Characterizing the rationalization of spaces. - MathOverflow

Characterizing the rationalization of spaces.

2010-07-08T16:48:14Z

In the category of rational spaces, loop spaces split as products of Eilenberg-Mac Lane spaces and SUSPENSIONS split as wedges of (rational) spheres. I wonder if anything of the following form is true:

(*) Any functor $F$ from spaces to spaces which splits suspensions and loop spaces as above must factor through the rationalization.

EDIT 1: Greg raises some fine questions, but I stand by my wording. This is a question that arises from curiosity, not because I need it for anything, so I'd be happy with "anything like" the given statement.

EDIT 2: At least for simply-connected spaces, rationalization commutes with loop and suspension. But, it seems to me that the power of the property is that the suspension of any F-space splits and the loops of any F-space splits. So I would go with:
the suspension of any rational space splits as a wedge of rational spheres and the loops of any rational space splits as a product of rational Eilenberg-Mac Lanes spaces.

Thus, we'd be looking for functors to some model-esque category with some relatively manageable list of objects whose products exhaust the homotopy types of loop spaces and whose wedges exhaust the homotopy types of suspensions.

Answer by Jeff Strom for Characterizing the rationalization of spaces.

2012-05-11T01:41:01Z

I have an answer.

Look at $f$-localization functors $L_f$. The restriction of $L_f$ to simply-connected spaces is rationalization if and only if the following three conditions hold:

$L_f(S^2)$ is nontrivial and simply-connected
$L_f$ commutes with cofiber sequences of simply-connected finite complexes
if $X$ is a simply-connected finite complex, then for large enough $k$, $\Sigma^k L_f(X)$ splits as a wedge of copies of $L_f(S^n)$ for various values of $n$.

Details can be found here: http://arxiv.org/abs/1205.2140

Answer by Fernando Muro for Characterizing the rationalization of spaces.

2012-05-11T07:34:56Z

I think the following is a trivial counterexample, which may lead you to reflect about your question:

\begin{align*} F\colon Spaces & \longrightarrow Spaces\\ X&\;\mapsto\;\bigvee_{H_1(X,\mathbb{F}_2)}S^1 \end{align*}

This functor takes any space to a wedge of several circles, one circle for each element in the homology group ${H_1(X,{\mathbb{F}}_{2})}$. Such wedges are both suspensions and Eilenberg-MacLane spaces. Obviously this functor does not factor through rationalization, since there are spaces $X$ and $Y$ with $X\simeq _{\mathbb{Q}} Y$ but $|H_1(X,\mathbb{F}_2)|\neq |H_1(Y,\mathbb{F}_2)|$.

Of course, you can replace $H_1$ with $H_n$ for any $n$ if you wish to work with simply connected spaces.