On the proof of the simplicial decomposition theorem My question is on the proof of the simplecial decomposition theorem as found in the book by Fuchs and Fomenko in page 29. Specifically, I am struggling with the proof that the mapping the authors construct is a simplicial map, i.e. taking simplexes onto simplexes, and more precisely in this case, vertices to vertices.
The authors describe the setting.
We are given a mapping $f:K\longrightarrow L$, between finite simplicial complexes, and they denote by $L'$ the (first) barycentric subdivision of $L$. Then they let $K'$ be some $r$'th barycentric subdivision of $K$ and they define the map $f':K'\longrightarrow L$ by defining it first on the vertices $w'$ of $K'$. My issue is with their definition.
For a vertex $w'$, they set $f'(w')$ to be any vertex $v$ such that $f(w')\in St'(v)$, where $St'(v)$ is $v$'s star in $L'$. I assume, although not explicitly written, that $v$ is a vertex of $L$, for if otherwise $f'$ would not be a simplicial mapping between $K'$ to $L$.
My question is: why would there be a vertex $v$ of $L$ that satisfies $f(w')\in St'(v)$? Take for example $L$ to be a triangle (single 2-d simplex), then for each of its vertices, $St'(v)$ contains a quarter triangle, and the area in the middle of the triangle is not covered by any of $L$'s 3 vertices' stars in $L'$. Then how is a vertex $w'\in K'$ such that $f(w')$ is inside the middle of the triangle mapped to a vertex of $L$?
 A: In your example where $L$ is a single 2-d simplex, for each of the vertices $v \in L$, it is not accurate to refer to the set $St'(v)$ as a "quarter" of a triangle. But you can instead think of $St'(v)$ as a "third" of a triangle. If we denote the three vertices of $L$ as $v_1,v_2,v_3$ then, as seen here, each triangle of the first barycentric subdivision $L'$ contains exactly one of the $v_1,v_2,v_3$. Also, each vertex $v_i$ is contained in exactly two of the triangles of $L'$, and the union of the two triangles of $L'$ containing $v_i$ is exactly $St'(v)$. Thus we have
$$L = St'(v_1) \cup St'(v_2) \cup St'(v_3)
$$
In general, for any simplicial complex $L$, as $v$ varies over the vertices of $L$ we have
$$L = \bigcup_v St'(v)
$$
So for any vertex $w'$ of $K'$, since $f(w') \in L = \bigcup_v St'(v)$, it follows that there exists $v$ a vertex of $L$ such that $f(w') \in St'(v)$.
Perhaps you were thinking of $L''$, the second barycentric subdivision, when you wrote "the area in the middle of the triangle is not covered by any of $L$'s 3 vertices' stars in $L'$".
