Hypersurface orientable if it admits a smooth normal vector field Let $X$ a codimension 1 smooth submanifold of the n-dimensional smooth manifold $Y$. Assume $Y$ is oriented. We want to show that $X$ is orientable if and only if it admits a global smooth normal vector field (in Y).
How can we prove this? I have no idea how to even begin...
 A: Hint:  For $p\in X$, let $U_p$, with coordinates $(x_1,...,x_n,t)$, be a slice chart around $p$ (meaning around $p$, $X$ corresponds to points where $t=0$).
Now, given your normal vector field $V$, orient $X\cap U_p$ by declaring the ordered basis $\{\partial_{x_i}\}$ to be positively oriented iff the ordered basis $\{\partial_{x_i}, v\}$ is positively oriented in $Y$.
Conversely, if $X$ is oriented, define $V = \partial_t$.
I'll leave it to you to prove that all this works.
A: Here's an alternative proof using some facts about the first Stiefel-Whitney class $w_1$.
We have a short exact sequence of vector bundles on $X$:
$$0 \to TX \to i^*TY \to \nu \to 0$$
where $i : X \to Y$ is the inclusion, and $\nu$ denotes the normal bundle of $X$ in $Y$. Therefore $w_1(i^*TY) = w_1(TX) + w_1(\nu)$. As $Y$ is orientable, $w_1(TY) = 0$ so $w_1(i^*TY) = i^*w_1(TY) = 0$ and hence $w_1(TX) = w_1(\nu)$. So $X$ is orientable if and only if $w_1(\nu) = 0$, but as $\nu$ is a line bundle ($X$ has codimension one), this is equivalent to $\nu$ being trivial. Therefore $X$ is orientable if and only if $\nu$ has a nowhere-zero section (i.e. $X$ admits a nowhere-zero normal vector field).
