Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background

For a topological space $X$, let $R(X)$ be the category of retractive spaces over $X$. An object of this category is a space $Y$ equipped with maps $s: X \to Y$ and $r: Y \to X$ such that $r\circ s$ is the identity. These are called structure maps. A morphism of $R(X)$ is a map of underlying spaces which commutes with the structure maps.

Let $\text{Top}$ be the category of spaces with the model structure given by weak homotopy equivalence, Serre fibration and Serre cofibration. Quillen showed that $R(X)$ inherits the structure from $\text{Top}$ where a map of $R(X)$ is a weak equivalence/fibration if and only if it is so after applying the the forgetful functor $R(X) \to \text{Top}$.

In this model structure one can characterize a cofibration $A \to Y$ as follows:

(i) up to isomorphism, $Y$ is obtained from $A$ by sequentially attaching cells, where a cell is an object of the form $X \amalg D^n$ and the attaching is done with respect to a morphism from $X \amalg S^{n-1}$. In this way the pair $(Y,A)$ is a relative cell complex over $X$.

(ii) More generally, we must also allow retracts of the pairs $(Y,A)$ appearing in (i).

Observation

If $f: X \to X'$ is a map, then pullback along $f$ defines a functor$^\flat$ $$ f^* : R(X') \to R(X) $$ which fails to be exact. Even if $f$ is a fibration this functor can fail to be exact since it need not preserve cofibrations. For example if $X' = *$ is a point, and $Y$ a based cell complex (which is a cofibrant object of $R(*)$), then $f^*(Y)$ has underlying space $X \times Y$, and although I don't at the moment have an easy explanation as to why this needn't be cofibrant, I am pretty sure it isn't in general.

Questions

I am looking for an alternative model structure on $R(X)$, having the same weak equivalences (= weak homotopy equivalence) as above, but which has the property that base change along fibrations preserves cofibrations. I have a candidate for what the cofibrations should be. I'd like to know if it can be completed to a model structure.

A cofibration $A \to Y$ in this conjectured model structure is an inclusion which is up to isomorphism gotten by starting with $A$ and attaching "cells," but now a cell is something of the form $X\times D^n$ and we are doing the attaching along a morphism out of $X\times S^{n-1}$. More generally, we need to allow retracts of these kinds of inclusions $A \to Y$.

Question 1: If we define the fibrations by the lifting property with respect to the trivial cofibrations, does this form a model structure on $R(X)$?

Question 2: If the answer to Question 1 is yes, then can we explicitly pin down the fibrations?

Remark

In the model structure defined by Quillen, the quotient space (= cofiber) $Y/s(X)$ arising from a cofibrant object $Y$ of $R(X)$ is a cofibrant object of $\text{Top}$.

In the model structure I'm proposing, if $Y$ is cofibrant, then the fibers of the structure map $r: Y \to X$ will be cofibrant spaces

${}^\flat$A minor point: It's not really a functor, but it is a $2$-functor, since base change is defined only up to unique isomorphism. In particular, the base change of a composite is uniquely isomorphic to the iterated base changes of each constituent. So be it...

share|improve this question
add comment

1 Answer

John, your question is an advertisement for Johann Sigurdsson's thesis and our book ``Parametrized homotopy theory'', http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf, which is where the results of his thesis appear. Section 6.1 there explains many of the pitfalls of the obvious model structure that you start with in your question, which we call the q-model structure. Following James, we call your retractive spaces "ex-spaces''. Johann's thesis gives a new model structure which we call the $qf$-model structure, starting with a $qf$-model structure on spaces over the given base space $X$, rather than the $q$-model structure on spaces, and going from there to ex-spaces. The rest of Chapter 6, pages 100-108, is devoted to proofs and explanations of that model structure and its properties. That is probably too lengthy to summarize here. The model structure does have the same weak equivalences, and it is cofibrantly generated. As you expect, the subtlety is in the precise definition of the generating cells, and that is really subtle and entirely due to Johann. Of course, you must restrict to quite special generating cells, and this does raise problems, forcing us to introduce some variant model structures in Chapter 7. The identity functor is a Quillen equivalence from the $qf$-model structure to the $q$-model structure, so the homotopy category is what you want. Incidentally, we generalize to the equivariant context in Chapter 7.

Edit: In light of the comments, here are my grounds for skepticism, from first principles.
Given an adjunction $(L,R)$ between categories $\mathcal C$ and $\mathcal D$ and a cofibrantly generated model structure on $\mathcal C$ with generating sets of cofibrations and acyclic cofibrations $I$ and $J$, the natural way to try to construct a model structure on $\mathcal D$ is to let $R$ create the weak equivalences and fibrations and to take $LI$ and $LJ$ as generating sets. Formally, the maps that satisfy the RLP wrt $LI$ are then the acyclic maps that satisfy the RLP wrt $LJ$. I know of no examples where the same conclusion holds defining acyclicity in terms of a different class of weak equivalences. The relevance is that (modulo basepoint details) John starts with the adjunction $(r^*,Sec(X,-))$ between ex-spaces over $X$ and based spaces. His proposed fibrations are created by $Sec(X,-)$, and his proposed generating sets are $r^*I$ and $r^*J$. However, his proposed weak equivalences are not the weak equivalences created by $Sec(X,-)$ but rather the maps that are weak equivalences on total spaces.

share|improve this answer
    
Peter: I can't see how Johann's structure does what I want. I would like the set of generating cofibrations to be (or at least contain) the inclusions $X \times S^{n-1} \subset X \times D^n$ and $X \times D^n \times 0\subset X\times D^n \times I$. If this is correct, then it seems to me that the fibrations should be morphisms $E \to B$ of $R(X)$ such that the induced map of section spaces $\text{sec}(E\to X) \to \text{sec}(B \to X)$ is a Serre fibration. Perhaps this doesn't work. By the way, I don't see objects of the form $X \times S^{n-1}$ appearing in your Chapter 7. –  John Klein Aug 21 '13 at 17:53
    
Presumably you mean Chapter 6. My recollection (from a decade ago) is that we gave up on your choice: obviously that would be the first thing one thinks of. You are starting from the adjunction in Example 2.1.8, and I'm pretty sure we discarded that idea for good reason. But it has been a while. Warning 6.1.8 may be relevant. –  Peter May Aug 21 '13 at 18:54
    
I haven't checked carefully, but I cannot see what goes wrong: let's define the fibrations as above. Then define the cofibrations via the lifting property. Then show that attaching $X \times D^n$ to an object $U$ along a morphism $X\times S^{n-1} \to U$ gives a cofibration $U \to U'$. This is easy I think. Induction shows that repeated sequential attaching of this kind forms a cofibration. Next one should show the mod-cat factorizations exist. It seems to me this can be done as usual, by the gluing construction with respect to each of the types of cofibrant generators (cf. Dwyer-Spalinski). –  John Klein Aug 21 '13 at 21:07
    
John, I think you should provide details rather than argue in public. I would be happy to have you be right, in fact I would be very happy, but this is an area where long experience shows that healthy skepticism is justified. One place to expect trouble is proving that your proposed acyclic cell complexes are in fact acyclic. That is not formal. Details or nothing please! –  Peter May Aug 22 '13 at 2:58
    
Yes Peter. I know I have to generate a complete argument to convince you (as well as myself). –  John Klein Aug 22 '13 at 4:08
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.