TITLE: Prove that$ H_x (X)$ does not depend on the choice of local parametrization.
QUESTION [5 upvotes]: Suppose that $X$ is a manifold with boundary and  $x∈∂X$. Let $ϕ:U→X$ be a local parametrization with $ϕ (0)=x$ where $U$ is an open subset of $H^k$. Then $dϕ_0:R^k→T_x (X)$ is an isomorphism. Define the upper half space $H_x (X)$ in $T_x (X)$ to be the image of $H^k$ under $dϕ_0, H_x (X)=dϕ_0 (H^k )$. Prove that$ H_x (X)$ does not depend on the choice of local parametrization.
I tried to construct another local parametrization say $\omega:V→X$ where $V$ is also an open subset of $H^k$ and $\omega (0)=x$. Since both $U$ and $V$ are subset of $H^k$, $U \cap V$ is also subset of $H^k$. I found a hint that tell me to consider $ \phi (U) \cap \omega(V)$ but I can't see how this can help me.

REPLY [7 votes]: Since you are using G&P (exercise 2.1.7*), here is a solution in that flavor which doesn't use parametrizing local curves.
Let $\phi:U\to X$ and $\psi:W\to X$ be local parametrizations about a point $x\in\partial X$ where $U$ and $W$ are open subsets of $H^k$ with the usual $0\mapsto x$ for $\phi$ and $\psi$. Then, by shrinking neighborhoods if need be, $g=\psi^{-1}\circ\phi$ is a diffeomorphism $U\cong W$. Let $G$ be an extension of $g$ to an open subset of $\mathbb{R}^k$, $G:U'\to W$. By definition $dg_0=dG_0$ (p.59). Since $\phi$ and $\psi$ map boundary to boundary (exercise 2.1.2), they must map (strict) upper half space to $X^\circ$. This gives us that $G$ maps $H^k$ to $H^k$. Now observe that since $G$ is smooth, the limit $\lim_{t\to 0}\frac{G(tv)}{t}$ exists and equals $dG_0(v)$ for all $v\in\mathbb{R}^k$. So, in particular, for $v\in H^k$ since $G$ maps $H^k$ to $H^k$ and $H^k$ is a closed set, we have that
$$dG_0(v)=\lim_{t\to 0^+}\frac{G(tv)}{v}\in H^k$$This shows that $dG_0(H^k)\subset H^k$. But since the chain rule still works (p.59) we have that $dG_0=dg_0=d\psi^{-1}_x\circ d\phi_0$ and therefore we have shown that $d\phi_0(H^k)\subset d\psi_0(H^k)$. Defining $G$ in the reverse order gives the other inclusion that we seek and thus gives $d\phi_0(H^k)= d\psi_0(H^k)$. This guarantees that the definition $H_x(X)=d\phi_0(H^k)$ is well defined.