$ is not open in $\mathbb{R}^n$.
