I want to solve Prob. 10 of Chapter 8 from Milnor's Topology From a Differentiable Viewpoint.
The problem states that:
Let $M\subseteq \mathbf R^k$ be a smooth manifold of dimension $m$. Show that $$TM=\{(x,v)\in M\times \mathbf R^k\mid v\in T_xM\}$$ is a smooth manifold of dimension $2m$ in $\mathbf R^k\times \mathbf R^k$.
Milnor's book talks only about manifolds which are already subsets of Euclidean spaces. Any subset of a Euclidean spaces is either a manifold or it's not. So we do not have to give a smooth structure to $TM$ unlike we do in the study of abstract smooth manifolds.
ATTEMPT: There is a natural thing to try.
Let $g:U\subseteq \mathbf R^m\to W\cap M$ be a parameterization of $M$.
A natural candidate for a parameterization on $TM$ is the function $$g^*:U\times\mathbf R^m\to \mathbf R^k\times\mathbf R^k$$ defined as $$g^*(x,v)=(g(x),dg_x(v))$$ Clearly $g^*$ is a smooth function. Also, it can be seen that $\text{rank} \big(dg^*_{(x,v)}\big)=2m$ for all $(x,v)\in U\times\mathbf R^m$.
The image $g^*(U\times \mathbf R^m)$ of $g^*$ is contained in $TM$.
So only if we could show that the image $g^*(U\times \mathbf R^m)$ of $g^*$ is open in $TM$ then we'd be done but here I am stuck.
Can somebody help?
Thanks.
To show that $g^*(U\times \mathbb R^m)$ is open in $TM$, it suffices to show that $$g^*(U\times \mathbb R^m) = TM \cap (W\times\mathbb R^k),$$ because $W\times\mathbb R^k$ is open in $M\times\mathbb R^k$, and $TM$ has the subspace topology it inherits as a subset of $M\times\mathbb R^k$.
It follows from your definitions that $g^*(U\times \mathbb R^m) \subseteq TM \cap (W\times\mathbb R^k)$. To prove the reverse inclusion, let $(w,v)$ be an arbitrary element of $TM \cap (W\times\mathbb R^k)$. This means $w\in W$ and $v\in T_wM$. Since $g\colon U\to W$ is bijective, there is a point $x\in U$ such that $w=g(x)$. Since $dg_x\colon \mathbb R^m\to T_wM$ is bijective, there is a vector $y\in \mathbb R^m$ such that $v=dg_x(y)$. Thus $(w,v) = g^*(x,y)$.