I am trying to understand the following lemma: If $f,g:I\rightarrow X$ are paths s.t. $f(1)=g(0)$ then the 1-chain $f*g-f-g$ is a boudary.
Proof: On the standard 2-complex (should it say simplex?) $\Delta_2$ put $f$ on the edge $(e_0,e_1)$ and $g$ on the edge $(e_1,e_2)$. Then define a singular 2-simplex $\sigma:\Delta_2\rightarrow X$ to be constant on the perpendicular lines to the edge $(e_0,e_2)$ (HOW?). This results in the path $f*g$ being on the edge $(e_0,e_2)$ (WHY?). Therefore $\partial \sigma=g-(f*g)+f$
I know that $\partial \sigma=\partial^0 \sigma-\partial^1 \sigma-\partial^2 \sigma$ and that $\partial^0 \sigma(x,y,z)=\sigma(0,y,z)$ $\partial^1 \sigma(x,y,z)=\sigma(x,0,z)$ $\partial^2 \sigma(x,y,z)=\sigma(x,y,0)$
I also drew a picture and still dont get it.
This is one of three previous lemma to the Hurewicz theorem, any help would be appreciated.