Suppose that $Z$ is a hyperspace in the oriented manifold $Y$, as sub manifold of codimension $1$. Prove that following statement are equivalent Suppose that $Z$ is a hyperspace in the oriented manifold $Y$, as sub manifold of codimension $1$. Prove  that following statement are equivalent
a) $Z$ is orientable
b)  There exists a smooth field of normal vector $\vec {n}(z)$ along $Z$ in $Y$
c)  The normal bundle $N(Z,Y)$ is trivial
d)  $Z$ is globally definable by an independent function; that is there is a smooth function θ on a neighborhood of $Z$ such that $θ^{-1} (0)=Z$, and dθ is nonzero at every point of $Z$
I want to prove them in order a) => b) =>c) => d) => a)
For a) => b)
Assume that $Z$ is orientable, then $Z$ can be given an orientation. Any orientation $Z$ induce a boundary orientation on $\partial Z$. At every point $z\in \partial Z$, $T_z(\partial Z)$ has codim 1 in $T_z(Z)$. Therefore there are precisely 2 univector in $T_z(Z)$ that are perpendicular to $T_z(\partial Z)$. Otherword if $h:U\to Z$ is a local parametrization around $z$, $U$ being openin $H^k$ and $h(0)=z$ then $(dh_0)^{1}:T_z(Z)\to R^k$ carries one unit normal vector into $H^k$ (the inward) and carries the other into $-H^k$ (the outward). The smoothness of orientation guarantee that there exist such function $h$, so there exists a smooth field of normal vector $\vec {n}(z)$ along $Z$ in $Y$
for b) =>c), I think it's kinda trivial, I'm not sure I should say anything about this part.
Form one of my previous exercise I have proved that $N(Z,Y)$ is trivial if and only if  there exists a set of $k$ independent global defined function $g_1 , \dots, g_k$ for $Z$ on some set $U$ in $Y$. So this cover c) => d)
I'm not sure how I should link d) back to a). 
 A: If $Z$ is orientable, as you mentioned, b),c) follow easily, since a choice of a local orientation of the tangent spaces of $Z$ and $Y$ induce locally an orientation on the normal bundle of $Z$, but since $Z$ and $Y$ are globally orientable, so is $N$, hence $N$ is isomorphic to a trivial line bundle. This also yields "c) implies d)", since you can find a tubular neighborhood $U$ of $Z$, i.e. a bicollared neighborhood such that $Z\times (-1,1) \to N(Z,Y) \to U \subset Y$ is a diffeomorphism. Now define a function $U \to \mathbb R$ such that $Z \times (-1,1) \to N(Z,Y) \to (-1,1)$ is precisely the projection onto the second factor, which gives d).
Now for "d) implies a)" just note that this pullback also pulls back an orientation. I.e. if you look closely at the regular value theorem you see that since $0 \in \mathbb R$ is a regular value of $U \to \mathbb R$, the preimage of $0$ is an orientable submanifold (which is actually oriented if you choose orientations for the other spaces involved).
