The singular chain homotopy formula

If $n=0$ and $\sigma :{\mathrm{\Delta }}^0\to X$ is a singular $0$-simplex, then [L2] gives $\mathrm{\partial }{\mathrm{\Pi }}_0={\iota }_1-{\iota }_0$. Composing with $H\circ (\sigma \times {\mathrm{id}}_I)$ and using [L1] and [L3] yields $\mathrm{\partial }{P}_H(\sigma )=g_{\mathrm{\#}}(\sigma )-f_{\mathrm{\#}}(\sigma )\mathrm{.}$

Assume $n\ge 1$ and let $\sigma :{\mathrm{\Delta }}^n\to X$ be a singular $n$-simplex. Compose the boundary formula [L2] with the continuous map $H\circ (\sigma \times {\mathrm{id}}_I):{\mathrm{\Delta }}^n\times I\to Y\mathrm{.}$ By [L1] and [L3], the image of ${\mathrm{\Pi }}_n$ is $P_H(\sigma )$, the image of ${\iota }_1$ is $g_{\mathrm{\#}}(\sigma )$, and the image of ${\iota }_0$ is $f_{\mathrm{\#}}(\sigma )$.

Again using [L1], [L3], and [L4], the image of the side-prism term ${\sum }_{j=0}^n(-1{)}^j({\delta }_j\times {\mathrm{id}}_I{)}_{\mathrm{\#}}{\mathrm{\Pi }}_{n-1}$ under $H\circ (\sigma \times {\mathrm{id}}_I)$ is exactly $P_H(\mathrm{\partial }\sigma )$. Therefore step 1.2 becomes $\mathrm{\partial }{P}_H(\sigma )=g_{\mathrm{\#}}(\sigma )-f_{\mathrm{\#}}(\sigma )-P_H(\mathrm{\partial }\sigma ),$ or equivalently $g_{\mathrm{\#}}(\sigma )-f_{\mathrm{\#}}(\sigma )=\mathrm{\partial }{P}_H(\sigma )+P_H(\mathrm{\partial }\sigma )\mathrm{.}$

Singular simplices generate $C_n(X;\mathbb{Z})$, and the coefficient-$G$ version is obtained by tensor extension, so step 2.1 holds on all chains for every $n\ge 1$, while step 1.1 handles degree $0$. Hence the stated identities hold in all degrees.