most recent 30 from http://mathoverflow.net 2013-05-24T04:48:33Z http://mathoverflow.net/feeds/question/86489 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86489/is-there-a-free-cohomology-ring-space-functor Will Sawin 2012-01-23T21:59:27Z 2012-01-24T00:27:51Z

<p>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 $X$.</p> <p>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.</p> <p>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.</p>

http://mathoverflow.net/questions/86489/is-there-a-free-cohomology-ring-space-functor/86497#86497 Peter May 2012-01-24T00:27:51Z 2012-01-24T00:27:51Z

<p>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.</p>