Show that the outward unit normal is smooth vector field Exercise 2.1.8 of Guillemin and Pollack asks us to prove that if $X^m \subset \mathbb{R}^n$ is an embedded submanifold with boundary, then the outward unit normal to $\partial X$ is a smooth function $\partial X \to \mathbb{R}^n$. 
By a partition of unity we can get a smooth vector field consisting of outward-pointing vectors. Then intuitively, we want to project from $T_x(X)$ onto $T_x(\partial X)^\perp \subset \mathbb{R}^n$ at each $x \in \partial X$, and then normalize. What's troubling me is demonstrating that the projection is a smooth operation. Normally I'd just like to take slice charts for $X$ and $\partial X$, but these don't necessarily preserve orthogonality. 
(I think this would all be OK if I was comfortable thinking about the induced Riemannian metric on $X$ and $\partial X$ from $\mathbb{R}^n$, but I haven't gotten that far yet.)
 A: We know that if $X \subset \mathbb{R}^N$ is a $k$-dimensional manifold, then $T_x(X)$ is a $k$-dimensional linear subspace of $\mathbb{R}^N$. Moreover, $\partial X$ is a $(k-1)$-dimensional manifold submanifold, so $T_x(\partial X)$ is a $(k-1)$-dimensional subspace of $T_x(X)$. So the orthogonal complement $T_x(\partial X)^\perp$ in $T_x(X)$ has dimension 1, and thus contains exactly 2 vectors of unit length, which we will call $u$ and $-u$. If $\phi$ is a parametrization of $x \in \partial X$ with $\phi(0) = x$, then $T_x(X) = \text{Im}\,d\phi_0$, so there must exist some vector $w$ in $\mathbb{R}^k$ such that $d\phi_0(w) = u$. By linearity, we have that $d\phi_0(-w) = -u$, so that if $u \in H_x(X)$, then $w \in H^k$, and $-w \notin H^k$, so that $-u \notin H_x(X)$. Similarly, $-u \in H_x(X)$ implies $u \notin H_x(X)$.
Let $\vec{n}(x)$ be the outward pointing normal. We first note that in the proof that $\partial X$ is a submanifold of $X$, we parametrized $\partial X$ by $\partial \phi: \partial H^k \to \partial X$. Hence, $T_x(\partial X)$ is the span of $d\phi_{\phi^{-1}(x)}(e_1)$, $\dots$ , $d\phi_{\phi^{-1}(x)}(e_{k-1})$, where $e_1, \dots, e_k$ are the standard basis vectors for $\mathbb{R}^k$. In particular, as $d\phi_p$ depends smoothly on $p$, and $\phi^{-1}(x)$ depends smoothly on $x$, then $T_x(\partial X)$ is the span of $k-1$ vectors that depend smoothly on $x$.
Now, we have that $d\phi_{\phi^{-1}(x)}(-e_k)$ also depends smoothly on $x$ for the same reasons, and by the definition and well-definedness of $H_x(X)$, we know that $d\phi_{\phi^{-1}(x)}(-e_k)$ is a vector in $T_x(X)$ that does not lie in $H_x(X)$. If $d\phi_{\phi^{-1}(x)}(-e_k)$ is orthogonal to $T_x(\partial X)$, we are done, but generally that will not be the case. If not, then we apply Gram-Schmidt:$$\vec{n}'(x) = d\phi_{\phi^{-1}(x)}(-e_k) - {{d\phi_{\phi^{-1}(x)}(-e_k) \cdot d\phi_{\phi^{-1}(x)}(-e_1)}\over{d\phi_{\phi^{-1}(x)}(-e_1) \cdot d\phi_{\phi^{-1}(x)}(-e_1)}}d\phi_{\phi^{-1}(x)}(-e_1) - \dots$$$$- {{d\phi_{\phi^{-1}(x)}(-e_k) \cdot d\phi_{\phi^{-1}(x)}(-e_{k-1})}\over{d\phi_{\phi^{-1}(x)}(-e_{k-1}) \cdot d\phi_{\phi^{-1}(x)}(-e_{k-1})}}d\phi_{\phi^{-1}(x)} (-e_{k-1}).$$Then take $\vec{n}(x) = \vec{n}'(x)/|\vec{n}'(x)|$. This is a smooth function in $x$, as each of the $d\phi_{\phi^{-1}(x)}(-e_j)$ are smooth functions of $x$.
A: The projection can be obtained as follows: choose a smooth set of vector fields $X_i$, $1\le i < m$ spanning  $T \partial X$, and use Gram Schmitt to orthonormalize them -- I will use the same letter to denote the result, for simplicity. Then consider any smooth vector field $Y$ in $TX$, transversal to $\partial X$ along $\partial X$, and consider
$$Z:= Y-\sum_i <X_i, Y>X_i$$ 
This will be normal to $\partial X$ in $TX$ and everywhere nonzero. Then normalize $Z$. All operations are obvously smooth (why?), so the result is smooth.
This is of course a sketch, you still need to show that all the ingredients I used exist, for example, but that should not be too difficult.
