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 $L$?
Best Answer
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 $L'$".