This is exercise 1.2.2 on Guillemin and Pollack's Differential Topology
If $U$ is an open subset of the manifold $X$, check that $$T_x(U) = T_x(X) \text{ for } x \in U.$$
I am fairly confused with this problem, because I found the implicit definition of tangent space at Guillemin and Pollack's Differential Topology comes not very handy for this problem. In the book, tangent space is defined by $d\phi_0$, which is not directly defined:
By definition, the tangent space of $X$ at $x$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.
So here's my attempt:
$T_x(X)$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.
Because $U$ is an open subset of manifold $X$, so the best linear approximation at $x$ is the same as that of $X$ around $x$. Hence, $T_x(U)$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow U$ at 0. $\phi: V \rightarrow U$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$. Therefore, $T_x(U) = T_x(X) \text{ for } x \in U.$