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Apr 4, 2022 at 18:49 comment added Sam Nead @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.
Apr 4, 2022 at 7:42 history edited Sam Nead CC BY-SA 4.0
Fixed paren
Apr 4, 2022 at 7:04 history edited Sam Nead CC BY-SA 4.0
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Apr 4, 2022 at 4:26 comment added Michael Albanese 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 $D_g$.
Mar 31, 2022 at 2:38 history edited Sam Nead CC BY-SA 4.0
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Mar 31, 2022 at 2:27 history answered Sam Nead CC BY-SA 4.0