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I'm working with some result in Do carmo, Riemannian Geometry. Which consits in to find all isometries in $\mathbb{H}^n$, but in the case $\mathbb{H}^n$, $n \geq 3$, the autor said the comformal map $h$ maps $\partial \mathbb{H}^n$ to $\partial \mathbb{H}^n$. In a first place I think that statement was due some special property of comformal maps. But I find this

Does a conformal map take boundaries to boundaries?

So, I don't know how prove this affirmation. Some hint?

Thanks in advance!

Zacarias89.
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  • What do you mean by conformal in your question? What dimension? Do you assume connected domain?.... – Moishe Kohan Feb 27 '23 at 02:07
  • @Moishe Kohan. Comformal map, means the map which preserve angles, the domain in this case is $\mathbb{R}^n$ , just that $h$ maps $\mathbb{H}^n$ to $\mathbb{H}^n$. The dimension is $n\geq 3$. – Zacarias89. Feb 27 '23 at 02:25
  • No, your domain cannot be $R^n$, it should be the 1-point compactification of $^n$. Then all conformal transformations are diffeomorphisms (actually, Moebius transformations), hence, map boundaries of subsets of $R^n\cup\infty$ to boundaries of the images. – Moishe Kohan Feb 27 '23 at 12:01

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