The unit tangent bundle for submanifold $M^{m}\subset \mathbb{R}^{n}$ is a (2m-1)-dim submanifold This is homework so no answers please
Here is the problem: Show that $UM:=\{(x,v)\in T\mathbb{R}^{n}:x\in M^{m}, v\in T_{x}M^{m},|v|=1\}$ is a  (2m-1)-dim submanifold of $T\mathbb{R}^{n}$.
My attempt is ridiculously long, so I was wondering if I can get hints for a shorter one.
Also, is there a rigorous way to shrink open sets? At location (***), I need to shrink an open set $\pi^{-1}(U)$, to fit its image into another open set V. The problem is that both sets $\pi^{-1}(U)$, V are arbitrary and so I am not sure how to do rigorous shrinking.
Here is my attempt:
The goal is to show $UM$ satisfies the (2m-1) local slice criterion in $T\mathbb{R}^{n}$ for chart $(W,f)$. We will build f as a composition of charts.
$\blacktriangleright$ Since $M^{m}$ is an embedded submanifold, for each $p\in M$ there exists chart $(U,\phi)$ in $\mathbb{R}^{n}$ containing it s.t. $\phi(U\cap M)=\{(x_{1},...,x_{n})\in U: x_{m+1}=...=x_{n}=0\}$. The associated map for the tangent bundle is:
$\widetilde{\phi}:\pi^{-1}(U)\to \mathbb{R}^{2n}$ defined as $\widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=(x_{1},...,x_{n},v_{1},...,v_{n})$, where $\pi:T\mathbb{R}^{n}\to \mathbb{R}^{n}$. 
Therefore, for $p\in M$ we have $\widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m},0,...,0)$.
$\blacktriangleright$ The idea is to compose $\widetilde{\phi}$ by diffeomorphic map $(v_{1},...,v_{m})\mapsto (v_{1},...,v_{m-1},0)$, where $|(v_{1},...,v_{m})|=1\Leftrightarrow (v_{1},...,v_{m})\in \mathbb{S}^{m-1}$, and so we get a chart on $\pi^{-1}(U)\cap UM$.
The $\mathbb{S}^{m-1}$ is an embedded (m-1)-dim submanifold  of $\mathbb{R}^{m}$ and so we have some $(V,\psi)$ s.t. $\psi(V\cap \mathbb{S}^{m-1})=\{(x_{1},...,x_{m})\in V: x_{m}=0\}$. Then define $g:\mathbb{R}^{n}\times V\times \mathbb{R}^{n}\to \mathbb{R}^{n}$ as: 
$$g(x_{1},...,x_{n},v_{1},...,v_{n})=(x_{1},...,x_{n},\psi(v_{1},...,v_{m}),v_{m+1},...,v_{n})$$.
The map g is a diffeomorphism as it is the identity in $\mathbb{R}^{n}\times \mathbb{R}^{n}$  and the diffeomorphic chart $\psi$ on V.
$\blacktriangleright$ We will show that the desired map is $f:=g\circ \widetilde{\phi}$ and $W:=\pi^{-1}(U)\cap UM$.\
Well-defined: g's domain is $\mathbb{R}^{n}\times V\times \mathbb{R}^{n}$, so we need to make sure $\pi_{(n+1,n+m)}(\widetilde{\phi}(\pi^{-1}(U) ))\subset V$, where $\pi_{(n+1,n+m)}:\mathbb{R}^{2n}\to \mathbb{R}^{m}$ be the projection onto $n+1,...,n+m$ coordinates.  (***)
The map $g\circ \widetilde{\phi}:\pi^{-1}(U)\to \mathbb{R}^{2n}$ is a composition of diffeomorphisms and thus a coordinate chart on $\pi^{-1}(U)$. Now we need to show where it sents $UM\cap \pi^{-1}(U)$.
