Compatibility of Atlases is Equivalence Relation

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Theorem

Let $M$ be a topological space.

Let $d$ and $k$ be natural numbers.

Let $\mathcal A$ denote the set of all $d$-dimensional 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 $\mathcal G$:

$\mathcal F \sim \mathcal G$ if and only if $\mathcal F$ and $\mathcal G$ are $C^k$-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.

$\Box$


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 $\mathcal G \sim \mathcal F$.

$\Box$


Transitive

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