# Orientation on a Manifold

Let M be an (n-1)-manifold in R^n . Let M(e) be the set of end-points of normal vectors (in both directions) of length e and suppose e is small enough so that M(e) is also an (n-1)-manifold. Show that M(e) is orientable (even if M is not)

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What definition of orientability do you want to use? FYI The manifold $M(e)$ is a special-case construction of what people like to call "the orientation cover" of a manifold. – Ryan Budney Nov 9 '10 at 23:26

Each point $p$ of $M(\epsilon)$ comes from a point $q$ in $M$. Consider the vector $X$ field on $M(\epsilon)$ which on $p$ takes the value $\vec{qp}$. Then $N$ is a non-zero normal field on $M(\epsilon)$ (maybe one needs to consider the projection of $N$ onto the normal line to $M(\epsilon)$ at each point, but that projection is surely non-zero and, moreover, to normalize this projection)
Dear Mariano, What if $M \subset \mathbb{R}^2$ is the union of two circles whose radii differ by $2\varepsilon$? In that case there are points $p \in M(\epsilon)$ which come from two points $p_1,p_2 \in M$. Also, how do you know that the resulting vector field is defined continuously? – user1337 Jun 10 '15 at 16:55
@user1337 I am assuming that $e$ is small enough so that doees not happen. – Mariano Suárez-Alvarez Jun 10 '15 at 22:28
Sorry to bother you again, but can you show that $\epsilon$ can be chosen so that it doesn't happen? (provided $M_\epsilon$ is indeed a manifold, of course). Also, could you please elaborate on why the said projection is non-zero? Thanks – user1337 Jun 13 '15 at 15:15