Derivations in $UM\cap \pi^{-1}(U)$ are tangent to M and so $\widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m},0,...,0)$. Since $(v_{1},...,v_{m})\in \mathbb{S}^{m-1}$, we get $\psi(v_{1},...,v_{m})=(v_{1},...,v_{m-1},0)$ and so 
$$g\circ \widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=g(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m},0,...,0)=$$
$$=(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m-1},0,0,...,0)$$.
Thus, $g\circ \widetilde{\phi}(UM\cap \pi^{-1}(U))=\{(x_{1},...,x_{n},v_{1},...,v_{n})\in g\circ \widetilde{\phi}(\pi^{-1}(U)): x_{m+1}=...=x_{n}=v_{m}=...=v_{n}=0\}$.
Thanks
 A: Let $F:M\to \Bbb R^n$ be the inclusion map and $dF: TM\to T\Bbb R^n$ the smooth map induced by $F$,
then we have $$dF:TM\to T\Bbb R^n,$$
$$(x,v)\mapsto (x,v).$$
$\forall x\in M$, we choose a smooth chart containing $x$ on $M$,
then $dF$ has the following coordinate representation in terms of natural coordinates for $TM$ and $T\Bbb R^n$:
$dF(x^1,\cdots,x^m,v^1,\cdots,v^m)=(F^1(x),\cdots,F^n(x),\frac{\partial F^1}{\partial x^i}(x)v^i,\cdots,\frac{\partial F^n}{\partial x^i}(x)v^i).$
We composite $dF:TM\to T\Bbb R^n$ with $T\Bbb R^n\to \Bbb R$ defined by $(x,v)\mapsto |v|^2$, then
we get $\Phi: TM\to \Bbb R$ defined by $(x,v)\mapsto |v|^2$. Correspondingly, $\Phi$ has the following coordinate representation:
$\Phi(x^1,\cdots,x^m,v^1,\cdots,v^m)=\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x)v^i)^2$.
Suppose $\Phi(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i)^2=1$.
Because we have
$\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n\frac{\partial F^k}{\partial x^1}(x_0)(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i),$
$\qquad \qquad \qquad \vdots$
$\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n\frac{\partial F^k}{\partial x^m}(x_0)(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i),$
then $v_0^1\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)+\cdots+v_0^m\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i)^2=2$,
so at least one of $\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m),\cdots,\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)$ is not equal to $0$, then $(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)$ is a regular point of $\Phi$ such that $\Phi(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=1$, hence $\Phi^{-1}(1)$ is a regular level set.
By Corollary 5.14(Regular Level Set Theorem) of Introduction to Smooth Manifolds by Lee, $UM=\Phi^{-1}(1)$ is an embedded $(2m-1)$-dimensional submanifold of $TM$, thus an embedded $(2m-1)$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$.
A: It's much easier to avoid the use of coordinate charts altogether and using the properties of level sets of smooth functions.
Hint Denote by $g$ the metric induced on $M$ by pulling back the Euclidean metric on $\Bbb R^n$ via the inclusion map $M \hookrightarrow \Bbb R^n$. We can identify $g$ with the smooth map
$$\hat g : TM \to \Bbb R , \qquad (p, X) \mapsto g_p(X, X)$$ that maps a vector $X$ to the square of its norm.

Additional hint By definition, $UM$ is the level set $\hat g^{-1}(1)$. So, if $1$ is a regular value of $\hat g$, that is, that $\hat g$ has constant rank $1$ on $UM$, then $UM$ is an embedded submanifold of $TM$ of codimension $1$.

Remark Notice that we only used the embedding to identify the metric on $M$. So, the embedding is irrelevant in the sense that the argument applies just as well to an abstract Riemannian manifold.
A: So, actually $(x,v)\in TM^m $ a manifold of dimension $2m $. Define a function $f:TM^m \rightarrow \mathbf {R} $ as $f (x,v)=|v|$; assuming everything is embedded in some big euclidean space. Check if pre-image theorem works.
A: The solution is wrong. 
$v_1^2+\dots+v_m^2=1$ is not equivalent to $|v|=1$, because the choice of the chart is random.
