Proving that equivalent atlases form an equivalence relation? I know that this question was already asked here, but I'm afraid I cannot really follow the answers (which are given by the OP her- or himself). So, here is a definition of equivalent atlases:

Two atlases $\mathcal A$ and $\mathcal B$ on $M$ are called equivalent if $\mathcal A \cup \mathcal B$ is an atlas on $\mathcal M$.

And here a definition of atlas:

Let $M$ be a second countable Hausdorff topological space. An $n$-dimensional smooth atlas on $M$ is a collection of maps $$\mathcal A = \left\{ \left(\varphi_i, U_i\right) \mid i\in A\right\}, \quad \varphi_i: U_i\rightarrow \varphi_i(U_i)\subset \mathbb R^n,$$ such that all $U_i \subset M$ are open, all $\varphi_i$ are homeomorphisms, and

*

*$\{U_i, i\in I\}$  is an open covering of $\mathcal M$

*$\varphi_i\circ \varphi_j^{-1}: \varphi_j\left(U_i\cap U_j\right)\rightarrow \varphi_i\left( U_i\cap U_j\right)$ are smooth for all $i, j\in I$.



Let $\mathcal A = \{(\varphi_i, U_i)\mid i\in A\}$, $\mathcal B = \{(\psi_i, V_i)\mid i\in B\}$ and $\mathcal C = \{(\chi_i, W_i)\mid i\in C\}$ be atlases on the same set $M$.

*

*Now, if $\mathcal A$ is an atlas, then obviously, $\mathcal A\cup \mathcal A = \mathcal A$ is also an atlas.

*$\mathcal A\cup \mathcal B$ is an atlas if and only if $\mathcal B \cup \mathcal A$ is an atlas.

*The tricky part is proving transitivity. Let $\mathcal A \cup \mathcal B$ be an atlas (i.e. the transition functions $\varphi_i\circ \psi_j^{-1}: \psi_j\left(U_i \cap V_j\right)\rightarrow\varphi_i\left(U_i\cap V_j\right)$ are smooth $\forall i\in A$, $j\in B$), and let $\mathcal B\cup \mathcal C$ be an atlas (i.e. the transition functions $\psi_j\circ \chi_k^{-1}: \chi_k\left(V_j \cap W_k\right)\rightarrow\psi_j\left(V_j\cap W_k\right)$ are smooth $\forall j\in B$, $k\in C$). For $\mathcal A\cup \mathcal C$ to be an atlas, we need to show that
$$\varphi_i\circ \chi_k^{-1}: \chi_k\left(U_i \cap W_k\right)\rightarrow \varphi_i\left(U_i \cap W_k\right)\quad \text{smooth}\ \forall i\in A, k\in C.$$

My idea to prove this was sth like this:
$$\varphi_i\circ \chi_{k}^{-1} = (\varphi_i\circ\psi_j^{-1})\circ \left(\psi_j \circ \chi_k^{-1}\right) \forall i\in A, j\in B, k\in C.$$
However, this raises the question of the well-definedness of the RHS. The codomain of $\chi_k^{-1}$ is $W_k$, whereas the domain of $\psi_j$ is $V_j$. How can we resolve this?
 A: Transitivity
Let $(\varphi, U)\in \mathcal A$ and $(\chi, W)\in \mathcal C$ be charts with non-empty intersection, i.e. with $U\cap W\neq\emptyset$. We must show that the transition maps
$$ \varphi\circ\chi^{-1}\Big|\chi(U\cap W)\;:\;\chi(U\cap W)\to\varphi(U\cap W)$$
and its inverse
$$ \chi\circ\varphi^{-1}\Big|\varphi(U\cap W)\;:\;\varphi(U\cap W) \to \chi(U\cap W) $$
are smooth.
For the former (the latter is analogous), it suffices to show that, for each $x'\in\chi(U\cap W)$, the map $\varphi\circ\chi^{-1}\Big|\chi(U\cap W)$ is smooth in a suitable open neighborhood of $x'$. Putting $x=\chi^{-1}(x')\in U\cap W$, consider a chart $(\psi, V)\in \mathcal B$ containing $x$ in its domain. We can write
$$ \varphi\circ\chi^{-1}\Big|\chi(U\cap V\cap W) = \left[(\varphi\circ\psi^{-1})\Big|\psi(U\cap V\cap W)\right]\circ\left[(\psi\circ\chi^{-1})\Big|\chi(U\cap V\cap W)\right], $$
and we are done since $\chi(U\cap V\cap W)$ is an open neighborhood of $x'$ contained in $\chi(U\cap W)$, and the maps in square brackets are smooth since they are (restrictions of) transition maps of charts of equivalent atlases, by hypothesis.
