commutative, and the fourth we have already shown. So all the action takes place around the middle equality. We have shown that when $\beta^{*}\omega$ is transported to SO(3), it becomes the $d$ of something. So we may use Stokes’ theorem to go from left to right.

We can also see this easily from the swaddling map construction- let us choose a $\beta$ mapping, pasting two discs onto the boundaries of $C_{2}(S^{1})$, making it a compact space. Let us choose our discs such that corresponding points on $D_{1}$ and on $D_{0}$ map to antipodal points on $S^{2}$ via $\beta$. Cutting out a small neighbourhood of the centres of the discs, we may glue a cylinder between them, connecting them into a shape isomorphic to the cylinder on which our $\phi$ map was defined. Now every $\beta$ map can be smoothly extended to a $\sigma$ map, because the two discs with the narrow tube connecting them is homotopically a cylinder.

But as Tahl Novik observed, going from right to left in this set of equalities we have to watch out, because $\pi_{1}(SO(3))=\mathbb{Z}/2$, and for a path belonging to the non-trivial homotopy class of $SO(3)$, there can exist no pre-image via a $\sigma$ 