3-manifold equivalent of a fundamental polygon Can the idea behind the fundamental polygon of gluing instructions for edges be extended to 3-manifolds by gluing faces of a cube together? 
Specifically, take some fundamental polygon, "extrude it" and introduce some equivalence relation for the faces. Would this cube be homeomorphic to a fibration of $[0,1]$ or $S^1$ over the surface described by the original polygon? Does this process have a name?
 A: It sounds like you are simply asking for the concept of a fundamental polyhedron: start with a polyhedron, and then give instructions for gluing faces in pairs, using gluing maps which respect vertices and edges.
For example, the 3-torus $S^1 \times S^1 \times S^1$ can be obtained from the fundamental polyhedron $[0,1] \times [0,1] \times [0,1]$ by gluing opposite faces by translation: $(x,y,0) \sim (x,y,1)$; and $(x,0,z) \sim (x,1,z)$; and $(0,y,z) \sim (1,y,z)$. This example fits your idea of "extruding".
For a famous example which does not fit your "extruding" idea, start with a dodecahedron, which has 12 pentagonal sides. Arrange those sides in opposite pairs. Viewing each opposite pair with a birds-eye view, map the nearest face to the opposite by pushing it downwards and giving it a $\frac{1}{10}$ clockwise rotation. The result is a 3-manifold called the Poincare homology sphere, said to be the counterexample which led Poincare to correct his wrong conjecture (every closed 3-manifold with the homology of the 3-sphere is homeomorphic to the 3-sphere) to his right conjecture (every closed 3-manifold with the homotopy type of the 3-sphere is homeomorphic to the 3-sphere).
Unlike gluing side pairs of a fundamental polygon, which always produces a 2-manifold, gluing side pairs of a fundamental polyhedron is not guaranteed to produce a 3-manifold: the image under gluing of a vertex might not have the correct local topology for a 3-manifold.
