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 $dD_{F}+D_{F}d=f-g$. 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 $\Delta ^{q}\times I$. Let $e_{i}$ be the vertex of the $q$-simplex with only the $i^{th}$ coordinate nonzero. Let $e_{i}^{0}=(e_{i},0)\in \Delta ^{q}\times I$ and $e_{i}^{1}=(e_{1},1)\in \Delta ^{q}\times I$.This is defined as:

$D=\sum _{i=0}^{q}(-1)^{i}S(e_{0}^{0},\ldots ,e_{i}^{0},e_{i}^{1},\ldots ,e_{q}^{1})$

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

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

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