Given a topological space and two maps from $[0,1]=|\Delta^1|$, say $f$ and $f'$ to $X$, a homotopy using an element of the 2-simplices of the singular set of $X$, such that its edges are given by the diagram below
This diagram denotes a simplex $\sigma$ such that $d^2 \sigma=id$, $d^0(\sigma)=f$, and $d^1(\sigma)=f'$. The second figure illustrates how this corresponds to a homotopy between $f$ and $f'$.
My question is if I consider the 'space' whose points are chain complexes, so that $f$ and $f'$ are maps of chain complexes$ C^* \to D^*$, then why does the existence of such a 2 simplex guarantee that $f$ and $f'$ are chain homotopic?
(More precisely I consider the natural transformations from the canonical natural transformation $Hom(*,[2]) \to Kom$ given by mapping the vertex $0$ to the chain complex $C$, $1$ to $C$, $0 \to 1$ to the identity morphism from $C \to C$, $1 \to 2$ to the morphism $f$, $0 \to 2$ to the morphism $f'$, etc. )
If I naively proceed using my own intuition of what the boundary of a 2-simplex should be and call the 2-simplex $\sigma$. $\sigma$ by definition is a map from $C \to D$. Let $f$ and $f'$ be maps from $C \to D$, I get that $\partial \sigma:= d^0(\sigma)-d^1(\sigma)+d^2(\sigma)=f-f'+Id$.
How can I use this to show that $f$ and $f'$ are chain homotopic?
Answer:
The textbook definition(crash course on $\infty$-categories, gaitsgory seminar) of the boundary map appears to have been written so that $\sigma$ defines a chain homotopy. Thus $\partial \sigma+f \circ \partial =f'-f \circ Id=f'-f$, if $d(\sigma)$ is defined to be $\partial \sigma + f \circ \partial$.