$U(1)$ principal bundle over $\mathbb{S}^1$ In this question, the accepted answer uses the fact that the only $U(1)$ principal bundle over $\mathbb{S}^1$ is the trivial bundle. I'm not quite familiar with the classification of bundles, so I don't quite understand why this is true. Any pointers or sketch of the proof would be appreciated.
 A: Because such a thing corresponds to a (homotopy class of a) map $S^1 \to BU(1)$, and $BU(1)$ is simply-connected, meaning that every continuous map from the circle is homotopic to every other one.
Another way to see it is via the clutching function $c\colon \{\pm 1\}\to U(1)$, and continuous deformation of clutching functions give isomorphic bundles. So you can deform $c$ to be constant at 1, hence giving the trivial bundle. An even smaller construction is this: for any (topological) group $G$, you can pull back a principal $G$-bundle on $S^1$ along the surjective exponential map $[0,2\pi]\to S^1$. Every $G$-bundle on an interval has a section as it is trivialisable. Evaluate this section at $0$ and $2\pi$, then see where these two points go under the map back to the total space of the bundle. They are in the same fibre, so differ by an element of $G$. Continuously deforming your original section traces out a path in $G$. The original bundle can be reconstructed up to isomorphism from this one element, and points in the same path components of $G$ give rise to isomorphic bundles. As $U(1)$ is connected, only one bundle up to isomorphism is possible.
Another way to see it is because smooth/topological principal $U(1)$-bundles on $X$ are classified up to isomorphism by elements of the cohomology group $H^2(X,\mathbb{Z})$. For any manifold $X$ of dimension $n$, and $k>n$, $H^k(X,\mathbb{Z})=0$, so for $X=S^1$, $H^2(S^1,\mathbb{Z})$ has only one element.
