My question is part of a question which has been asked a few years ago here: Barycentric subdivisions of simplices yield a simplicial complex, but has not been answered to my satisfaction:
The barycenter $b_\sigma$ of an $n$-simplex $\sigma$ with vertices $v_0, \ldots, v_n$ is the point $$\frac{1}{n+1}(v_0 + \cdots + v_n).$$ Show that:
(a) For any strictly increasing sequence $\sigma_0 \subset \sigma_1 \subset \ldots \subset \sigma_k$ of faces of $\sigma$, the barycenters $b_{\sigma_0}, b_{\sigma_1}, \ldots, b_{\sigma_k}$ form the vertices of a $k$-simplex.
(b) These simplices form a simplicial complex.
I already managed to proved (a) in all details, but I did not succeed with (b).
We have to show that if $\sigma_1, \sigma_2$ lie in the barycentric subdivision, then $\sigma_1 \cap \sigma_2$ is a face of both $\sigma_1$ and $\sigma_2$. The answer given to the original question is intuitively right: The intersection of the convex hulls of two sets of barycenters is exactly the convex hull of their intersection. But this is not true in general for all sets! However, it seems to be true in this special case.
Does anybody know a technically clean proof for this fact or a different approach which leads to a proof of (b)?