Framed bordism class of the circle has order $2$ As the question title says, how do I see that the framed bordism class of the circle has order two?
 A: This is just an amplification of the comments.  
Framed bordism is about manifolds $M$ equipped with a framing of the stable normal bundle.  It takes some work to spell out carefully exactly what that means.  However, if we have an embedding $i\colon M\to\mathbb{R}^n$ then we get a bundle $\nu_i$ whose fibre at $x\in M$ is the orthogonal complement to $i_*(T_xM)$ in $\mathbb{R}^n$, and a trivialization $\phi$ of $\nu_i$ certainly gives rise to a stable normal framing.  Now suppose that $\partial W=M$ and that $j\colon W\to\mathbb{R}^n$ extends $i$, and that $\psi$ is a trivialization of $\nu_j$.  For any point $x\in M$ we have an inward-pointing unit vector $u_x$ that lies in $j_*(T_xW)$ but is orthogonal to $j_*(T_xM)$.  Suppose that for all such $x$, the ordered basis $\phi(x)$ consists of $\psi(x)$  followed by $u_x$.  In this case, the triple $(W,j,\psi)$ gives rise to a nullbordism of $(M,i,\phi)$.
The nontrivial one-dimensional framed bordism class is represented by the circle together with a suitable stable normal framing.  If we let $i_1$ denote the usual embedding of $S^1$ in $\mathbb{R}^2$, and let $\phi_1(z)$ denote the outward pointing unit vector at $i_1(z)$, then this gives the required framing.  We can also put $i_2(z)=(i_1(z),0)\in\mathbb{R}^3$ and $\phi_2(z)=(\phi_1(z),(0,0,1))$; the pair $(i_2,\phi_2)$ then represents the same stable normal framing as $(i_1,\phi_1)$.  
Now let $W$ be the top half of a torus which cuts across the $xy$-plane orthogonally, so $\partial W$ consists of two disjoint circles in the $xy$-plane, and let $j\colon W\to\mathbb{R}^3$ be the inclusion.  Then $\nu_j$ is one-dimensional and has an obvious "outward-pointing" trivialization $\psi$.  This gives a nullbordism of two copies of $(S^1,i_2,\phi_2)$, as required.
