Barycentric subdivisions and labeling of $(d-1)$-simplex I am trying to prove that it is always possible to label the vertices of the $k$-th barycentric subdivision of a $(d −1)$-dimensional simplex with labels
$1, 2, . . . , d$ such that each simplex present obtains all $d$ labels in its
vertices. 
For example, in the $2$ dimensional case, the first barycentric subdivision will yield $6$ inscribed triangles. 
I am thinking induction on both $k$ and $d$. Any thoughts?
 A: As corrected, this is fairly straightforward or impossible, depending on whether you want to have a certain condition satisfied: 


*

*If you want the original vertices to retain their labels (e.g., in a triangle with labels 1, 2, 3 on the vertices, the original vertices, after subdivision, must retain their labels), then it's impossible. Consider the 1D case of a segment with vertex labels 0 and 1: what label will you assign to the midpoint? 

*If you're willing to assign fresh labels to every vertex, then the rule is this:
Every vertex of the barycentric subdivision lies in some $p$-face of the original simplex, but no $p-1$-face. In a triangle, for instance, the original vertices lie in a 0-face of the triangle, but no $-1$-face. The midpoints of the edges lie in a 1-face, but in no $0$-face, and so on. (The edge midpoints lie in the $2$-face as well --- the whole triangle -- but since they also lie in a $1$-face, we know that $p$ is not 2.)
Alternatively expressed: aside from the original vertices, every subdivisions vertex $v$ lies in the interior of some $p$-face. For instance, the midpoints of the edges are in the interior of the edge, and while they also lie in the triangle, they lie on its boundary ... so for these edges, the value of $p$ is $1$. The midpoint of the triangle lies in the middle of the $2$-simplex, so for it, $p = 2$.
Now here's your labelling: Assign the label $0$ to each original vertex, and for each new vertex that lies in the interior of a $p$-face, assign the label $p$. 
Now every simplex of the subdivision contains all possible labels. In a triangle, for instance, the triangles of the subdivision each consist of an original vertex, an edge-midpoint, and the face-center, hence have labels 0, 1, 2. 
To answer the question you asked: you need to add 1 to each of the values I gave, i.e., assign labels 1, 2, 3 rather than 0, 1, 2. 
I don't know precisely how you've defined a simplex (geometrically? Abstractly?) or barycentric subdivision, so I cannot give a proof that this method works, but once you try it on a line-segment, triangle, and tetrahedron, you should be pretty convinced. :)
