# Characterizing the rationalization of spaces.

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.

• Do you mean that rationally a <i> suspension </i> splits as a wedge of <i> rationalized </i> spheres? And can you say more precisely what property you want F to have? Do you mean that F of a suspension should be equivalent to a wedge of spaces of the form F(Sphere)? Or may be F(wedge of spheres)? Would the constant functor that sends every space to the circle be an example? Jul 8, 2010 at 21:29
• I guess your constant functor is an example, and it does factor through rationalization. Jul 8, 2010 at 22:13
• Leave the question a little vague, sure, but Greg's initial question is about your initial statement, not your question. The rationalization functor applied to the loop space of X is always equivalent to a product of (rational) Eilenberg-MacLane spaces, yes? The rationalization functor applied to the <i>suspension</i> of X is always equivalent to a wedge of rationalized spheres, yes? Jul 9, 2010 at 1:06
• In particular, consider the rationalization of CP^\infty \vee CP^\infty. It is not a rational wedge of spheres. Please do change or <strike> your very first line and your initial question (*) (as you seem to be trying to do with EDIT 2, but that makes things hard to read without deleting/<strike>ing incorrect statements above.) By the way, I think the question you are getting at, as I understand it, is interesting and something along those lines seems likely to be true. Jul 9, 2010 at 6:27
• Sorry I couldn't see the think-o for so long. Fixed now. Jul 9, 2010 at 15:21

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:

1. $L_f(S^2)$ is nontrivial and simply-connected

2. $L_f$ commutes with cofiber sequences of simply-connected finite complexes

3. 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

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.

• This is not a localization and it does not commute with cofiber sequences. For example, apply it to $S^1 \to * \to S^2$. May 11, 2012 at 11:37
• That didn't seem to be a requirement of your question. May 11, 2012 at 11:40
• Fernando: a small point, to see that your construction is really a functor, it might be clearer to write it as $X\mapsto \Sigma (H_1(X;{\Bbb F}_2))_+$. May 11, 2012 at 12:24
• @John: nice remark! May 11, 2012 at 12:32
• @Jeff: Fernando is correct. You do not require the condition that the functor commutes with cofiber sequences. May 11, 2012 at 12:47