Homotopy of maps induces chain homotopy

Statement

Let $F:X\times I\to Y$ be a homotopy between $f,g:X\to Y$. In other words $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for all $x\in X$. Then, there is a chain homotopy $D_F$ from the singular complex of $X$ to the singular complex of $Y$ such that $d{D}_F+D_{F}d=f-g$. In fact, the map sending $F$ to $D_F$ is a homomorphism in the sense that if $H$ is the composite of $F$ and $G$, $D_H=D_F+D_G$.

Construction

To construct a chain homotopy, we first define a certain $(q+1)$-singular chain in ${\mathrm{\Delta }}^q\times I$. Let $e_i$ be the vertex of the $q$-simplex with only the $i^{th}$ coordinate nonzero. Let $e_i^0=(e_i,0)\in {\mathrm{\Delta }}^q\times I$ and $e_i^1=(e_1,1)\in {\mathrm{\Delta }}^q\times I$.This is defined as:

$D=\sum _{i=0}^q(-1{)}^{i}S(e_0^0,\dots ,e_i^0,e_i^1,\dots ,e_q^1)$

where $S$ of a tuple is the simplex with those as vertices. This clearly gives a singular chain in ${\mathrm{\Delta }}^q\times I$.

For a given homotopy $F:X\times I\to Y$ the map $D_F$ is defined as:

$\sigma \mapsto (\sigma \times id)\circ D$