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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.

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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? –  Gregory Arone Jul 8 '10 at 21:29
I guess your constant functor is an example, and it does factor through rationalization. –  Jeff Strom Jul 8 '10 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? –  Tom Goodwillie Jul 9 '10 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. –  Dev Sinha Jul 9 '10 at 6:27
Sorry I couldn't see the think-o for so long. Fixed now. –  Jeff Strom Jul 9 '10 at 15:21

2 Answers 2

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:

  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

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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.

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

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