Characterizing the rationalization of spaces. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:20:34Z http://mathoverflow.net/feeds/question/31078 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31078/characterizing-the-rationalization-of-spaces Characterizing the rationalization of spaces. Jeff Strom 2010-07-08T16:48:14Z 2012-05-11T07:34:56Z <p>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:</p> <p>(*) Any functor $F$ from spaces to spaces which splits suspensions and loop spaces as above must factor through the rationalization.</p> <p>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. </p> <p>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:<br> 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.</p> <p>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.</p> http://mathoverflow.net/questions/31078/characterizing-the-rationalization-of-spaces/96635#96635 Answer by Jeff Strom for Characterizing the rationalization of spaces. Jeff Strom 2012-05-11T01:41:01Z 2012-05-11T01:41:01Z <p>I have an answer.</p> <p>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:</p> <ol> <li><p>$L_f(S^2)$ is nontrivial and simply-connected</p></li> <li><p>$L_f$ commutes with cofiber sequences of simply-connected finite complexes </p></li> <li><p>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 $n$.</p></li> </ol> <p>Details can be found here: <a href="http://arxiv.org/abs/1205.2140" rel="nofollow">http://arxiv.org/abs/1205.2140</a></p> http://mathoverflow.net/questions/31078/characterizing-the-rationalization-of-spaces/96648#96648 Answer by Fernando Muro for Characterizing the rationalization of spaces. Fernando Muro 2012-05-11T07:34:56Z 2012-05-11T07:34:56Z <p>I think the following is a trivial counterexample, which may lead you to reflect about your question:</p> <p>\begin{align*} F\colon Spaces &amp; \longrightarrow Spaces\\ X&amp;\;\mapsto\;\bigvee_{H_1(X,\mathbb{F}_2)}S^1 \end{align*}</p> <p>This functor takes <strong>any</strong> 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 $|H_1(X,\mathbb{F}_2)|\neq |H_1(Y,\mathbb{F}_2)|$. </p> <p>Of course, you can replace $H_1$ with $H_n$ for any $n$ if you wish to work with simply connected spaces.</p>