Exercise 5-15 from Spivak's Calculus on Manifolds I came across the following question (Exercise 5-15) in Spivak's Calculus on Manifolds and am not sure how to solve it.
Let $M$ be an $(n-1)$-dimensional manifold in $\mathbb{R}^n$. Let $M(\epsilon)$ be the set of end-points of normal vectors (in both directions) of length $\epsilon$ and suppose $\epsilon$ is small enough so that $M(\epsilon)$ is also an $(n-1)$-dimensional manifolds. Show that $M(\epsilon)$ is orientable (even if $M$ is not). What is $M(\epsilon)$ if $M$ is the Mobius strip?
I have thought about this problem for quite some time now and can't seem to figure it out. Also, how do we know that $M(\epsilon)$ is a manifold for sufficiently small $\epsilon$? Note that I am currently using Lee's Introduction to Smooth Manifolds, so any of the tools presented in that text can be used here.
UPDATE: This seems to be almost the same thing as the oriented double covering presented in Lee's text. Can that be applied here?
 A: Presumably $M$ must be compact, since there are non-compact submanifolds of $\Bbb R^n$ for which $M(\epsilon)$ need not be a manifold for any $\epsilon > 0$ (for example, $\left\{\left(x, \sin\frac 1x\right)\ \middle|\ x > 0\right\}$ in $\Bbb R^2$).
But if $M$ is compact, then the curvature of $M$ is bounded and disparate parts of $M$ have a minimum separation. So if $\epsilon$ is small enough, the offset of $M$ by $\epsilon$ will neither pinch nor intersect itself.
$M(\epsilon)$ offsets $M$ by $\epsilon$ in either direction. If $M$ is orientable, then $M(\epsilon)$ will be two disjoint copies of $M$ (near copies - there is some distortion from the offset), which will each be orientable as well.
If $M$ is not orientable, then we can see what happens by considering the Möbius strip. When you offset it, the two sides meet at the twist, so $M(\epsilon)$ is connected. But whereas $M$ has only one twist, $M(\epsilon)$ has two twists, one on each side of the twist in $M$. A strip with two twists is orientable. Think of sliding a normal vector of length $\epsilon$ along the Möbius strip. The tip of the vector is a point of $M(\epsilon)$. If you slide the vector all the way around the strip, you come back to the starting point, but with the normal pointing in the opposite direction. This is why the strip is not orientable. But the tip of the vector in $M(\epsilon)$ has not returned to its starting point. It now lies on the part of $M(\epsilon)$ opposite of where it started. To get it back to its starting point, you have to move the vector around the strip a second time, so it points in the same direction again, as does the normal to $M(\epsilon)$ at its tip. This is why $M(\epsilon)$ remains orientable. The same thing holds for any unorientable manifold.
