# Simplicial maps commute with face maps (that delete vertex)?

Let $f: K\to L$ be a simplicial map. And let $d_i([v_0,\dots,v_n]=[v_0,\dots,\widehat{v_i},\dots,v_n]$, i.e. delete the $i$th vertex of the simplex $[v_0,\dots,v_n]$.

Is it true that for any simplex $\sigma\in K$, we have $f(d_i(\sigma))=d_i(f(\sigma))$?

Thanks for any help.

Are you working with simplicial complexes or simplicial sets? With simplicial sets, the answer is certainly yes. With simplicial complexes, the answer is no: let $K=L=1-\text{simplex}$ with vertices $0$ and $1$. Then the map interchanging the vertices is simplicial but does not commute with the face maps.
If you work with simplicial complexes in which the vertices are ordered and the maps must preserve the ordering, then I think $f d_i = d_i f$ for such maps.
• Just to ask why in your example the map interchanging the vertices is simplicial but does not commute with the face maps? I tried the possibilities like $f(d_1[v_0,v_1])=f([v_0])=v_1$ and $d_1f([v_0,v_1])=d_1([v_1,v_0])=v_1$. It seems to work? Thanks. – yoyostein Jun 9 '17 at 4:10
• When you apply $d_1$ to the face $[v_1, v_0] = [v_0, v_1]$, you should get $v_0$, not $v_1$. Or are you viewing $[v_0, v_1]$ as being different from $[v_1, v_0]$? In a simplicial complex, it is often the convention that the unordered vertices determine each simplex, in which case $[v_0, v_1] = [v_1, v_0]$. If one keeps track of the orientation (= ordering of vertices), then it is common to only consider simplicial maps which preserve this ordering, so the $f$ under consideration would not work. – John Palmieri Jun 9 '17 at 15:38
• And maybe your question really comes down to: how do you define the face maps $d_i$? The domain should be the set of $n$-simplices, so what is that set? Does it keep track of orientations, etc.? – John Palmieri Jun 9 '17 at 16:06