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.

`$\{a_i\}$`

the cohomology of $Y$ is free, in the category of algebras with Steenrod operations satisfying the instability relations, on the generators`$\{y_i\}$`

. – Tyler Lawson Jan 24 '12 at 12:39