Barycentric subdivision of an edge and triangle

The edge has nonempty faces $a,b,ab$, abbreviating braces. Its only length-two chains are $a<ab$ and $b<ab$, so the two subdivided edges meet at $b_{ab}=(a+b)\mathrm{/}2$. There are three singleton chains and one empty chain. Thus its numbers of faces in dimensions $-1,0,1$ are $(1,3,2)$.

The triangle has face labels $a,b,c,ab,ac,bc,abc$. The maximal chains are $a<ab<abc$, $b<ab<abc$, $a<ac<abc$, $c<ac<abc$, $b<bc<abc$, and $c<bc<abc$. The comparable pairs are six vertex-edge pairs, three vertex-triangle pairs, and three edge-triangle pairs, giving twelve edges. The seven singleton chains and one empty chain give face counts $(1,7,12,6)$. Barycenters place these six triangles inside the original triangle and the barycentric homeomorphism identifies their union with it. For a vertex-free complex there is only the empty chain, so the counts reduce to $(1,0,0,0)$.