Compatibility of Atlases is Equivalence Relation

Theorem
Let $M$ be a locally Euclidean space of dimension $d$.

Let $\mathcal A$ denote the set of all atlases of class $\mathcal C^k$ on $M$.

Define a relation $\sim$ on $\mathcal A$ by putting, for any two $\mathcal C^k$-atlases $\mathcal F$ and $\mathcal G$:


 * $\mathcal F \sim \mathcal G$ $\mathcal F$ and $\mathcal G$ are compatible.

Then $\sim$ is an equivalence relation on $\mathcal A$.

Proof
It is to be shown that $\sim$ is reflexive, symmetric and transitive.

Reflexive
Let $\mathcal F \in \mathcal A$ be a $C^k$-atlas.

That $\mathcal F$ is compatible with itself is precisely condition $(2)$ of the definition of $C^k$-atlas.

Symmetric
Let $\mathcal F$ and $\mathcal G$ be $C^k$-atlases and suppose that $\mathcal F \sim \mathcal G$.

Let $\left({U, \phi}\right) \in \mathcal F$ and let $\left({V, \psi}\right) \in \mathcal G$ be charts.

Then by hypothesis:


 * $\phi \circ \psi^{-1}: \psi \left({U \cap V}\right) \to \phi \left({U \cap V}\right)$

is a $C^k$ mapping.

Because $\phi$ and $\psi$ are homeomorphisms, we have that:


 * $\psi \circ \phi^{-1}: \phi \left({U \cap V}\right) \to \psi \left({U \cap V}\right)$

is also a homeomorphism, and in particular continuous.

By the Inverse Function Theorem, $\psi \circ \phi^{-1}$ is also a $C^k$ mapping.

Since the charts were arbitrary, we conclude that $\mathcal G \sim \mathcal F$.

Transitive
Let $\mathcal F$, $\mathcal G$ and $\mathcal H$ be $C^k$-atlases, and suppose that $\mathcal F \sim \mathcal G$ and $\mathcal G \sim \mathcal H$.