Units of triangulation in general 3-manifold

Question

A caveat to the following question is that this might be a super stupid question for topologists :). Sorry that I am quite new to the field.

What I understand regarding triangulation for a surface (2-manifold) is that we can cover the surface with triangles of any shape, like the torus in the following figure, it is a triangulated torus.

Recently I had gone through a paper regarding the triangulation of 3-manifold:

It said every 3-manifold is triangulable (Theorem 3).

Based on these, I got 2 questions:

1. For a 2-manifold, we can recognize the "atomic" element in the triangulation as 2d triangle. Is it true that basic element in the triangulation of 3-manifold being tetrahedron?
2. (Concept Check) Is the triagulation in Moise's paper the same definition as that for the triangulation of a surface (that I understand)?

It will be perfect if someone can give a visualization of what it mean by triangulated 3-manifold with boundary in 3d space.

Answer 1

1. Yes, the fundamental units of a 3-manifold triangulation are tetrahedra.

2. Yes, almost -- you have to generalize the rules about "incidence" of adjacent tetrahedra carefully. For a surface, two triangles that intersect (1) might be identical, (2) might intersect in a segment that's an edge of each one, or (3) might intersect in a point which is a vertex of each one. For a 3-manifold, two tets that intersect (1) might be identical, (2) might intersect in a polygon that is a triangular face of each tet, (3) might intersect in a segment that's an edge of each tet, or (4) might intersect in a point which is a vertex of each tet.

An example of a triangulated (topological) manifold-with-boundary in 3-space:

Let $C$ be the unit cube $[0, 1] \times [0, 1] \times [0, 1]$; let $A = (0,0,0)$ and $Z = (1,1,1)$.

$C$ can be partitioned into six "cells" by "order" as follows: consider the rule $x \ge y \ge z$. Certain points $(x,y,z) \in C$ satisfy this rule. There are five other possible orders, like $y \le z \le x$ or $x \le z \le y$, etc.

You can spent a little time drawing to see that each of the cells I've described is a tetrahedron in 3-space. All six cells share the edge $AZ$. The intersection of any two cells is either $AZ$, or (in certain cases, like $C_{x \le y \le z} \cap C_{y \le x \le z}$), it's a triangle. In this example, the triangle consists of all triples $(x,y,z)$ such that $x = y$ and $x < z$ and $0 \le x,y,z,\le 1$.

So the cells constitute a triangulation of the cube.