6
$\begingroup$

The question is motivated by Eckmann-Hilton duality and certain flaws of the homotopy category of CW-complexes. Unfortunately, I do not know the formalism of model categories, so excuse me if it is a basic fact concerning them.

First consider a circle $S^1$ with a fixed point. All spaces are supposed to be connected and have a fixed point, and I will suppress it in notations. Then there are its $(n-1)$-suspension $S^n:=\Sigma^{n-1} S^1$ and its $(n-1)$-delooping $K(\mathbb Z, n):=\Omega^{-(n-1)}S^1$ (I believe that one can construct it geometrically). From them we can build $n$-th homotopy groups $\pi_n(X):=[S^n, X]$ and $n$-th (integral) cohomology groups $H^n(X):=[X, K(\mathbb Z, n)]$, where $[A, B]$ means a set of homotopy classes of maps from $A$ to $B$, and give them a natural group structure, see Fomenko, Fuchs Homotopical topology, $\S$1.4.

As suspension $\Sigma$ and looping $\Omega$ are adjoint, one has $H^i(S^n) \simeq \mathbb Z$ for $i=n$ and $0$ otherwise, and $\pi_i(K(\mathbb Z, n)) \simeq \mathbb Z$ for $i=n$ and $0$ otherwise (we ignore $\pi_0$ and $H^0$). Moreover, in the category of CW-complexes $K(\mathbb Z, n)$ is defined by this condition up to homotopical equivalence, but $S^n$ is not -- there is Poincaré homology sphere $\widehat{S^3}$.

A quick idea is that it happens because CW-complexes are constructed from $S^n$, so one should work in the category of spaces constructed from $K(\mathbb Z, n)$, with something like Postnikov towers instead of CW-complexes' skeletons. There are certain conditions on CW-complexes, the main one is that the quotients of skeletons are bouquets of spheres; of course, one may ask that the homotopy fibers of Postnikov tower are products of $K(\mathbb Z, n)$, that is that the homotopy groups are torsion-free, but the resulting category looks very small, without spheres.

So a question is whether there is a nice category that is "Eckmann-Hilton dual" to CW-complexes? A perfect case would be if one can dualize any statement about CW-complexes and get a true statement about objects in this category, and if it contains spheres and $K(\mathbb Z, n)$, so that we have homotopy and cohomology groups. It is wrong for CW-complexes themselves: for example, the pullback of a fibration by a cofibration is always a fibration, but the pushforward of a cofibration by a fibration is not always a cofibration.

Update. According to the article May The Dual Whitehead Theorems from Dan Ramras' comment, another candidate to "Echmann-Hilton dual" category to CW-complexes is simple spaces, that is whose $\pi_1(X)$ acts trivially on $\pi_n(X)$ (in particular, it acts by conjugation on itself, so is abelian). Simple CW-complexes are those for which Postnikov towers are usually defined.

$\endgroup$
6
  • 1
    $\begingroup$ Isn't $\Omega^{n-1} S^1$ the looping (and not delooping) of $S^1$? I believe the delooping would more traditionally be written as $B^{n-1}S^1$. $\endgroup$ Commented Nov 24, 2016 at 16:13
  • 1
    $\begingroup$ I am not sure I understand your question: in a "nice enough" category of spaces every space has a Postnikov tower (although saying that a Postnikov tower is "constructed" by Eilenberg-MacLane spaces might be overstating things). Can you elaborate on what kind of construction would you want? $\endgroup$ Commented Nov 24, 2016 at 16:13
  • $\begingroup$ Najib, Denis, thank you, I modified the question. $\endgroup$
    – evgeny
    Commented Nov 24, 2016 at 16:25
  • $\begingroup$ @evgeny It is still quite vague. At least to me a CW-complex is a kind of structure you put on a space to help computations. You can construct the category of Postnikov towers and it will be equivalent to the usual category of spaces (at least in a suitably weak sense). However it does not feel "dual" to the category of CW complexes in any meaningful sense. Is this the kind of answer you are after? $\endgroup$ Commented Nov 24, 2016 at 16:50
  • 2
    $\begingroup$ You may find Peter May's article "The Dual Whitehead Theorems" interesting in this context. It's at math.uchicago.edu/~may/PAPERS/47.pdf $\endgroup$
    – Dan Ramras
    Commented Nov 24, 2016 at 18:33

2 Answers 2

2
$\begingroup$

Repeating my comment above as answer:

You may find Peter May's article "The Dual Whitehead Theorems" interesting in this context. It's at math.uchicago.edu/~may/PAPERS/47.pdf

$\endgroup$
5
$\begingroup$

The answer is yes. You mention model categories, so I guess what you have in mind is right Bousfield localization. Given a nice enough model category $M$ (spaces or simplicial sets qualify), and a set of objects $K$, the right Bousfield localization $R_K(M)$ is new model structure where the objects of $K$ are the new cells, and a map $f:X\to Y$ is a weak equivalence (a $K$-colocal equivalence) if for every $A\in K$, the map of simplicial sets $map(A,X)\to map(A,Y)$ is a weak equivalence. These are the maps seen to be weak equivalences by $K$.

This model structure has the same fibrations as $M$. The cofibrant objects are precisely the $K$-colocal objects, i.e. objects $W$ such that $map(W,X)\to map(W,Y)$ is a weak equivalence for all $K$-colocal equivalences $W$. If every object of $M$ is fibrant (e.g. Top) then $R_K(M)$ is cofibrantly generated, with the same generating trivial cofibrations $J$ as $M$, and with generating cofibrations defined as $\overline{\Lambda(K)} = J \cup \{A^*\otimes \partial \Delta[n]\to A^*\otimes \Delta[n] | A\in K, n\geq 0\}$, where $A^*$ is a cosimplicial resolution of $A$. Morally, this is saying that the objects of $K$ are the new cells, and cofibrant replacement in this new model structure entails building things out of the objects in $K$.

In your case, taking $M = Top$ and $K = \{K(\mathbb{Z},n)\}$ will accomplish what you asked for. All the facts I wrote above are in Hirschhorn's book. A source for lots of examples is my paper (with Donald Yau): Right Bousfield Localization and Operadic Algebras

$\endgroup$
1
  • 1
    $\begingroup$ Now that the original question has been edited, this does not appear to be a full answer. I am not quite sure what the last paragraph of the new question is asking. $\endgroup$ Commented Nov 24, 2016 at 16:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .