An extension of PL triangulation of $S^k$ Let $\mathcal{T}$ be a piecewise linear triangulation of the sphere $S^k$. I am wondering is it possible to get an extension of $\mathcal{T}$ to a piecewise linear triangulation $\mathcal{T}'$ of the disk $B^{k+1}$ (so that $\mathcal{T}'$ coincides with $\mathcal{T}$ on the boundary of $B^{k+1}$) that satisfies one more condition:
if $u,v\in V(\mathcal{T}')\setminus V(\mathcal{T})$ and for two distinct $k$-dimensional simplices $A,B\in \mathcal{T}$ such that $\{u\}\cup A, \{v\}\cup B$ are in $\mathcal{T}'$, then $u\neq v$.
For example, given a triangle on $x,y,z$ that is a triangulation of $S^1$, for the extension, I don't want to just add $w$ in the center to barycentric subdivision the triangle to get an extension triangulation of $B^2$. (While we can add 3 more vertices $a,b,c$ and the new simplices are $xya,xzb,yzc,xab,zbc,yac,abc$.)
I feel the requirement of the extension can be satisfied: for example, for each $k$-dimensional simplex $A\in \mathcal{T}$, we add a new vertex $v_A$, (where in last example, $a$ is for $xy$, $b$ is for $xz$, etc) and then we try to fill with more simplices. While I don't know how to prove it rigorous.
 A: After adding those new vertices $v_A$ and the corresponding simplex $\{v_A\} \cup A$, one for each $k$-simplex $A$ of $S^k$, the part that is left untriangulated is homeomorphic to $B^{k+1}$ with a boundary triangulation that is simplicially isomorphic to a subdivision of the given triangulation of $S^k$, namely the subdivision in which each $k$ simplex is subdivided by adding an interior vertex and coning from that vertex to the boundary of the $k$-simplex.
So what you could do now is to simply cone off that subdivision, with a cone point $o$ (at the center of $B^{k+1}$, say). The result will still be a PL triangulation of $B^{k+1}$. For instance, since the link of $o$ is simplicially isomorphic to a subdivision of the given triangulation of $S^k$, that link is a PL $k$-sphere (here is where PL topology comes in to play: every subdivision of a PL triangulation is PL). You'll also have to argue that the links of the $v_A$ vertices are PL $k$-spheres, but I suspect that this can be checked after first describing those links in a completely direct manner that is expressed (perhaps) as a subdivision of the given triangulation of $S^k$.
So in your $B^2$ example, instead of adding a 2-simplex $abc$, you would instead add the new vertex $o$, new edges $oa$, $ob$, $oc$, $ox$, $oy$, $oz$, and new 2-simplices $oax$ etc.
