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Suppose given a well-generated triangulated category with a compatible symmetric monoidal structure, $\mathcal{T}$ (in the sense of Neeman). Is it clear that the image of a localization functor will also be well-generated? If not in true in general, is it easy to prove with respect to the $X$-acyclics (i.e. objects which tensor with $X$ to zero) for some element $X$?

The main thing I'm worried about I suppose (though the second axiom for well-generated-ness is not obvious to me either) is the requirement for the generators to be $\alpha$-small for some cardinal $\alpha$. For instance, the $MU$-localization of the stable homotopy category has no small objects. But perhaps they are $\alpha$-small for some $\alpha$? My understanding of such issues is still somewhat superficial.

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    $\begingroup$ The quotient of a well generated category by a localizing subcategory generated by a set is again well generated. All localizing subcategories are generated by sets if we assume some set-theoretical axioms. $\endgroup$
    – Fernando Muro
    Commented Oct 2, 2012 at 22:21
  • $\begingroup$ Thanks @Fernando! Is this somewhere in Neeman's book? Also, do you happen to know which set-theoretic principles must be assumed? $\endgroup$
    – Jonathan Beardsley
    Commented Oct 2, 2012 at 22:28
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    $\begingroup$ Why does the $MU$-localization of the stable homotopy category have no small objects? $\endgroup$
    – Akhil Mathew
    Commented Oct 3, 2012 at 11:28
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    $\begingroup$ Hey @Akhil I don't really know why, but the proof/explanation is in appendix B (specifically Cor. B.13) of Hovey and Strickland's "Morava K-theoriesand Localisation" which I believe can be found here: hopf.math.purdue.edu/Hovey-Strickland/kn.pdf This is also discussed briefly in Example 3.5.4(a) of Hovey, Palmieri and Strickland's "Axiomatic Stable Homotopy Theory" (math.rochester.edu/people/faculty/doug/otherpapers/…) although I'm sure there are people on this site who can explain it perhaps intuitively. $\endgroup$
    – Jonathan Beardsley
    Commented Oct 3, 2012 at 12:58
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    $\begingroup$ @Akhil no problem! I'm pretty sure I was sitting behind you at the Dan Quillen memorial conference! :-) $\endgroup$
    – Jonathan Beardsley
    Commented Oct 13, 2012 at 3:12

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