The free group $F_2$ acts faithfully on the circle $S^1$ by homeomorphisms. Someone told me that given any such action on $S^1$, we can lift it to a faithful action $F_2\curvearrowright \mathbb{R}$ by homeomorphisms. Why isn't there an obstruction? It seems to me like we should have to worry about lifts composing correctly.
2026-03-28 14:05:37.1774706737
Lifting Free Group Actions on $S^1$ to $\mathbb{R}$
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