Cap naturality and projection formula

If $n\ge p$, the left side of the first chain identity is $f_{\mathrm{\#}}(a(f\sigma [0,\dots ,p])\sigma [p,\dots ,n])$, equal to $a(f\sigma [0,\dots ,p])(f\sigma )[p,\dots ,n]$. This is exactly $a\cap f_{\mathrm{\#}}\sigma$, since postcomposition commutes with face restriction. If $n<p$, both sides are zero. Linearity gives the chain identity for every chain.

If $n\ge p+q$, evaluating the left side of the second identity yields $a(\sigma [0,\dots ,p])b(\sigma [p,\dots ,p+q])\sigma [p+q,\dots ,n]$. Capping first by $a$ leaves the chain $a(\sigma [0,\dots ,p])\sigma [p,\dots ,n]$; capping by $b$ gives the same scalar and back face by $R$-linearity. If $n<p$, the inner cap on the right is zero; if $p\le n<p+q$, its remaining degree $n-p$ is less than $q$, so the outer cap is zero. In both cases the left side also vanishes. Thus the equality holds on every simplex and extends bilinearly.

For cocycles and cycles, [F1] and [F2] make every operation in steps 1.1–1.2 well-defined on the corresponding quotient classes. Passing the chain identities to classes proves the formulas. The case $p=0$ is initial-vertex multiplication; the case $p+q=n$ retains the last vertex with no sign, and zero degrees in either factor need no alteration. Empty spaces, zero inputs/ring, point spaces and degenerate simplices use the same formulas. All maps and products are explicit, without AC.