5
$\begingroup$

This is more or less a followup of this question. There, it was established that (it is well known that) the homotopy category of topological spaces is not concrete, in other words, there is no faithful functor

$\bf{HoTop}$ $\to$ $\bf{Set}$ .

In this blog post, Akhil Mathew explains that this lack of "concreteness" is due, more or less, to the objects of HoTop having a proper class of subobjects (or of quotients).

But... what if one "does not believe" in proper classes? If I'm not mistaken, every "set vs proper class" phenomenon can be seen, in a suitable foundation that assumes the existence of Grothendieck universes, as a "small universe vs large universe" phenomenon. The non-concreteness of HoTop, translated in this language, would mean that there is no faithful functor

$\bf{Ho(U}$-$\bf{Top)}$ $\to$ $\bf{U}$-$\bf{Set}$

where $\bf{U}$ is a Grothendieck universe and $\bf{U}$-$\bf{Set}$ (resp. $\bf{U}$-$\bf{Top}$) is the category of $\bf{U}$-small sets (resp. $\bf{U}$-small topological spaces). Now, what can happen if $\bf{U}\in\bf{V}$ for $\bf{V}$ a larger Grothendieck universe? My question is

Are there faithful functors $\bf{Ho(U}$-$\bf{Top)}$ $\to$ $\bf{V}$-$\bf{Set}$ ?

The same question, of course, can be asked for any other non-concrete category.

$\endgroup$
1
  • 5
    $\begingroup$ Any small category is concrete by just taking the coproduct of all the Hom functors. $\endgroup$ Commented Dec 12, 2014 at 17:03

2 Answers 2

6
$\begingroup$

By the Yoneda embedding, any category is a full subcategory of its category of presheaves. For this to be valid, the collection of objects of the category has to be a set. So if you enlarge the universe to make this the case then you can make any category concrete.

That being said, I "don't believe in" (the usefulness of) any of this terminology, either the vocabulary associated with "sets" or that of "concrete" categories.

$\endgroup$
8
$\begingroup$

As already noted above, any category can be considered concrete after a base change to a suitably large universe. However, doing so would be completely missing the point of concreteness. The underlying topos of sets should really be considered not as a first-class object, but rather as a background reality within which you are doing mathematics. Some things are describable and studyable within this "reality", but some are hopelessly out of reach. Concreteness serves as a (vague) borderline between these cases. Basically, a concrete category is a one that can be reasonably studied once you fix the collection of objects (=sets) that interest you. It's not a strict rule, of course. For example, both of the categories $Set^{op}$ and $Rel$ (sets as objects, relations as morphisms) are not concrete, but are good enough to study. The reason is that they can be constructed from a concrete one in a simple way.

Note that any algebraic category or a category of sheaves on a small site is concrete. In a sense that is a general example: some theory defined via a list of first-order axioms. It's not exhaustive of course, but it is what usually happens in practice.

Going back to $Ho(Top)$, the theory essentially states that it is hopelessly complicated from the PoV of classical set theory. And it really is: it doesn't (usually) have limits and colimits, and they are badly behaved even when they exist, and doing algebra in this category, or studying its parametrized versions is a painful and fruitless experience. You can enlarge the universe, but this new universe on its own will be hopelessly complicated wrt your base set theory (at least as complex as the category you are embedding), so the net gain is zero, except for a formal concreteness property which no one really cares about. The moral is, you need some genuinely new ideas to do homotopy theory. You must either get a much better understanding of passage to the homotopy category (=invent model categories), or find some related, but entirely different and better behaved object (=higher categories). Both of these approaches work, with my love for the latter.

$\endgroup$
2
  • 10
    $\begingroup$ In fact, $\mathbf{Set}^\mathrm{op}$ and $\mathbf{Rel}$ are concrete. First, note that the former embeds in the latter in an obvious way. Then note that $\mathbf{Rel}$ can be embedded in $\mathbf{Set}$ by sending each set $X$ to $2^X$. $\endgroup$
    – Zhen Lin
    Commented Dec 13, 2014 at 6:13
  • 1
    $\begingroup$ $\mathbf{Set}^\mathrm{op}$ is equivalent to the category of complete atomic Heyting algebras (you may read "complete atomic Boolean algebras" here, if you're willing to work in classical logic) and therefore concrete as a subcategory of a concrete category. $\endgroup$ Commented Apr 30, 2017 at 19:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .