Simplex and face map Let $F_{i,p}: \Delta_{p-1} \rightarrow \Delta_{p}$ denote the $i$-th face map i.e. the map that maps $e_0 \mapsto e_0$,$\dots$,$e_i \mapsto e_{i+1}$,$\dots$,$e_{p-1} \mapsto e_p$.
Let's consider $\Delta_2$ (the triangle) and let its vertices be $e_0, e_1, e_2$.
Now I want to write down $F_{0,2}$:
$F_{0,2}(e_0, e_1) = e_1 e_2$, OK
$F_{0,2}(e_1, e_2) = e_2 (?)$
$F_{0,2} (e_0, e_2)= e_0 (?)$
Can someone tell me what happens in the other two cases? It seems the definition of face map can't cope with those cases. Thanks for your help!
 A: I think what causes the confusion here is that you have already identified the $\Delta_{p-1}$ as a subspace of $\Delta_{p}$ but the whole purpose of the face map is to make this identification formal.
So, suppose you have a simplex $\Delta_{p-1} = (f_0, \ldots, f_{p-2})$ in your hand and you are trying to find a way to map it to a simplex $\Delta_p = (e_0, \ldots, e_{p-1}$ standing your table (I am being overly talkative here just to ellucidate that these two objects are completely distinct). Now, pick $0 \leq i \leq p-1$ and consider the object $\tilde \Delta_{p-1} = (e_0, \ldots, e_{i-1}, e_{i+1}, e_{p-1})$. This is again a $(p-1)$-simplex and we have a natural identification $F_i : \Delta_{p-1} \to \tilde \Delta_{p-1} \subset \Delta_p$.
For example, let $(f_0, f_1)$ be a line. You can map this to the triangle $(e_0, e_1, e_2)$ in three different ways by setting aside one vertex $e_i$, $0 \leq i \leq 2$ which leaves you with ${3 \choose 2} = 3$ possibilities for the choice of the two other vertices defining one the triangles's faces.
