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There is one sentence I don't understand in some paper.

"A simply connected and conformally flat three mainifold can be conformally immersed into $S^3$" by the means of a developing map.

Is any reference about this short argument? Maybe it is a direct consequence from definition. Could anyone explain a little bit to me?

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    $\begingroup$ The general story is explained in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, but the argument is precisely Petrunin's below. $\endgroup$
    – Ben McKay
    Commented Nov 28, 2018 at 17:00

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Since $\mathbb{S}^3$ is conformally flat, we can think that any point of our manifold $M$ admits a neighborhood that is conformally equivalent to an open set in $\mathbb{S}^3$.

If two such neighbohoods overlap then the corresponding gluing map between corresponding open sets in $\mathbb{S}^3$ is a composition of inversions (Liouville's theorem).

So after applying a composition of inversions to one of the open sets you can assume that the gluing map is identity. This way you can extend the parametrization of a neighborhood to an immersed neighborhood of any path.

The composition of inversions at a neighborhhod of the end point of path depends continuously on the path and therefore has to be the same for homotopic paths. Since $M$ is simply connected it gives a well defined conformal immersion $M\hookrightarrow \mathbb{S}^3$.

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