The completed cohomology ring in degrees divisible by four

Let $B$ be a space (a topological space; no finite-dimensionality or CW assumption is made). The completed fourfold-graded cohomology group of $B$ with rational coefficients is the product ${\widehat{H}}^{4\ast }(B;\mathbb{Q}):={\prod }_{j\ge 0}{H}^{4j}(B;\mathbb{Q}),$ the set of all sequences $a=(a_0,a_1,a_2,\dots )$ with $a_j\in H^{4j}(B;\mathbb{Q})$ (Singular cohomology ring). Addition is componentwise, $(a+b{)}_j:=a_j+b_j,$ and multiplication is the convolution of the graded cup product, $(a\cdot b{)}_n:={\sum }_{i+j=n}{a}_i⌣b_j\mathrm{.}$

The product is well defined: for fixed $n$ the sum has exactly $n+1$ terms, so no infinite sum occurs in any component, and each $a_i⌣b_j$ lies in $H^{4n}(B;\mathbb{Q})$ because cup product adds degrees. The unit is $\mathbf{1}:=(1,0,0,\dots ),$ with $1\in H^0(B;\mathbb{Q})$ the unit of Singular cohomology ring. For a continuous map $f:B\to C$ the pullback is $f^{\ast }:{\widehat{H}}^{4\ast }(C;\mathbb{Q})\to {\widehat{H}}^{4\ast }(B;\mathbb{Q}),f^{\ast }(a{)}_j:=f^{\ast }(a_j),$ the componentwise pullback of the ordinary cohomology rings.

The ring laws and the naturality assertions are not part of this definition: they are proved in The completed fourfold-graded cohomology ring is natural and satisfies the ring laws ↗, the lemma named in justified_by. The construction is a product of abelian groups and a prescribed formula; nothing is selected, so no choice principle is used.