I have been reading "An Introduction to Computational Topology" by Herbert Edelsbrunner and John Harer and they give the following exercise question in the second chapter of the book on "Surfaces":
An irreducible triangulation is one in which every edge contraction changes its topological type. Prove that the only irreducible triangulation of $\mathbb{S}^2$ is the boundary of the tetrahedron, which consists of four triangles sharing six edges and four vertices.
The define edge contraction on pg. 52 by
Let $a$ and $b$ be two vertices and $ab$ the connecting edge in (a triangulation) $K$. By the contraction of $ab$ we mean the operation that identifies $a$ with $b$ and removes duplicates from the triangulation. Calling the new vertex $c$, we get the new triangulation $L$ from $K$ by removing $ab$, $abx$, and $aby$; substituting $c$ for $a$ and for $b$ wherever they occur in the remaining set of vertices, edges, and triangles; removing resulting duplications making sure $L$ is a set.
Proving that the boundary of the tetrahedron is an irreducible triangulation is pretty straightforward, however, I cannot figure out how to prove that it is the ONLY irreducible triangulation.
I think the best approach that I've been able to come up with is to use that the Euler Character of $\chi(\mathbb{S}^2) = 2$ and possibly use some facts related to the Classification of Surfaces. Specifically, if $K$ is an irreducible triangulation of $\mathbb{S}^2$ different than that of the boundary of the tetrahedron, then it must have at least $6$ faces or more. Otherwise, its Euler Characteristic will change or it would be identically the boundary of the tetrahedron. Furthermore, any edge contraction of $K$ to $K'$ will still preserve that $K'$ is connected. If I can show that $K'$ must still be orientable then I think should be finished.
Any help is much appreciated!