TITLE: Diffeomorphism preserves open set?
QUESTION [6 upvotes]: This question might be elementary, but I am a little confused. Does diffeomorphism preserve open sets?
Suppose I have two coordinate charts $(U,\varphi)$, $(V,\psi)$ and atlas $\mathcal{A}=\{(U_{\alpha},\varphi_{\alpha})\}$ such that the two charts are compatible with $\mathcal{A}$. Then clearly $\varphi_{\alpha}(U\cap U_{\alpha})$ and $\varphi_{\alpha}(V\cap U_{\alpha})$ are open. How can I conclude then that $\varphi_{\alpha}(U\cap V\cap U_{\alpha})$ is also open?
I know that $U\cap V\cap U_{\alpha}$ is open, so is the openness preserved by $\varphi_{\alpha}$?

REPLY [3 votes]: Diffeomorphisms are homeomorphisms because differentiability implies continuity. So we should prove, in order to answer at your question above, that homeomorphisms are open maps. Let $\phi:X \rightarrow Y$ be an homeomorphism between two topological spaces $X,Y$. Since $\phi$ is an homeomorphism then consider $g:Y \rightarrow X$ be the inverse map of $\phi$ which is a continuous map because we assumed $\phi$ to be an homeomorphism (namely a bicontinuous map between topological spaces). Then for every open set $V \subset X$ $g^{-1}(V) \subset Y$ is open in $Y$, but we can write $g^{-1}(V)=(\phi^{-1})^{-1}(V)=\phi(V)$ $\Rightarrow$ $\phi(V)$ is open in $Y$ , this holds for all open set $V$ and proves that $\phi$ is an open map because brings open sets into open sets.