An exercise in differential topology Problem: Given a smooth submanifold $M\subset\mathbb{R}^k$, show that the tangent bundle space $$TM=\{(x,v)\in M\times\mathbb{R}^k:v\in TM_x\}$$ is also a smooth manifold. Show that any smooth map $f:M\rightarrow N$ gives rise to a smooth map $$df:TM\rightarrow TN$$ where $$d(\text{identity})=\text{identity},d(g\circ f)=(dg)\circ(df).$$
This is a exercise in "Topology from the differentiable viewpoint" by John Milnor. My question is, the smooth manifold $M$ is a subset of $\mathbb{R}^k$, whose dimension may be another integer, saying $n(<k)$. Then how can the tangent vector $v$ lie in $\mathbb{R}^k$?

One way to define the tangent space to a submanifold $M\subset \Bbb R^k$ at some point $x$ is to consider the set of all derivatives at $t=0$ of all smooth curves $f:\Bbb R\to M$ such that $f(0)=x$. As such, the tangent space $T_xM$ is indeed defined as a subspace of $\Bbb R^k$ (whose dimension is the dimension of $M$).

As an example, the tangent space to the standard $2$-sphere in $\Bbb R^3$ at some point $x$ "is" the subspace of $\Bbb R^3$ defined by all vectors that are orthogonal to $x$. It is isomorphic to $\Bbb R^2$, but it is still defined as a subspace of $\Bbb R^3$.
