Let $X$ be a topological space. A free cohomology ring space is a space $Y$ and a map $X \to Y$ such that the $\mathbb Z/2$ cohomology of $Y$ is a polynomial ring with generators $a\_1,...,a\_n$, and the pullbacks of the generators along the maps form a basis for all the cohomology groups of $X$.


This definition may seem kind of, or extremely, strange, which is perhaps why I had to use a word salad title. The motivating example that interests me is the map $G\_n^1 \to G\_n^\infty$, for real or complex Grassmanians. (The map is induced by an embedding $\mathbb R^{n+1}\to \mathbb R^{\infty}$.) In either case, the latter is a free cohomology ring space of the former.


What I would like to know is if such a relationship could be made natural, that is, that there is a functor that takes a space $X$ to a space $Y$ and a map with this property that forms the appropriate commutative diagram. So far I have been unable to find one, and proving that a functor does not exist is probably beyond my command of category theory or topology.



Is there a free cohomology ring space functor?
Not all polynomial algebras over $\mathbf{Z}/2$ on generators of chosen degrees are realizable
as the mod $2$ cohomology of a space, but any set of generators of any such (connected) 
polynomial algebra forms the basis for the mod $2$ cohomology of a space $X$ (it can be
chosen to be a wedge of spheres). Therefore there can be no such functor.


