Singular cohomology ring

For a space $X$ and commutative unital coefficient ring $R$, put $H^{\ast }(X;R)={⨁}_{n\ge 0}{H}^n(X;R),[\phi ]⌣[\psi ]=[\phi ⌣\psi ]\mathrm{.}$ The groups and equality of representatives are those of Singular cohomology with coefficients. The multiplication descends through both quotient maps by Cup product Leibniz identity, including simultaneous changes. Extend to finite sums of homogeneous classes by distributivity; only finitely many summands occur, so each product belongs to the displayed direct sum.

This is the singular cohomology ring. To check the ring assertion before using it, let $\phi ,\psi ,\eta$ have degrees $p,q,r$. On an arbitrary $(p+q+r)$-simplex both parenthesizations of their cochain cup product evaluate to $\phi (\sigma [0,\dots ,p])\psi (\sigma [p,\dots ,p+q])\eta (\sigma [p+q,\dots ,p+q+r])\mathrm{.}$ Restriction to a face and then a face of that face is restriction to the listed vertex block, and multiplication in $R$ is associative. Thus cochain associativity holds exactly and descends to classes. Bilinearity gives the two distributive laws. Addition, additive inverses and zero come from the abelian quotient groups. Products of degrees $p,q$ have degree $p+q$.

Its multiplicative unit is the class of the cochain $\mathbf{1}\in C^0(X;R)$ assigning $1_R$ to every singular vertex. On any edge, $\delta \mathbf{1}=1_R-1_R=0$. The front/back formula gives $\mathbf{1}⌣\phi =\phi =\phi ⌣\mathbf{1}$ on every simplex. This includes all components at once: degree-zero cochains need not have finite support on the set of vertices or components.

If $X$ is empty or $R=0$, the ring is the zero ring and its unit equals zero, as allowed by our unital-ring convention. The formula for three factors works also when any degree is zero and for degenerate simplices; for one-point $X$ it has the same unit. Nothing here chooses a representative for every class: the quotient operation is defined by representative independence. No AC is used.