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S Apr 1, 2022 at 10:06 history suggested Daniel Asimov CC BY-SA 4.0
"the universal covering" —> "the universal cover of a compact manifold"
Mar 31, 2022 at 22:37 review Suggested edits
S Apr 1, 2022 at 10:06
Mar 31, 2022 at 16:41 vote accept Nick L
Mar 31, 2022 at 15:07 answer added Vitali Kapovitch timeline score: 35
Mar 31, 2022 at 14:04 comment added HJRW One simple observation is that, since the number of (Freudenthal) ends is an invariant of the fundamental group, if so then $\pi_1(M)$ has two ends, from which it follows that it has a finite-index subgroup isomorphic to $\mathbb{Z}$. Passing to the corresponding finite-sheeted cover, you may as well assume that $\pi_1(M)\cong\mathbb{Z}$.
Mar 31, 2022 at 12:49 history asked Nick L CC BY-SA 4.0