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Let $\mathbb{R^2}$ be equipped with the standard topology, and let $\mathbb{H^2}$ be the upper half plane (containing the x-axis), equipped with the subspace topology. Let $U$ be an open neighborhood of $0$ in $\mathbb{R^2}$, $V$ an open neighborhood of $0$ in $\mathbb{H^2}$. How can I show that $U$ and $V$ are not diffeomorphic? My guess is that I should proceed by contradiction, and suppose that there is a diffeomorphism from U onto V. I feel like there will be an issue with the pre-image of a semi open ball centered at $0$, which should be open in $\mathbb{R^2}$. But I struggle to prove that properly.

Emilien
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  • Note that $V\setminus{0}$ is simply connected, but $U\setminus {p}$ is not. –  Mar 05 '16 at 00:50
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    http://math.stackexchange.com/questions/1394591/exercise-2-1-1-differential-topology-by-guillemin-and-pollack/1394725#1394725 Duplicate of this question. In general a map with nonsingular derivative is an open map (as a map $\Omega \to \Bbb R^n$) by the inverse function theorem and a neighborhound of a boundary point in $\Bbb H^n$ is not an open set in $\Bbb R^n$. – PVAL-inactive Mar 05 '16 at 00:51
  • @PVAL You linked to your answer instead of the question. –  Mar 05 '16 at 00:53
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    @JohnMa well the accepted answer on that question is ridiculous so I linked to a better one. – PVAL-inactive Mar 05 '16 at 00:54

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