Exercise 1.18 (Maximal atlases and compatibility)

Proof. (a) We first demonstrate that every smooth atlas $A$ for $M$ is contained in a unique maximal smooth atlas $\bar{A}$.

• (existence) Let $\bar{A}$ be the set of all smooth charts on $M$ which are compatible with $A$¹. Then $\bar{A}$ obviously covers $M$ since it at least contains $A$, i.e., $A\subseteq \bar{A}$. To demonstrate that $\bar{A}$ is smooth, we need to demonstrate that $\psi \circ {\phi }^{-1}:\phi (U\cap V)\to \psi (U\cap V)$ is smooth for any two charts $(U,\psi )$, $(V,\phi )$ in $\bar{A}$. In our case, it is easier to show that $\psi \circ {\phi }^{-1}$ is smooth in any neighbourhood of $\phi (U\cap V)$.

  Let, therefore, $p\in U\cap V$, or equivalently $\phi (p)\in \phi (U\cap V)$ be arbitrary. Let $(W_p,𝜃)$ be a smooth chart around $p$ from the original atlas $A$. By construction, the transition maps

  $$\begin{array}{lll}\hfill \psi \circ {𝜃}^{-1}:𝜃(U\cap W_p)\to \psi (U\cap W_p)& & \hfill \\ \hfill 𝜃\circ {\phi }^{-1}:\phi (W_p\cap V)\to 𝜃(W_p\cap V)& & \hfill \end{array}$$

  are both smooth. Thus, the composition

  $$\begin{array}{lll}\hfill \psi \circ {\phi }^{-1}=(\psi \circ {𝜃}^{-1})\circ (𝜃\circ {\phi }^{-1}):\phi (U\cap W_p\cap V)\to \psi (U\cap W_p\cap V)& & \hfill \end{array}$$

  must be smooth as well at the neighborhood $\phi (U\cap W_p\cap V)$ of $\phi (p)$. Since we can perform this in some neighborhood $U\cap W_p\cap V$ of every point $p\in U\cap V$, $\psi \times {\phi }^{-1}$ must be smooth everywhere on $\phi (U\cap V)$.

• (maximality) To demonstrate that $\bar{A}$ is maximal, consider any atlas $A^{\prime }$ that is smoothly compatible with $\bar{A}$. In particular, $A^{\prime }$ must be also smoothly compatible with $A$ since $A\subseteq \bar{A}$. But every chart compatible with $A$ is contained in $\bar{A}$ by costruction; thus, $A^{\prime }\subseteq \bar{A}$.
• (uniqueness) Now suppose that there is another maximal atlas ${\bar{A}}^{\prime }$. By a similar argument, ${\bar{A}}^{\prime }$ must be compatible with $A$ and thus is contained in $\bar{A}$ by construction, i.e, ${\bar{A}}^{\prime }\subseteq \bar{A}$. By maximality, ${\bar{A}}^{\prime }=\bar{A}$.

(b) Now we demonstrate that two smooth atlases for $M$ determine the same smooth structure if and only if their union is a smooth atlas.

• ( $⟹$) Suppose that two atlases $A_1$ and $A_2$ determine the same maximal smooth atlas $\bar{A}$. Since $A_1\subseteq \bar{A}$ and $A_2\subseteq \bar{A}$, any chart of $A_1$ is smoothly compatible with any chart in $A_2$ because $\bar{A}$ is a smooth atlas. This is just another way of saying that $A_1\cup A_2$ is a smooth atlas.
• ( $⟸$) Suppose that the union $A_1\cup A_2$ of two atlases $A_1$ and $A_2$ is smooth and let $\bar{A}$ be its maximal smooth atlas. Since $A_1\subseteq A_1\cup A_2\subseteq \bar{A}$, by maximality and uniqueness of maximal smooth atlas, $\bar{A}$ must also be the maximal smooth atlas of $A_1$. A similar argument shows that $\bar{A}$ is the maximal smooth atlas of $A_2$.