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 $σ$ , let $P_{i} (q)$ be the singular $(q + 1)$ -simplex in $X \times I$ obtained by Fill this in later