Orientable double-cover of the Mobius strip Let C0 be the cylinder, M1 the Mobius band (a cylinder with 1 twist), and C2 a cylinder with 2 twists (each embedded in R3).  
Although C0 and C2 are diffeomorphic, intuitively one should think that C2 is a better choice for the oriented double-cover of M1, as it should preserve the notion of the "twisting" of M1.  Clearly C0 is diffeomorphic to C2, but the two are not ambient isotopic within R3.  
How specifically can we state "C2 is the better choice for the oriented double-cover of M1, as opposed to C0"?  Is there an intrinsic way of doing so? Is there some way to phrase this in terms of cobordism?  
 A: I don't know whether there's a way to phrase it in terms of (co)bordism, but you might observe that the standard embedding of $C_1$ has a normal bundle; the tubular nhd theorem says that that bundle, too, can be embedded so that that the 0-section is exactly $C_1$. Taking the image of $[-1, 1]$ in each fiber, you get a 3-manifold with boundary; if you take the boundary in the fiber direction, you get an embedding of the double cover. Here's a picture, before and after taking the double-cover (crudely drawn, but it's the best I can do at 5 AM):
Before: 
After:
Now you can take that second surface, and claim (or try to claim) that it's ambient isotopic to the twice-twisted cylinder, but not to the untwisted cylinder. Experimenting with a piece of paper confirms this. (The isotopy to the twice-twisted cylinder is easy. To prove there's no isotopy to the untwisted cylinder, look at the linking number between the top and bottom edge of the cylinder. That's invariant under isotopy, but in the untwisted case it's zero, and in the double-twist case, it's two.)
