7
$\begingroup$

Can we construct the category of spectra (or maybe just its homotopy category) from the category of pointed topological spaces using some kind of localization combined with other categorical constructions?

[The first part of the original question was wrong for a trivial reason pointed out by Reid Barton.]

Improve this question
$\endgroup$
1
  • $\begingroup$ I don't know the answer, but I have a related question. What if we let $f$ be the wedge of the maps $X\to \Omega \Sigma X$ (representing suspension) for all countable CW complexes X, and then apply Bousfield/Farjoun localization $L_f$? It seems to me that, for the purposes of mapping in finite complexes, we have inverted the suspension operation. $\endgroup$
    – Jeff Strom
    Mar 1, 2010 at 0:46

1 Answer 1

Reset to default
9
$\begingroup$

[Removed a paragraph relating to an earlier version of the question]

You can construct Spectra categorically by adjoining an inverse to the endofunctor Σ of Top as a presentable (∞,1)-category. Inverting an endofunctor is a very different operation than inverting maps! It's like the difference between forming ℤ[1/p] and ℤ/(p).

Here is one way to verify the claim. To invert the endomorphism Σ of Top we should form the colimit, in the (∞,1)-category Pres of presentable categories and colimit-preserving functors, of the sequence Top → Top → ... where all the functors in the diagram are Σ. A basic fact about Pres is that we can compute such a colimit by forming the diagram (on the opposite index category) formed by the right adjoints of these functors, and taking its limit as a diagram of underlying (∞,1)-categories [HTT 5.5.3.18]. The functors in the limit cone will have left adjoints which are the functors to the colimit in Pres. In our case we obtain the sequence Top ← Top ← ... where the functors are Ω, and the limit of this sequence is precisely the classical definition of (Ω-)spectrum: a sequence of spaces Xn with equivalences Xn → ΩXn+1.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Minor quible. Don't you only get connective spectra by starting with Top and inverting the suspension functor? Also a question: If you start with Top and invert the loops functor do you also get the category of (connective) spectra? $\endgroup$
    – Chris Schommer-Pries
    Feb 24, 2010 at 4:39
  • 3
    $\begingroup$ It's even worse than that, you only get the category of suspension spectra. Also, I had some other argument written here about inverting the loop functor which suffered from me accidentally getting an adjunction on the wrong side. Remember, kids: no Math Overflow late at night. $\endgroup$
    – Tyler Lawson
    Feb 24, 2010 at 5:23
  • 3
    $\begingroup$ I should have mentioned that I'm working in the world of presentable (∞,1)-categories. I'm pretty sure my new statement is correct. $\endgroup$
    – Reid Barton
    Feb 24, 2010 at 5:28
  • $\begingroup$ Is the operation of adjoining an inverse to an endofunctor explained somewhere in Lurie's papers or anywhere else? $\endgroup$
    – Dmitri Pavlov
    Feb 24, 2010 at 5:42
  • $\begingroup$ I don't know of a specific place where it is written down, but invertibility of an endofunctor is an (∞,1)-categorical (as opposed to (∞,2)-categorical) notion, so it's directly analogous to the situation in classical algebra. $\endgroup$
    – Reid Barton
    Feb 24, 2010 at 6:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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