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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!

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It's not clear what your problem is. The $0$-th face of $\Delta_2$ maps $\Delta_1=(e_0,e_1)$ onto the segment $(e_1,e_2)$, and that's all. – Andrea Mori Jul 29 '11 at 8:47
The triangle has 3 faces, can I only apply the $i$-th face map to face $i$? That makes more sense but it's not explicitly stated in the book. – Rudy the Reindeer Jul 29 '11 at 8:58
The triangle has 3 "faces" (edges, indeed). The zeroth is the one just described, the first maps $(e_0,e_1)$ to $(e_0,e_2)$ and the second maps $(e_0,e_1)$ to $(e_0,e_1)$. – Andrea Mori Jul 29 '11 at 9:21
up vote 1 down vote accepted

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.

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