Let $M$ be $S^3$ with $k$ disjoint open balls $D^3$ removed.
Can we classify all principal circle bundles over $M$ such that the total space is simply-connected?
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$\begingroup$ Let $e \in H^2$ be the Euler class. By combining the homotopy long exact sequence + Gysin sequence, I think that the situation is $\pi_1 = \mathbb{Z}$ exactly when $e=0$, $\pi_1 = \mathbb{Z}_n$ exactly when $e$ is divisible by $n$, $\pi_1 = 0$ exactly when $e$ is primitive. $\endgroup$– Nick LCommented Mar 13 at 14:04
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Let $e \in H^{2}(B)$ be the Euler class.
The relevant portion of the homotopy long exact sequence is $$0 \rightarrow \pi_{2}(E) \rightarrow \pi_{2}(B) \rightarrow \mathbb{Z} \rightarrow \pi_{1}(E) \rightarrow 0.$$
The relevant portion of the Gysin sequence is $$H^{0}(B) \rightarrow H^{2}(B) \rightarrow H^{2}(E),$$ where the first map is multiplication by $e$, second is induced by the projection to the base. By Hurewicz, the natural map $\pi_{2}(B) \rightarrow H_{2}(B)$ is surjective.
Combining these facts gives $\pi_{1}(E)= \mathbb{Z}, \mathbb{Z}_n, 0$ when $e$ is $0$, divisible by $n$ and primitive (respectively). Note that different $e$ might give diffeomorphic manifolds, it depends how the diffeomorphism group of $B$ acts on $H^{2}(B)$, this should be possible to work out.
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$\begingroup$ The diffeomorphism group acts by permutation on the boundary components, hence by the standard representation of the symmetric group - more precisely the action of $S^k$ on vectors in $\mathbb Z^k$ that sum to zero. $\endgroup$ Commented Mar 13 at 17:29
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$\begingroup$ Though, the permutation on the boundary components does not determine the diffeomorphism up to isotopy. Because (say) S^3 minus the interiors of five disjoint 3-disks is a 3-disk minus the interiors of four disjoint 3-disks, and two of the boundary 2-spheres can be twisted any number of times with respect to the other two. $\endgroup$ Commented Mar 13 at 19:58