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 need a homomorphism from the set of the group of $q$-singular chains of $X$ to the group of $(q-1)$-singular chains of $Y$. To define such a homomorphism, we need to define it only on singular simplices (since it'll extend uniquely by linearity).

Here's how we do this. Given a singular $q$-simplex $\sigma$, compose <math>\sigma<math> with the following mapFill this in later