# Compatible atlas induced the same topology.

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!

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.