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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.

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    $\begingroup$ What would be more likely is that you can get a space/spectrum that is free as a module over the steenrod algebra that relates to your space. $\endgroup$
    – Sean Tilson
    Commented Jan 24, 2012 at 3:26
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    $\begingroup$ Note that while it doesn't work over $Z_2$ by Peter May's and Tom Goodwillie's answers it does work over $\mathbb Q$ if instead of an actual map $X\to Y$ you are willing to settle for its homotopy class. just take $Y$ to be the product of $K(H^n(X,\mathbb Q),n)$ with natural map $X\to Y$ coming from $H^n(X,H^n(X,\mathbb Q))=[X, K(H^n(X,\mathbb Q),n)]$. This correspondence is clearly functorial in $X$. $\endgroup$
    – Vitali Kapovitch
    Commented Jan 24, 2012 at 4:38
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    $\begingroup$ For connected $X$ with a choice of basepoint, you can take $Y$ to be the infinite symmetric product $SP^\infty X$, and then the natural inclusion $X \to SP^\infty X$ gives a functorial construction with gives you the desired map on rational cohomology back. If, instead of the infinite symmetric product (the free abelian monoid on $X$) you take the free $\mathbb{Q}$-vector space (which has a canonical choice of topology), you get the same result and it provides a functorial point-set version of Vitali Kapovitch's construction. $\endgroup$
    – Tyler Lawson
    Commented Jan 24, 2012 at 12:35
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    $\begingroup$ If, instead of either of these, you construct the free $\mathbb{Z}/2$-vector space on $X$, you get a space $Y$ where the map back on mod-2 cohomology has the following property. There exists a lift of the generators of the cohomology of $X$ to $Y$, and given any lift $\{a_i\}$ the cohomology of $Y$ is free, in the category of algebras with Steenrod operations satisfying the instability relations, on the generators $\{y_i\}$. $\endgroup$
    – Tyler Lawson
    Commented Jan 24, 2012 at 12:39
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    $\begingroup$ (In both of these I'm assuming that the homology of $X$ is finitely generated in each degree. Otherwise, you have to be a lot more careful.) $\endgroup$
    – Tyler Lawson
    Commented Jan 24, 2012 at 12:40

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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.

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  • $\begingroup$ Which ones aren't, and why? A reference would of course work as well. $\endgroup$
    – Will Sawin
    Commented Jan 24, 2012 at 2:01
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    $\begingroup$ A polynomial ring on one generator in dimension $3$ is impossible. In fact, the squaring map $H^3\to H^6$ in mod $2$ cohomology must be zero if $H^5=0$ because (1) for $x\in H^n$ with mod $2$ coefficients the Steenrod square $Sq^nx$ is the cup product $x\cup x$ and (2) $Sq^3=Sq^1\circ Sq^2$. $\endgroup$
    – Tom Goodwillie
    Commented Jan 24, 2012 at 2:23
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    $\begingroup$ The problem of exactly which polynomial rings can be realized was known as the Steenrod problem, and has a complete solution. It turns out that e.g., for algebras over Z/2 the only ones possible are those arising form classifying spaces of compact Lie groups, and one extra one with generators in degree 15, 14, 12, and 8 related to the exotic 2-compact group DI(4). See "The Steenrod problem of realizing polynomial cohomology rings. J. Topology 1 (2008), 747-760" and "The classification of 2-compact groups. J. Amer. Math. Soc. 22 (2009), 387-436" for much more info. $\endgroup$
    – Jesper Grodal
    Commented Jan 30, 2012 at 10:41

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