Degreewise mod-two cohomology of the universal real Thom prespectrum

Assume AC, inherited from the shared bundle and Thom construction. Use the real levels $T_r=\mathrm{Th}({\gamma }_r)$ and first-coordinate structure maps ${\alpha }_r:S^1\wedge T_r\to T_{r+1}$ of the shared Thom prespectrum. For $q\in \mathbb{Z}$ put $A_r(q)={\tilde{H}}^{r+q}(T_r;{\mathbb{F}}_2)$ and define the backwards bonding map

${\rho }_r(q)={\sigma }^{-1}{\alpha }_r^{\ast }:A_{r+1}(q)⟶A_r(q),$

where $\sigma$ is reduced cohomology suspension with the sphere coordinate first. Define

${\widehat{H}}^q(TO;{\mathbb{F}}_2)=\{(x_r{)}_{r\ge 0}\in {\prod }_{r\ge 0}{A}_r(q):{\rho }_r(q)x_{r+1}=x_r\text{ for every }r\},{\widehat{H}}^{\ast }(TO;{\mathbb{F}}_2)={⨁}_{q\in \mathbb{Z}}{\widehat{H}}^q(TO;{\mathbb{F}}_2)\mathrm{.}$

The conditions are linear, so this is a well-defined vector space with componentwise operations. It is an inverse-limit prespectrum invariant; identifying it with represented spectrum cohomology would require a separate comparison theorem.

The normalized Thom isomorphisms give $A_r(q)\cong H^q(BO(r);{\mathbb{F}}_2)$, with $u_r$ the normalized rank-$r$ Thom class. The bonding-map computation and eventual constancy are proved in Stable universal Thom cohomology is eventually constant in every degree ↗: all terms vanish for $q<0$, and for $q\ge 0$ projection to any rank $r\ge q$ identifies the compatible tuples with the weight-$q$ polynomial Thom module. In particular, writing $U=(u_r)$,

${\widehat{H}}^{\ast }(TO;{\mathbb{F}}_2)\cong {\mathbb{F}}_2[w_1,w_2,\dots ]U,\mathrm{\mid }{w}_i\mathrm{\mid }=i,\mathrm{\mid }U\mathrm{\mid }=0.$

This is a graded vector-space and characteristic-polynomial-module identification, justified by that lemma. It gives no ring structure from reduced finite-level cup products. Each weight is finite-dimensional.