# Show that the maps are chain homotopic

Let $\Delta _{2}$ be a 2-simplex, $I=\left [ 0,1 \right ]$. Given are two maps $i_{0}:\Delta _{2}\rightarrow \Delta _{2}\times I$, defined by $x \mapsto (x,0)$ and $i_{1}:\Delta _{2}\rightarrow \Delta _{2}\times I$, defined by $x \mapsto (x,1)$. Need to show that the induced chain maps are chain homotopic.

Since the two inclusions are homotopic, they map the 2-simplex to the upper and lower triangle of the prism $\Delta _{2}\times I$. The induced chain maps are $(i_{0})_{\#}:C(\Delta _{2})\rightarrow C(\Delta _{2}\times I)$ and $(i_{1})_{\#}:C(\Delta _{2})\rightarrow C(\Delta _{2}\times I)$. By definition of chain homotopy we have to look for a prism operator $P_{n}:C_{n}(\Delta_{2}) \rightarrow C_{n+1}(\Delta_{2} \times I)$ such that $\partial P+P\partial =(i_{1})_\#-(i_{0})_\#$. I am struggling to show formally the chain homotopy.

Can anybody help me, please? I appreciate any help and comments. Thank you very much!

• Are you working with singular or simplicial theory ? Commented Apr 26, 2016 at 21:21
• Simplicial theory Commented Apr 26, 2016 at 21:23
• Ok, so you have a triangulation of $\Delta_2\times I$ ? Commented Apr 26, 2016 at 21:25
• Basically, if $\sigma$ is an $n$ simplex then $P_n(\sigma)=\sigma \times I$, or what ever chain represents the triangulation of that set. Commented Apr 26, 2016 at 21:28
• I guess it is the same like the triangulation for the singular case? The three tetrahedra? Commented Apr 26, 2016 at 21:30

So we can actually build up the answer inductively here: let's label the simplices of $\Delta_2$ as $v_0, v_1, v_2$, and the corresponding vertices of the top and bottom faces of $\Delta_2 \times I$ as $t_0, t_1, t_2, b_0, b_1, b_2$.
So let's define $P$ on $C_0(\Delta_2)$. Here $\partial P + P\partial = \partial P$ (as $\partial$ vanishes on 0-simplices) and we want $\partial P([v_i]) = [t_i] - [b_i]$, so let $P([v_i]) = [t_i, b_i]$.
Go up a dimension and define $P$ on $C_1(\Delta_2)$. For concreteness, let's figure out $P([v_0, v_1])$. We have $$\partial P([v_0, v_1]) = [t_0, t_1] - [b_0, b_1] - P([v_0]) + P([v_1]) = [t_0, t_1] - [b_0, b_1] - [t_0, b_0] + [t_1, b_1]$$ This implies we want $P([v_0, v_1]) = [t_0, t_1, b_0] + [t_1, b_1, b_0]$ (this is much clearer if you draw a little picture! The four edges we have give you a square on the side of $\Delta_2 \times I$, and we've split it up into two triangles.)