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Let $M$ be a closed, connected, oriented 3-manifold. If $M$ is prime, then we know what the universal cover of $M$ looks like: it is either $S^3, \mathbb{R}^3$ or $S^2 \times \mathbb{R}$ depending on whether the 3-manifold is spherical, aspherical or $S^2 \times S^1$.

If $M$ is non-prime, what does its universal cover look like? It must be a simply connected, non-compact 3-manifold (without boundary), but I do not know whether they are well-understood.

Perhaps something more concrete: what is the universal cover of the connected sum of lens spaces?

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    $\begingroup$ For sums of lens spaces, usually the universal covers are the complements of Cantor sets in $S^3$. There are a few exceptions, but that describes most of them. You construct the cover explicitly, by thinking of the universal cover of a punctured lens space as a multiply-punctured sphere, then realizing you can do all the connect-sum operations in one ambient $S^3$. $\endgroup$
    – Ryan Budney
    Commented Mar 30, 2022 at 17:52
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    $\begingroup$ This general construction describes all the universal covers, including the exceptional cases, like $\Bbb RP^3$ sum $\Bbb RP^3$, whose universal cover is $\Bbb R \times S^2$. $\endgroup$
    – Ryan Budney
    Commented Mar 30, 2022 at 18:12

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Here is a version of Ryan's answer (in the comments).

Suppose that $M$ and $N$ are non-trivial lens spaces. Note that $M \# N$ is covered by $D_g$, the connect sum of $g$ copies of $T = S^2 \times S^1$, for some $g > 0$. (Also, $g = 1$ if and only if $M = N = \mathbb{RP}^3$).

Suppose that we are not in that special case. Then $g > 1$ and $D_g$ covers $D_2 = T \# T$. So it suffices to understand the universal cover of $D_2$. The manifold $D_2$ is obtained by doubling a genus two handlebody $U_2$. Thus the universal cover of $D_2$ is obtained by taking the universal cover of $U_2$, and doubling. However, the universal cover of $U_2$ is homeomorphic to a closed ball, minus a Cantor set from its boundary. Thus the universal cover of $D_2$ is, as claimed, a copy of $S^3$ minus a Cantor set from its equatorial two-sphere.

This is somewhat similar to my answer here.

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    $\begingroup$ Regarding the claim that $M\#N$ is covered by $D_g$: a closed orientable 3-manifold $Y$ with no aspherical factors in its prime decomposition has $\pi_1(Y)\cong F_l\ast Q_1\ast\dots\ast Q_k$ where $F_l$ is a free group of rank $l$ and $Q_1,\dots,Q_k$ are finite. The kernel of the natural homomorphism $\varphi:F_l\ast Q_1\ast\dots\ast Q_k\to Q_1\times\dots\times Q_k$ has finite index and is free of finite rank by the Kurosh subgroup theorem. So there is a finite covering $Z\to Y$ with $\pi_1(Z)\cong F_g$ for some $g$. The only closed orientable 3-manifold with fundamental group $F_g$ is $D_g$. $\endgroup$
    – Michael Albanese
    Commented Apr 4, 2022 at 4:26
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    $\begingroup$ @MichaelAlbanese - I had the following "direct" proof (avoiding groups and Kurosh) in mind. Namely, suppose that $A$ and $B$ are balls in $M$ and $N$. Then the universal cover of $M - A$ is a punctured sphere, as is the universal cover of $N - B$. We now take the correct number of copies of these and glue everything in a bipartite fashion. $\endgroup$
    – Sam Nead
    Commented Apr 4, 2022 at 18:49

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