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. 



Does geometrization of Alexandrov 3-spaces follow from that of 3-orbifolds?
In principle, yes. Note, however, that topologically singular Alexandrov 3-spaces are homeomorphic to non-orientable orbifolds. We could not find an appropriate reference for the geometrization in the non-orientable case (where topological singularities are present). To the best of my knowledge, the geometrization statements in the literature are for orientable orbifolds.


