Homotopical algebra is not concrete

Question

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

• there is a homotopical category $(\bf Top, \cal W)$, such that the localization ${\bf Top}\!\!\left[{\cal W}^{-1}\right]$ is not ($\bf Set$-)concrete.

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?

Answer 1

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.