Homotopic maps induce equal maps in singular cohomology

For $n\ge 1$ define $K^n:C^n(Y;G)\to C^{n-1}(X;G)$ by $K^n\phi =\phi P_{n-1}$, and set $K^n=0$ for $n\le 0$. For $n\ge 0$, the composite maps on a cochain satisfy ${\delta }_X^{n-1}{K}^n\phi +K^{n+1}{\delta }_Y^n\phi =\phi P_{n-1}{\mathrm{\partial }}_{X,n}+\phi {\mathrm{\partial }}_{Y,n+1}{P}_n=\phi (g_{\mathrm{\#},n}-f_{\mathrm{\#},n})\mathrm{.}$ When $n=0$ the first term is zero and the identity is exactly [F1]'s degree-zero formula. Thus $g^{\mathrm{♯}}-f^{\mathrm{♯}}=\delta K+K\delta$ with the stated positive sign.

If $\phi$ is a cocycle, the last summand $K\delta \phi$ vanishes; hence $g^{\mathrm{♯}}\phi -f^{\mathrm{♯}}\phi =\delta K\phi$ is a coboundary. It follows on [F2] quotients that $g^{\ast }[\phi ]=f^{\ast }[\phi ]$. In degree zero this is equality of the actual cocycles, since $K^0=0$. In negative degrees both induced maps are the unique map of zero groups.

Step 2.1 proves equality in every degree. Empty source or target, zero coefficients, and a point cause no exception: where the maps exist, the same prism formula applies; zero cochains give zero maps. Constant homotopies need not have zero unnormalized prism, but step 1.1 still yields the correct equality. This proof composes explicitly given homomorphisms and uses no extension of homomorphisms, primitive selection or AC.