Proposition 1.17

Part A: Every smooth atlas \(\mathcal{A}\) for \(M\) is contained in a unique maximal smooth atlas, called the smooth structure determined by \(\mathcal{A}\).

Note that \(\mathcal{A}\) is itself only a collection of smooth charts \((U, \varphi)\) covering all point of \(M\), such that all charts are smoothly compatible. Let \(\overline{\mathcal{A}}\) be the collection of all charts defined on \(M\) which are smoothly compatible with all elements of \(\mathcal{A}\). We claim that \(\overline{\mathcal{A}}\) is the desired, unique, maximal, smooth atlas.

Obviously, \(\mathcal{A} \subset \overline{\mathcal{A}}\). Given \((U, \varphi)\) and \((V, \phi)\) of \(\overline{\mathcal{A}}\), both are compatible with every element of \(\mathcal{A}\). In the case that \(U \cap V = \emptyset\), these charts are compatible. Otherwise, if they intersect non-trivially, choose some \(x \in U \cap V\), and we can choose a chart \((W, \theta)\) with \(x \in W\), such that

\[(\varphi \circ \theta^{-1}) : \theta(W \cap U) \rightarrow \varphi(W \cap U) \ \ \ \text{and} \ \ \ (\phi \circ \theta^{-1}) : \theta(W \cap V) \rightarrow \phi(W \cap V)\]

are diffeomorphisms. It follows that \((\varphi \circ \theta^{-1}) \circ (\phi \circ \theta^{-1})^{-1} = \varphi \circ \phi^{-1}\) is a diffeomorphism between \(\phi(W \cap V \cap U)\) and \(\varphi(W \cap V \cap U)\).

Now, given \(\phi(x) \in \phi(U \cap V)\), note that we can choose open \(W\) with \(x \in W\), such that \((\varphi^{-1} \circ \phi)\) is a diffeomorphism on \(\phi(x) \in \phi(V \cap W) \cap \phi(U \cap V)\), so \((\varphi^{-1} \circ \phi)\) is a diffeomorphism in a neighbourhood around each point in the domain. It follows that it is a diffeomorphism on the whole domain: we have differentiability at each point, and clearly the map is a bijection with smooth inverse.

Suppose \(\mathcal{B}\) were an smooth atlas containing \(\mathcal{A}\): then every chart in \(\mathcal{B}\) must be smoothly compatible with every chart of \(\mathcal{A}\), so \(\mathcal{B} \subset \overline{\mathcal{A}}\), and \(\overline{\mathcal{A}}\) is maximal. This fact also proves uniqueness.

Part B: Two smooth atlases determine the same smooth structure if and only if their union is a smooth atlas.

In the case that the union of two atlases \(\mathcal{A}_1\) and \(\mathcal{A}_2\) is a smooth atlas, the two atlases must be smoothly compatible, and thus \(\mathcal{A}_2 \subset \overline{\mathcal{A}_1}\) and \(\mathcal{A}_1 \subset \overline{\mathcal{A}_2}\). By uniqueness of smooth structures, \(\overline{\mathcal{A}_1} = \overline{\mathcal{A}_2}\). The converse is trivial.