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Other resolutions of can be obtained by blowing-up the divisors of the special fiber in a different order. Given any order on , we denote
The refinement of the fan of corresponding to is such that the skeleton as subspaces in the Berkovich space of ; it is independent on the chosen order so that we simply denote this subspace by . However, the models and induce in general different simplicial subdivisions and different retractions onto . For instance, the only edge in the interior of is , which indeed depends on the chosen order.
Here below we illustrate the skeletons and the Berkovich retractions in a couple of examples.
The blow-up of along the toric strata and yields a refinement of the fan which coincides with the fan of , constructed at the end of SectionΒ 4.3. It follows that the model dominates all resolutions independently on the order, hence all Berkovich retractions , and factors through .
Our goal is to construct a map , composing the Berkovich retraction with a collapse of the additional 3-cell and a combinatorial retraction
such that, given any vertex in ,
the restriction of over (the is taken with respect to the first barycentric subdivision, as in DefinitionΒ 3.2.1) is for any order on , i.e. any order where the index is the biggest. This guarantees that around each , the map is the Berkovich retraction induced by a small resolution where the strict transform of is isomorphic to , so that we are in the set-up of CorollaryΒ C.
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The retraction . We identify again the skeleton with the polyhedron in described in SectionΒ 4.3. On the convex hull of and , the retraction is given as follows
(4.4.1)
Here is a pictorial description for certain values of :
We extend the definition of to by symmetry along the medians of the triangles and . In particular, we note that the image of is the graph in of DefinitionΒ 3.2.1.
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The combinatorial retraction . We define the collapse as the projection of the additional -cell of onto along the -direction. We call the combinatorial retraction of the skeleton onto .
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Finally, we check that over . As the preimage of is disjoint from , we have to prove that . By symmetry of , it is enough to check this for . Over we have ; there, the expression of determined in Eq.Β 4.3.1 coincides with the definition of in Eq.Β 4.4.1, hence we conclude.