Lie groups in metric space

Question

es $V_{x_0} \subset F$. I thought of saying that by taking the union of all these open sets given by the local diffeomorphism on every point, then I can say in a way I currently ignore that the inverse of $f$ is $C^1$ but I don't know how to do it and get confused easily...

Answer 1

Yes, that's exactly right. Let $g:F\to E$ be the inverse function of $f$. All we have to do is show $g$ is $C^1$. Now, take a point $y_0\in F$, and let $x_0:=g(y_0)\in E$. Since $f$ is by assumption a local diffeomorphism, there exist open neighborhoods $U_{x_0}$ of $x_0$ in $E$, and $V_{y_0}$ of $y_0$ in $F$, such that $f:U_{x_0}\to V_{y_0}$ is a diffeomorphism (draw a diagram to keep all this information organized). Well, what is the inverse of this restritced function? It is precisely the restriction $g:V_{y_0}\to U_{x_0}$. So, we have shown that $g$ is $C^1$ on the open set $V_{y_0}$. Lastly, $y_0\in F$ was arbitrary, so it follows $g$ is $C^1$ on the entire set $F$ (recall that continuity and differentiability are all local properties; that's why it is enough to verify these things point by point).