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I'm not sure if this is standard, but we'll call the property that every weak homotopy equivalence is an honest homotopy equivalence the Whitehead property (from Whitehead's theorem for CW complexes).

Then the question: Is there any (nontrivial) category of spaces that is cartesian-closed and has the Whitehead property? If not, is there some counterexample we can construct to show that these properties are mutually exclusive?

I am familiar with the category of m-cofibrant spaces, but it is not cartesian closed (even though there is the weaker result allowing us to take loop spaces etc.).

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This is only a partial answer Harry, but, to my knowledge, all Cartesian-closed categories of topological spaces that I know of arise as the monocoreflective hull (in either Top or Hausdorff spaces etc.) of a productive class $C$ of generating exponentiable spaces. The requirement to be a productive class is that binary products of generators lie in the monocoreflective hull. Also, it is a well known theorem that a space is exponentiable if and only if it is core-compact. So, for example if $C$ is compact Hausdorff space, we get the category of compactly generated spaces as its monocoreflective hull. If you tried to use this machinery to produce a Cartesian-closed category of "whitehead spaces", I think it would fail: a natural choice would be compact Hausdorff m-cofibrant spaces for C- this should be a productive class (if I am not making a silly mistake), BUT, I don't think that being homotopy equivalent to a CW-complex is stable under colimits, as CW-complexes aren't. So, my guess of an answer would be "No." However, this is hardly a proof. By the way, what did you want to do with these spaces anyhow?

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  • $\begingroup$ Well, such a category spaces would be "optimally nice" if it existed (and it was equippable with a nice model structure, but it would certainly be important to show that this kind of category even exists before worrying about the model structure. Putting a model structure on a nonexistent category would be a pretty big waste of time.). $\endgroup$
    – Harry Gindi
    Commented May 11, 2010 at 5:29
  • $\begingroup$ Alright, but what is your ultimate goal? I mean, if you put a model structure on it, then it should only matter that every guy is weakly equivalent to an m-cofibrant guy. Why do you care about this whitehead property? Are you trying to represent the homotopy category of spaces by using a calculus of fractions or something? $\endgroup$
    – David Carchedi
    Commented May 11, 2010 at 10:13
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    $\begingroup$ An idea: start with just CW-complexes, and build the infinity-category right away which has 1-morphisms as continuous functions, 2-morphisms as homotopies, 3-morphisms as homotopies between homotopies etc... this infinity-category's objects all have the whitehead property, but it is equivalent to the one coming from model structure on spaces with weak homotopy equivalences as the weak equivalences. In particular, it will be an infinity-topos, so it will Cartesian-closed AS an infinity-category (this is slightly weaker than CC, but you shouldn't care if you want to do homotopy). $\endgroup$
    – David Carchedi
    Commented May 11, 2010 at 10:19

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