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 $\varphi_{\alpha}$?

  • 3
    $\begingroup$ Yes because an homeomrphism is an open map, so brings open set into open set. Let me know if you want to see a proof of the fact that homeomrphisms are open maps $\endgroup$
    – Salvatore
    Feb 27 '16 at 20:30
  • $\begingroup$ Hey @Salvatore I have some ideas but I am not very sure. Would you be able to write the answer down instead of as a comment so that I can acknowledge? Thanks! $\endgroup$
    – Everiana
    Feb 27 '16 at 20:34

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.


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