TITLE: Exercise about tangent space
QUESTION [5 upvotes]: I have just started of learning about Manifolds from Milnor's book.I am stuck on following exercise,give me some idea. 
Let $U$ be an open subset of a manifold $X$. Show that for any $x \in U$ the
tangent space at $x$ to $U$, i.e., $T_x(U)$ is same as $T_xX$.

REPLY [2 votes]: As you likely know, the topology of a manifold embedded in $\Bbb{R}^n$ is the subspace topology.  This means that a set $A$ is open in $X$ if and only if there is a set $B$ open in $\Bbb{R}^n$ such that $A = B \cap X$.
First note that $T_x U \subseteq T_x X$ in the most straightforward way:  Given a local diffeomorphism $f: W \to V$ with $W$ open in $\Bbb{R}^k$, $V$ open in $U$, and $f(0) = x$, we can identify $T_x U = \operatorname{im} df_0$.  But, we can also consider $V$ as an open set in $X$, so the same parametrization allows us to define $T_x X = \operatorname{im} df_0$.
If, on the other hand, we begin with $f: W \to V$ a local parametrization around $x \in X$, then we must restrict $f$ to $f' = f|_{f^{-1}(V \cap U)}$ so that its a diffeomorphism onto $V \cap U$.  The derivative has a local definition so $T_x U = df'_0 = df_0 = T_x X$.