> [!definition] > > Let $X$ be a [[Set|set]], $(U_i, \psi_i)_{i \in I}$ be an [[Atlas|atlas]] with $\psi_i: U_i \to E_i$, where each $E_i$ is a [[Banach Space|Banach space]]. Define > $ > S = \bracs{\psi_i^{-1}(V_i): V_i \text{ open in } E_i, i \in I} > $ > then $S$ forms a [[Base|base]] for some topology, and the [[Topological Space|topology]] generated by it is known as the topology on $X$ induced by the atlas. > > *Proof*. Since the $U_i$s cover $X$, for any $x \in X$, there exists $i \in I$ such that $x \in U_i = \psi_i^{-1}(E_i)$. > > Let $V_i$ be open in $E_i$ and $V_j$ be open in $E_j$, then $\psi_i^{-1}(V_i) \subset U_i$ and $\psi_j^{-1}(V_j) \subset U_j$. As $\psi_i \circ \psi_j^{-1}$ is a [$C^p$](Space%20of%20Continuously%20Differentiable%20Functions)-isomorphism, $W_j = \psi_i \circ \psi_j^{-1}(V_j)$ is open in $E_i$. With each $\psi_k$ being a bijection, > $ > \begin{align*} > \psi_i^{-1}(W_j) &= \psi_i^{-1}(\psi_i \circ \psi_j^{-1}(V_j)) = \psi_j^{-1}(V_j) \\ > \psi_i^{-1}(V_i \cap W_j) &= \psi_i^{-1}(V_i) \cap \psi_j^{-1}(V_j) > \end{align*} > $ > where $\psi_i^{-1}(V_i \cap W_j) \in S$. Therefore $S$ is a base for a topology. > [!theorem] > > Let $X$ be a set, $(U_i, \psi_i)_{i \in I}$ be an atlas. Then each $\psi_i: U_i \to E_i$ is a [[Homeomorphism|homeomorphism]] with respect to the final topology on $X$. > > *Proof*. Fix $i \in I$. Since $S$ contains the preimage of every open set in $E_i$ by $\psi_i$, $\psi_i$ is continuous. > > For any open set $E \subset X$, since $U$ is open, $U_i \cap E$ is also open. Therefore we can restrict our considerations purely to open sets contained in $U_i$. Let $\seqf{V_j}$ be a finite intersection of elements of $S$ contained in $U_i$, then by the above proof, there exists a collection of open sets $\seqf{W_j}$ in $E_i$ such that > $ > \bigcap_{j = 1}^nV_j = \psi_i^{-1}\paren{\bigcap_{j = 1}^nW_j} > $ > where $\bigcap_{j = 1}^nW_j = \psi_i\paren{\bigcap_{j = 1}^nV_j}$ since $\psi_i$ is a bijection. The preimage of finite intersections of elements of $S$ by $\psi_i^{-1}$ then is open in $E_i$. > > Now, since every open set on $U_i$ is a union of finite intersections of $S$, their preimage by $\psi_i^{-1}$ is also open. Therefore $\psi_i^{-1}$ is continuous, and $\psi_i$ is a homeomorphism.