Atlas Belongs to Unique Differentiable Structure

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

Let $\mathscr G$ be a pre-differentiable structure on $M$.

Then there exists a unique differentiable structure $\mathscr F$ on $M$ with $\mathscr G \subseteq \mathscr F$.

Proof
Let $\mathscr F$ be the set of co-ordinate systems:


 * $\mathscr F = \{(U,\phi) : \phi\circ \phi_\alpha^{-1}\text{ and }\phi_\alpha\circ \phi^{-1}\text{ are }C^k\text{ for all }\phi_\alpha \in \mathscr G\:\}$

Then by the definition of a pre-differentiable structure, $\mathscr G \subseteq \mathscr F$, so


 * $\bigcup_{(U,\phi) \in \mathscr F}U \supseteq \bigcup_{(U,\phi) \in \mathscr G}U = M$

Moreover, if $\phi_1, \phi_2 \in \mathscr F$, then for any $\phi_\alpha \in \mathscr G$,


 * $\phi_1 \circ \phi_2^{-1} = (\phi_1 \circ \phi_\alpha^{-1}) \circ (\phi_\alpha \circ \phi_2^{-1})$

This is the composition of two differentiable functions, so by Composition of Differentiable Functions is Differentiable it is $C^k$.

Finally, by construction, $\mathscr F$ satisfies part 3. of the definition of a differentiable structure.

If $\mathscr F'$ is another such structure, then because $\mathscr F$ is maximal, then $\mathscr F' \subseteq \mathscr F$.

By equally, $\mathscr F'$ is also maximal, so $\mathscr F \subseteq \mathscr F'$.

Therefore $\mathscr F = \mathscr F'$, and $\mathscr F$ is unique. We are done.