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I am posting a picture, containing a theorem and its proof from these notes on Lie groups.

enter image description here

I'm quoting one of the lines:

$UH$ is open in $G$ (which easily follows from inverse function theorem applied to the map $f:U\times H\to G$).

What exactly is meant by this? Is the argument that $UH=f^{-1}(G)$, and hence open? Where does the inverse function theorem come in?

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The differential of the map $f:U\times H\rightarrow G$ defined by $(u,h)\rightarrow uh$ is an isomorphism since $M$ is transversal to $H$, this justify the application of the local inverse theorem which implies that $f$ is a local diffeomorphism, hence $UH$ is open.

Tsemo Aristide
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  • Thanks! Could you write in more detail how the differential of $(u,h)\to uh$ can be seen to be an isomorphism at each point $(a,b)\in U\times H$? –  Jan 12 '20 at 15:34
  • Look the partial derivatives along $M$ and $H$, they show that the differential is surjective at each point. – Tsemo Aristide Jan 12 '20 at 15:36
  • Thanks. Could you also comment on the why $\tilde{U}\to U$ is a homeomorphism? –  Jan 12 '20 at 15:57
  • The inverse function thm applies because the multiplication map is a diffeomorphism. This follows by identifying $T_{(p,q)}(U\times H)$ with $T_p(U)\oplus T_q(H) $ via the map $(v,w)\mapsto di_U(v)+di_H(w)$ – Matematleta Jan 12 '20 at 16:07
  • This is due to the fact that $M$ is transverse so the restriction of the quotient to $U$ is a bijection to its image and $G/H$ is endowed with the quotient topology, if $V$ is an open subset of $M$, $V=W\cap M$ where $W$ is open, $p(W)=p(V)$ implies that $p(V)$ is open and the restriction of $p$ to $U$ is open this implies that it is an homeomorphism, since it is continuous. – Tsemo Aristide Jan 12 '20 at 16:08