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First, suppose that $g$ is a smooth map of an open set $U$ of $\mathbb{H}^k$ into $\mathbb{R}^l$. If $u \in \partial U$, the smoothness of $g$ means that it may be extended to a smooth map $\phi$ defined in an open neighborhood of $u$ in $\mathbb{R}^l$; define $dg_u$ to be the derivative $d\phi_u: \mathbb{R}^k \to \mathbb{R}^l$. We must show that if $\widetilde{\phi}$ is another local extension of $g$, then $d\phi_u=d\widetilde{\phi}_u$. Let $u_i$ be any sequence of points in $Int(U)$ that converge to $u$. Because $\phi$ and $\widetilde{\phi}$ both agree with $g$ on $Int(U)$, $d\phi_{u_i}=d\widetilde{\phi}_{u_i}$...

Why is there a sequence of points $u_i$ in $Int (U) $ converging to $ u $? I think that $ u $ is a limit point of $ U $, but I can not seem to find out more. Can anyone refer me a good book of analysis can explain the result or/and explain what is really happening in the statement?

Reference : Differential Topology (Guillemin and Pollack), page 59

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    $\begingroup$ Near $u$ things just look like $\{x \in \mathbb{R}^k : |x| < 1, x_k \geq 0\}$ with $u$ the origin. $\endgroup$ – Hoot Aug 14 '15 at 16:06
  • $\begingroup$ I know that this is an open set of $\mathbb{H}^k$, but how can this assertion answer my question? $\endgroup$ – user230283 Aug 14 '15 at 16:25
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By definition, for any p in the boundary, there is a neighborhood of p diffeomorphic to Hn. So if under this diffeomorphism, p maps to $(x_1,x_2,…,x_{n−1},0)$, then pick the sequence of points to be $(x_1,x_2,…,x_{n−1},1/k)$.

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  • $\begingroup$ Right, thank you @J.G $\endgroup$ – user230283 Aug 14 '15 at 16:41

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