29
$\begingroup$

There is this old result by Freyd that "homotopy is not concrete":

Freyd, Peter. "Homotopy is not concrete." The Steenrod Algebra and Its Applications: A Conference to Celebrate NE Steenrod's Sixtieth Birthday. Springer Berlin Heidelberg, 1970.

In 2017 we say that

The key result here is the following lemma, called "Isbell condition":

For $f\in {\cal A}/A$ an object of the slice category, let $C(f,B)$ be the class of pairs $u,v : A \to B$ such that $uf=vf$ as morphisms $\text{src}(f) \to A \to B$.

Define an equivalence relation $\asymp$ on $({\cal A}/A)_0$ that says $f\asymp g$ iff $C(f,B)=C(g,B)$ for every $B\in\cal A$, and let $\textsf{S}({\cal A}/A)$ the quotient $\left({\cal A}/A\right)_{0,/\asymp}$.

Isbell condition: $\cal A$ is concrete if and only if $\textsf{S}({\cal A}/A)$ is a set for every $A\in\cal A$.

If you follow the development of this, you find another, slightly older paper

Freyd, Peter. "On the concreteness of certain categories." Symposia Mathematica. Vol. 4. 1969.

that develops quite a bit this technology, and contains (Thm 4.1) a result that in 2017 we write as

  • the localization ${\bf Cat}\!\!\left[\text{Eqv}^{-1}\right]$ of the category of categories to equivalences (i.e. the homotopy category $\textsf{Ho}({\bf Cat}_\text{folk})$ is not concrete.

Now, following Christie's meta-theorem it's easy to wonder if there is a pattern, and maybe a proof, here.

Freyd's theorem is as old as Quillen's definition of a model category, so I doubt that Freyd ignored that you can ask the following question:

  • Let $\mho$ be a universe. If $\cal M \in {\bf Cat}$ is locally $\mho$-small and has a model structure, how often is the localization $\textsf{Ho}(\cal M)$ a ($\mho\text{-}\bf Set$-)concrete category?

(One could argue that this result really belongs to the world of homotopical categories and should be stated therein: it should, but a model structure is highly tamer to handle).

So:

  • Has anybody attacked this problem with modern technology?
  • Do you think that the above is a valuable question?
  • I believe it is, because
    1. Every category that breaks Isbell condition (let's call it a "non-Isbell category" for the sake of brevity) seems quite nasty. And yet its homotopy theory can be well-understood. Isbell condition itself is stated in terms of the set theory of $\cal A$ and it is (unsurprisingly) linked to $\cal A$ having "nice" factorization systems ("nice" here means proper+something; did somebody explicitly prove this, maybe even Freyd?). So one can "foresee" if $\cal M$ will have a non-concrete localization proving that there is no homotopy-nice factorization system on $\cal M$ (a factorization system on $\cal M$ is homotopy-nice if it is an homotopy FS in the sense of Bousfield, and the FS induced by it on $\textsf{Ho}(\cal M)$ is nice).
    2. All these categories seem to exist (but I will be happy to see you disproving me, especially in the nontrivial cases):
      • a concrete category whose localization is concrete
      • a non-concrete category whose localization is concrete
      • a non-concrete category whose localization is non-concrete
      • a concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is non-concrete
      • a non-concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is concrete
      • a non-concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is non-concrete
      • a concrete category $\cal M$ whose allocalization ${\cal M}[\text{All}^{-1}]$ is non-concrete
    3. There's an interesting problem: every category is a quotient of a concrete category (Kučera, JPAA 1971, link). Is every category a localization of a concrete one?
  • If $\textsf{Ho}(\cal M)$ is not concrete, there should be a property $P$ of $\cal M$ preventing $\textsf{Ho}(\cal M)$ to be Isbell. It seems obvious that every category $\cal N$ which is Quillen equivalent to $\cal M$ has $P$, and yet Freyd's technology seems to be rather context-specific, to the point that it's hard to believe that $P$ can be transported along the adjunction of a Quillen equivalence. Or maybe it is, under suitable assumptions?
$\endgroup$
  • $\begingroup$ Silly observation: In the first and third bullets of 2., you'd need some sort of non-triviality condition. And ditto for some of the others, else you can localise at the isomorphisms. $\endgroup$ – David Roberts Feb 4 '17 at 5:51
  • $\begingroup$ Also: I changed your SD direct link into a doi link, since that is more permanent. $\endgroup$ – David Roberts Feb 4 '17 at 6:00
  • 1
    $\begingroup$ I think a nice source of examples of "concrete homotopical algebra" comes from the hearts of $t$-structures on triangulated categories. I'm not sure, (someone please correct me if this is wrong!) but I suspect that if $C$ is a stable combinatorial model category with a reasonable $t$-structure, then the heart of the $t$-structure ought to be the homotopy category of another combinatorial stable model category, while at the same time being a locally presentable category, and thus concrete. There are also trivial examples, e.g. $(L,C,R)$ for a wfs $(L,R)$ on $C$ (so $ho(C)$ is trivial). $\endgroup$ – Tim Campion Feb 5 '17 at 18:55
  • $\begingroup$ Funny, I never tought about this. This is related to the fact that the heart of the $t$-structure is abelian, and hence equivalent to its $(\infty,1)$-category (in whatever model), right? I'm more comfortable seeing a $t$-structure as a factorization system, on stable quasicategories, as you may remember; so let's state what you said in different terms (cont.) $\endgroup$ – Fosco Feb 5 '17 at 19:09
  • $\begingroup$ (cont.) let $\cal C$ be a stable presentable quasicategory; and $t=({\cal C}^\ge, {\cal C}^<)$ a $t$-structure. Now (HA 1.3.5.23) the heart ${\cal C}^\heartsuit$ is presentable abelian and then "concrete" (but what's a "concrete" $(\infty,1)$-category???), and moreover equivalent to its "fundamental" category $ho({\cal C})$. That's truly inspiring (afaict, this answers a slightly different question) $\endgroup$ – Fosco Feb 5 '17 at 19:14
8
$\begingroup$

This stream of thought led to the following preprint that has been on the arXiv for a few days. https://arxiv.org/abs/1704.00303

This is not an act of self-promotion, but maybe some of the interested readers of this thread can find some development of these ideas.

Comments are welcome, and in fact encouraged.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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