# Homotopical algebra is not concrete

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?
• 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. – David Roberts Feb 4 '17 at 5:51
• Also: I changed your SD direct link into a doi link, since that is more permanent. – David Roberts Feb 4 '17 at 6:00
• 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). – Tim Campion Feb 5 '17 at 18:55
• 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.) – Fosco Feb 5 '17 at 19:09
• (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) – Fosco Feb 5 '17 at 19:14