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 $$dD_F + D_Fd = f - g$$. 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:

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