# Does geometrization of Alexandrov 3-spaces follow from that of 3-orbifolds?

Galaz-Garcia and Guijarro proved the geometrization of closed (compact, boundaryless) Alexandrov 3-spaces. Part of the strategy was to use the so-called ramified double cover $$\tilde{X}$$ of the space $$X$$. This ramified cover is a smooth $$3$$-manifold. Being this the case, the space $$X$$ would be isometric to a Riemannian $$3$$-orbifold.

I don't quite follow why then, it's not immediate that the geometrization of $$X$$ follows from the geometrization of $$3$$-orbifolds?

EDIT: Please note that My question is not whether geometrization can be deduced from "the geometrization of the ramified cover" but, once it has been deduced that the space itself is isometric to a Riemannian 3-orbifold, whether this information together with the geometrization of 3-orbifolds can be used to conclude.

• I don't follow either: since every three-manifold is a branched cover of $S^3$ why does not the geometrization theorem follow from the geometrization of $S^3?$ – Igor Rivin Sep 25 '18 at 23:05
• I think this would be missing the point. If I understand correctly you are implying that it is wrong to conclude the geometrization of the space from the geometrization of the cover. This is not what I'm asking. They use the cover to conclude that the space itself is a Riemannian 3-orbifold. My question is, why they cannot immediatly conclude from the geometrization of 3-orbifolds, a class of which apparently these spaces are elements of. Unless I'm completely missing the point. – Rp2s2 Sep 26 '18 at 1:14
• Ah, OK. I think it would be a good idea to include this excellent explanation of your question in the question itself... – Igor Rivin Sep 26 '18 at 2:05