Universal cover of a $3$-manifold is isometric to a cylinder

Question

Let $(M^3,g)$ be a compact, connected and orientable Riemannian $3$-manifold with boundary. Suppose that the universal cover of $M$ is isometric to the right cylinder $\tilde{M} = D \times \mathbb{R}$, where $D$ is a geodesic ball of the round unit sphere $\mathbb{S}^2$. Is it true to assert that $M$ is isometric to $D \times \mathbb{S}^1(r)$, where $\mathbb{S}^1(r)$ denotes the round circle of radius $r$, for some $r > 0$?

Answer 1

In this situation, you have to have $D=S^2$ since otherwise $\tilde{M}$ is incomplete which is impossible. The manifold $M$ need not even be homeomorphic to the product: It can be the total space of a nontrivial circle bundle over the projective plane.

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Since in the edit you indicated that $M$ is a manifold with boundary, if the boundary is nonempty, $M$ has to be homeomorphic to $D\times S^1$, where $D$ is the closed 2-disk. However, it need not be isometric to a product, just take an infinite cyclic group $G$ acting on $D\times R$ freely and generated by a skew motion (with nontrivial rotational component). Then $M:= (D\times R)/G$ is not isometric to a product.