I will answer the question of whether this also gives the classification cheaply.

No.

It gives the classification at the expense of proving that every surface group has a free cocompact action on the upper half plane. This in turn depends upon proving:

• Every surface has a triangulation (Rado's Theorem, and now you've done 90% of the work of the classification theorem)...

• Every triangulated surface has a compatible smooth structure...

• Every smooth surface has a compatible conformal structure...

and, finally

• Every conformal structure has a compatible hyperbolic metric (the uniformization theorem).

I will answer the question of whether this also gives the classification cheaply.

No.

It gives the classification at the expense of proving that every surface group has a free cocompact action on the upper half plane (or on the euclidean plane, or on the 2-sphere). This in turn depends upon proving:

• Every surface has a triangulation (Rado's Theorem, and now you've done 90% of the work of the classification theorem)...

• Every triangulated surface has a compatible smooth structure...

• Every smooth surface has a compatible conformal structure...

and, finally

• Every conformal structure has a compatible hyperbolic metric, euclidean metric, or spherical metric (the uniformization theorem).