Prove that$ H_x (X)$ does not depend on the choice of local parametrization. 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.
 A: 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.
A: I think a point of this exercise is that a boundary in some extent orients the tangentplane. The assertion would not be true for a boundaryless manifold, as take for example the sphere $S^1$. Choosing two parametrizations around the north pole $\phi(x)=(x,\sqrt{1-x^2})$ and $\psi(x)=(-x,\sqrt{1-x^2})$, $x\in(-1,1)$, we have $\phi[(-1,1)]=\psi[(-1,1)]$ but $d\phi_0(H^1)=H^1$ and $d\psi_0(H^1)=-H^1$.
Instead, let $\phi, \psi: V\rightarrow X$ be two parametrizations with $\phi(0)=\psi(0)=x$. We know that $d\phi_0$ and $d\psi_0$ are both isomorphisms from $\mathbb{R}^k$  to $T_x(X)$. Assume $v\in d\phi_0(H^k)$ but $v\notin d\psi_0(H^k)$. Then $d\psi^{-1}_x(v)\in \mathbb{R}^k-H^k$. Let $\alpha$ be a curve in $\mathbb{R}^k$ with $\alpha(0)=0$ and $\alpha'(0)=d\psi^{-1}_x(v)$. By definition, $\phi$ and $\psi$ can be extended to smooth functions $\Phi, \Psi$ on open nbhds of $0$ in $\mathbb{R}^k$, on which they are both still diffeomorphisms.
Note that $\alpha$ maps a short interval $(0,\epsilon)$ into $\mathbb{R}^k-H^k$, so $\Psi\circ\alpha$ maps $(0,\epsilon)$ to an open arc in the ambient space of $X$ that is disjoint from $X$. Finally, consider the map $g=\Phi^{-1}\circ\Psi\circ\alpha$ which is an arc in $\mathbb{R}^k$ with $g(0)=0$ and
$$
g'(0)=d\Phi^{-1}_x\circ d\Psi_0(d\Psi^{-1}_x(v))=d\Phi^{-1}_x(v)\in H^k,
$$
by our choice of $v$. The curve $g$ maps a short interval $(0,\epsilon)$ into $H^k$ whereas $\Psi\circ\alpha$ maps the same interval outside of $X$, but this would imply that the extension map $\Phi$ maps points in $H^k$ outside of $X$! This cannot be true since $\Phi|_{H^k}=\phi$ maps $H^k$ into $X$. We conclude that $d\phi_0(H^k)=d\psi_0(H^k)$.
A: This problem is very similar to one of mine. So I just take my solution and modify a little bit. Hope it helps.
Let $ω:V→X$ be a local parametrization with $ω(0)=x$ and $V$ is also a an open subset of $H^k$.  If we shrink both $U$ and $V$ small enough, we will have $ϕ(U)=ω(V)$. Then we have the map $g=ω^{-1} o ϕ:U→V$ is diffeomorphism. From this if we write $ϕ=ω o g$ then take derivative both sides , we have $dϕ_o=dω_o  o dg_o$. This mean that  $im(dω_o )⊂im(dϕ_o )$
Repeat the above process but swap $ϕ$ and $ω$ to each other, we will have $im(dϕ_o )⊂im(dω_o )$. So $im(dω_o )=im(dϕ_o )$, thus $dϕ_o (R^k)=dω_o (R^k)$ implies  $dϕ_o (H^k)=dω_o (H^k)$. From the definition of $H_x (X)$, we can conclude that $H_x (X)$ does not depend on the choice of local parametrization.
