I'm trying to understand the 2 spin structures on the circle. Since the frame bundle for the circle is just the circle itself, Spin structures on $S^1$ correspond to double covers of $S^1$. There are two choices: the connected double cover and the disconnected double cover.

From the point of view of Spin cobordism, we can view the circle as the boundary of the disk in the plane. The disk has a unique spin structure, and we can ask which spin structure this induces on the boundary.

Lawson/Michelson's "Spin Geometry" claims that this induces the spin structure coming from the double cover, but I'm having trouble seeing that. The frame bundle for the disk $D^2$ must be trivial, and thus isomorphic to $D^2\times SO(2) = D^2 \times S^1.$ There is a natural double cover given again by $D^2 \times S^1,$ and the map is just the identity on $D^2$ and $z \rightarrow z^2$ on the $S^1$ factor. 

To see what the induced spin structure on the boundary is, we must view the frame bundle of the boundary as sitting inside the frame bundle of $D^2\times S^1$ by fixing an outward normal vector field and then using it to complete any frame on $S^1$ to a frame on $D^2.$ To me, this seems to say that we view the frame bundle of $S^1$ (which is itself $S^1)$ as $S^1\times \{1\} \subset D^2 \times S^1,$ since once we fix one vector of a frame (in this case given by the normal) the other is entirely determined since we are in 2 dimensions. 

But now if we look at the inverse image of that in the double cover, we appear to get two disjoint copies of $S^1,$ i.e. the disconnected double cover. What am I doing wrong?

(This is crossposted on stack exchange as http://math.stackexchange.com/questions/245480/spin-structures-on-s1-and-spin-cobordism).