I want to prove a statement about the fixed points of simplicial maps.

If $f: |K|\to |K|$ is a simplicial map prove that the set of fixed points of $f$ is the polyhedron of a subcomplex of $K^1$ (where $K^1$ denotes the first barycentric subdivision of the simplicial complex $K$), though not necessarily of a subcomplex of $K$.

My ideas:

If a fixed point is a vertex of $K$ it is also a barycentre of a $0$-simplex in $K$.

If a fixed point lies in the interior of a simplex $S$ of $K$ then $f$ must take that simplex to itself. Since this induces a permutation on the vertices of $S$ I can prove that the barycentre $A_S$ is a fixed point too.

I can collect all the barycenters in a set $M$. Then $M$ is a subset of the vertices of $K^1$.

If I define $L$ to be the simplex that has the elements of $M$ as its vertices then each element of $L$ is a fixed point of $f$.

My problems with this construction are:

i) I don't think I can prove that each fixed point of $f$ is also in $L$.

ii) I don't see how to prove that $L$ is a simplex of $K^1$. If $M=\{A_0,...,A_k\}$ and $S_0,...,S_k$ are the corresponding simplexes I would have to prove that there is a permutation $\sigma$ of the integers $0,1,2,...,k$ such that

$S_{\sigma(0)} \lt S_{\sigma(1)} \lt ... \lt S_{\sigma(k)}$ where the notation $A \lt B$ means that the simplex $A$ has to be a face of the simplex $B$.

If I succeed in proving i) and ii) I can define the set $H=\{L\}$. By the above results this would be a subcomplex of $K^1$ such that the polyhedron $|H|$ is the set of fixed points of $f$. But I do not know how to prove i) and ii).