# Atlases on a Set

Recall from the Charts on a Set page that if $M$ is a set then an $m$-dimensional chart on $M$ is a pair $(U, \varphi)$ where $U \subseteq M$ and $\varphi : U \to \mathbb{R}^m$ is such that $\varphi (U) \subseteq \mathbb{R}^m$ is open and $\varphi$ is injective.

Furthermore, the $(U, \varphi)$ and $(V, \psi)$ on $M$ are said to be compatible if $\varphi (U \cap V) \subseteq \mathbb{R}^m$ is open, $\psi (U \cap V) \subseteq \mathbb{R}^m$ is open, and the transition map $\psi \circ \varphi^{-1} : \varphi (U \cap V) \to \psi (U \cap V)$ is a diffeomorphism from $\varphi (U \cap V)$ to $\psi (U \cap V)$.

We will now give another important definition in differential geometry.

Definition: Let $M$ be a set. An $m$-Dimensional Atlas on $M$ is a collection of $m$-dimensional charts on $M$, $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha}) : \alpha \in \Gamma \}$ (where $\Gamma$ is some indexing set) such that:a) $\mathcal A$ covers $M$, i.e., $\displaystyle{M = \bigcup_{\alpha \in \Gamma} U_{\alpha}}$.b) For all $\alpha, \beta \in \Gamma$, $(U_{\alpha}, \varphi_{\alpha})$ and $(U_{\beta}, \varphi_{\beta})$ are compatible. |

Definition: Let $M$ be a set and let $\mathcal A$ be an atlas on $M$. A chart $(U, \varphi)$ on $M$ is said to be Compatible with $\mathcal A$ if $(U, \varphi)$ is compatible with every chart in $\mathcal A$. |

We will now prove a simple lemma.

Lemma 1: Let $M$ be a set and let $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha}) : \alpha \in \Gamma \}$ be an atlas on $M$. Then a chart $(U, \varphi)$ is compatible with $\mathcal A$ if and only if $\mathcal A \cup \{ (U, \varphi) \}$ is an atlas on $M$. |

**Proof:**$\Rightarrow$ Suppose that $(U, \varphi)$ is compatible with $\mathcal A$. Then $(U, \varphi)$ is compatible with evey chart in $\mathcal A$. Since $\mathcal A$ is an atlas on $M$ we see that:

- Furthermore every atlas in $\mathcal A \cup \{ (U, \varphi) \}$ is compatible with one another, so $\mathcal A \cup \{ (U, \varphi) \}$ is an atlas on $M$.

- $\Leftarrow$ Conversely, suppose that $\mathcal A \cup \{ (U, \varphi) \}$ is an atlas on $M$. Then $(U, \varphi)$ is compatible with every chart in $\mathcal A$, so $(U, \varphi)$ is compatible with $\mathcal A$. $\blacksquare$

We now formula a similar definition to compare two atlases on a set $M$.

Definition: Let $M$ be a set and let $\mathcal A = \{ (U_{\alpha}, \varphi_{\alpha} : \alpha \in \Gamma \}$ and $\mathcal A' = \{ (U_{\alpha'}, \varphi_{\alpha'}) : \alpha' \in \Gamma \}$ be atlases on $M$. Then $\mathcal A$ is said to be Equivalent to $\mathcal A'$ if every chart in $\mathcal A$ is compatible with every chart in $\mathcal A'$. |