# Prove there exists a outward unit normal field on the boundary this manifold

Let $$M$$ be a compact subset of $$\mathbb{R}^3$$ with the standard orientation $$\mu =[e_1,e_2,e_3]$$ and let $$S = \partial{M}$$ is its smooth boundary with the induced orientation from $$M$$. Prove there exists a smooth normal unit field $$\textbf{n}(x) = (n_1(x),n_2(x),n_3(x))$$ for $$x \in S$$ such that if {$$v_1,v_2$$} is a basis for $$T_xS$$ with [$$v_1,v_2$$] = $$\partial{\mu}$$ then [$$v_1,v_2,\textbf{n}(x)$$] = $$\mu$$. I know that given $$v_1$$ and $$v_2$$, I can construct a unit vector $$v_3$$ orthogonal to $$v_1$$ and $$v_2$$ using the cross product. I think this would imply that $$n_1 = v_1$$, $$n_2 = v_2$$ and $$n_3 = v_3$$ but how does this imply exactly that [$$v_1,v_2,\textbf{n}(x)$$] = [$$v_1,v_2,v_3$$] = $$\mu$$?

By your description, I assume your $$M$$ is a smooth orientable $$3$$-manifold. Then I think you can drop the compactness hypothesis. Indeed, since your $$M$$ is a manifold with boundary $$\partial M=S$$, then $$S$$ is a $$2$$-manifold without boundary in $$\mathbb{R}^3$$, or a usual surface. As you pointed out, it is orientable as it can be put on the boundary orientation inherited from $$M$$. But you don't have to specify orientation, just now translate your orientability to charts. One well-known definition of orientation on a manifold is that you have an oriented atlas, i.e a compatible atlas $$\{(\phi,U)\}$$ such that whenever $$U\cap V\neq\emptyset$$ for a pair of carts $$(\phi,U)$$, $$(\psi,V)$$ in the atlas, $$\det(\psi\circ \phi^{-1})|_{\phi(U\cap V)}>0$$. Use this definition for your surface $$S$$. You can use parametrizations (inverses of charts) in this case, as they may be more useful as maps from open subsets of $$\mathbb{R}^2$$ into $$S\cap \mathbb{R}^3$$.
Whenever $$(x,A)$$ is parametrization of $$U=x(A)\subset S$$ and $$p\in U$$, define $$\eta(p)=\frac{x_u\times x_v}{|x_u\times x_v|}(x(p))$$, where $$u,v$$ are the coordinates in $$A\subset \mathbb{R}^2$$. This is well defined at $$p$$ because since $$x$$ is a parametrization, its differential is one-one and hence $$x_u\times x_v\neq 0$$. Clearly it is unit and normal to $$T_pS=span\{x_u, x_v\}$$. It's also smooth on $$U$$ by definition. So, you are only left with proving that it doesn't depend on the parametrization used around $$p$$.
For this, observe that if $$(y,B)$$ is any other parametrizations of $$S$$ with $$p\in x(A)\cap y(B)$$ and $$\eta^{'}(p)=\frac{y_u\times y_v}{|y_u\times y_v|}(y(p))$$, since $$\eta(p), \eta^{'}(p)$$ are both orthogonal to $$T_pS$$ and unit, it can only happen $$\eta(p)=\pm \eta^{'}(p)$$. The plus sign is going to happen precisely because $$(y_u\times y_v)(y(p))=x_u\times x_v (x(p))\cdot \det(y^{-1}\circ x)|_{x^{-1}(p)}$$, and this $$\det$$ is positive by the choice of your parametrizations, or charts.
$$\textbf{Added}$$: this last step you can prove it by observing that the $$dx_{x^{-1}(p)}, dy_{y^{-1}(p)}: \mathbb{R}^2\rightarrow T_pS\subset \mathbb{R}^3$$ are isomorphisms, the cross product in $$\mathbb{R}^3$$ is bilinear and antisymmetric, and that the only (up to scalars) bilinear and antisymmetric map in $$\mathbb{R}^2$$ is $$\det$$. It takes some details, but that's an idea to prove it. Another is just brute force calculation.