Can $S^2$ be homeomorphic to a simplicial complex with fewer than 3 two-simplices? I think I can see why $S^2$ is homeomorphic to a simplicial complex with four 2-simplices (for example, it can be obtained from the tetrahedron). Can $S^2$ be homeomorphic to a simplicial complex with three or two 2-simplices? I think not and tried proving this using Euler characteristic but can't come up with a proof. Any help is appreciated.
 A: One combinatorial way to do this is by using the Euler characteristic 
$$\chi(S^2) = V - E + F$$
as you suggested, where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number faces in your simplicial complex. Since $\chi$ is a topological invariant, this number is the same for each triangulation, and also for each CW-complex. For a $2$-sphere, we have $\chi(S^2)=2$, which we can see by triangulating $S^2$ as the boundary of a tetrahedron, or alternatively, by equipping it with the CW complex with only one $0$-cell and one $2$-cell.
Now each triangle has three edges, while any edge belongs to exactly two triangles in our complex, due to the local topology at a point in the interior of an edge (If $x$ is such a point, then it has a Euclidean neighborhood if and only if exactly two triangles meet at the edge, such a simplicial complex is a pseudomanifold without boundary). That means $2E=3F$. Moreover, each triangle is determined by three vertices, thus $F\le\binom V3$. Now
$$2 = \chi(S^2) = V-F/2$$
If $F=2$, then $V=3$, thus $F\le\binom 33=1$. That means we must have at least four faces in our triangulation.

You can avoid using the Euler characteristic (especially its homotopy invariance) by arguing as follows:
Start with a $2$-simplex $\sigma$. Any edge $e$ of $\sigma$ must belong to another $2$-simplex $\tau_e$. But that means $\tau_e\ne\tau_{e'}$ when $e'$ is another edge of $\sigma$, since otherwise $\tau$ would have all vertices in common with $\sigma$. So we need at least three further $2$-simplices.
Actually the $\chi(S^2)$ can be ignored completely in the first argument when we instead use the relation $V\le E$ which holds in any triangulation of a surface. Then $V\le E = 3F/2$ implies for $F=2$ that $V=3$, thus $F=1$. But $F$ must be even, so $F\ge 4$. In other words, no closed surface can be triangulated with less than four triangles.
