TITLE: Smallest number of $n$-simplices in a triangulation of the sphere
QUESTION [7 upvotes]: Let $X$ be a simplicial complex homeomorphic to $S^n$.
I proved that there must be at least $(n+2)$ vertices in $X$ and that there must be at least one $n$-simplex in $X$.
Now I want to prove that there are at least $(n+2)$ n-simplices in $X$.
My idea was to assume that there are fewer than $(n+2)$ n-simplices and then proving that the simplicial boundary map $\partial_n:C_n^{\Delta}(X)\to C_{n-1}^{\Delta}(X)$ is injective, contradicting $H_n^{\Delta}(X)=\Bbb Z$.
This quickly proved to be very messy and I am not sure if that's the best way to go about it.
I appreciate all help.

REPLY [4 votes]: As you said, there must be at least one $n$-simplex in $X$, call it $\sigma$. This simplex has $n+1$ faces $f_0,\dots, f_n,$ with the face $f_i$ being opposite to the vertex $v_i$ of $\sigma$. Due to the local topology of an $n$-manifold, each $f_i$ is the intersection of $\sigma$ and another $n$-simplex $\sigma_i$ which does not contain $v_i$. Since every other $\sigma_j$ intersects $\sigma$ in $f_j$, it contains $v_i$ and is thus distinct from $\sigma_i$. This shows that there are $n+2$ $n$-simplices $\sigma, \sigma_0,\sigma_1,\dots,\sigma_n$.