Given $X$ a manifold, two smooth atlas are equivalent if the union of them forms an smooth atlas.
I am trying to prove that this is a equivalence relation, but I am having troubles proving that $A\sim B$ and $B\sim C$ implies $A\sim C$, as I do not know how to use the atlas $B$ to "connect" them.
The punchline : Let $a,b,c$ be diffeomorphisms "between spaces so that the compositions we now do make sense". Then we have that: $$ac^{-1} = ab^{-1}bc^{-1}=(ab^{-1})(bc^{-1}) , ca^{-1}=cb^{-1}ba^{-1}=(cb^{-1})(ba^{-1})$$. We can, of course, freely invert, since $a,b,c$ are diffeomorphisms. Elaborating:
Consider the maps $ a : W_{j} \rightarrow U_j \subset \mathbb R^m ; b: W_k \rightarrow U_k \subset \mathbb R^m$ in the $C^r$ structures $ A,B$ respectively and let $W_{kj}$ be a (non-empty) intersection $W_k \cap W_j$. Then they are in the same $C^k -$ structure if the transition maps $ ab^{-1}, ba^{-1}$ are also $C^r$. Now, consider $d: W_n \rightarrow U_n \subset R^m$, so that the triple $W_j, W_n, W_k$ overlap. We have, by assumption, that $ bd^{-1}, db^{-1}$ are in the $D$ structure. We must use the fact that $ab^{-1}, ba^{-1}, bc^{-1}, cb^{-1} $ are also $C^r$ to show that $ac^{-1}, ca^{-1}$ are also $C^r$. But $ac^{-1} =ab^{-1}bc^{-1}$ , the composition of two $C^r -$ maps, one in the $A$ structure and the other in the $B$ structure, which ae assumed compatible, so then their composition is also $C^r$, which is what we wanted to show.