Let $M$ a set and $\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}$ an atlas we said that $A\subseteq M$ is open iif $\varphi_\alpha(A\cap U_\alpha)$ is open in $\mathbb{R}^n$ for all chart $(U_\alpha,\varphi_\alpha).$
We denote with $\tau_\mathcal{A}$ the topology definited by $\mathcal{A}$.
Now let $\mathcal{B}=\{(V_\alpha,\psi_\alpha)\}$ another atlas compatible with $\mathcal{A}$, that is $$\psi_\beta\circ \varphi^{-1}_\alpha\colon\varphi_{\alpha}(U_\alpha\cap V_\beta)\to\psi_\beta(U_\alpha\cap V_\beta),$$ is a diffeomorphism and $\varphi_{\alpha}(U_\alpha\cap V_\beta)$,$\psi_\beta(U_\alpha\cap V_\beta)$ are open sets of $\mathbb{R}^n.$
Denote with $\tau_\mathcal{B}$ the topology definited by $\mathcal{B}$.
I must prove that $\tau_\mathcal{A}=\tau_\mathcal{B}$.
Now, if $A\in\tau_\mathcal{A}$, then $\varphi_\alpha(A\cap U_\alpha)$ is open, therefore $$\varphi_\alpha(A\cap U_\alpha)\cap \varphi_\alpha(U_\alpha\cap V_\beta)=\varphi_\alpha(U_\alpha\cap V_\beta\cap A)$$ is open.
Now $$\psi_\beta(A\cap U_\alpha\cap V_\beta)=(\psi_\beta\circ\varphi^{-1})(\varphi_\alpha(A\cap U_\alpha\cap V_\beta))$$ is open since $\psi_\beta\circ\varphi_\alpha^{-1}$ is a diffeomorphism.
Question. I don't know how he shows at this point that $\psi_\beta(A\cap V_\beta)$ is open for all chart in $\mathcal{B}$.
Same hints?
Thanks!
Best Answer
Since $\psi_\beta$ is bijective with domain $V_\beta$, we have $$\psi_\beta(A\cap V_\beta)\ =\ \psi_\beta\big(\bigcup_\alpha (A\cap V_\beta\cap U_\alpha) \big) \ =\ \bigcup_\alpha\psi_\beta(A\cap V_\beta\cap U_\alpha)$$ is a union of open sets, hence open.