A category is concrete if there is a faithful functor . Most of the categories one initially encounters are in fact concrete: categories of groups, rings, modules, Lie algebras, and so on, and one can think of them as consisting of “structured sets” and “morphisms respecting that structure.” Every small category is concrete, because one can take the Yoneda embedding
followed by the product functor .
Nonetheless, not every category is concrete, and the following example shows that a very natural one is not:
Theorem 1 (Freyd) The homotopy category of pointed spaces is not concrete.
In other words, a homotopy type is somehow too complex to be encoded simply as a set with appropriate structure.
The idea of the proof is essentially the following. In a category of structured sets, a given object can only have so many subobjects, because a set has only so many subsets. But there are categories where an object may have an enormous collection of subobjects, because the definition of a subobject is purely arrow-theoretic. So a category where objects can have lots of subobjects is probably not concrete.
1. The categorical input
Let me now try to set down the argument in a bit more detail. The homotopy category has a special property: it is a pointed category, that is, the initial and terminal objects are the same, and called the zero object. This means that for any two objects , there is a canonical zero morphism (which can be realized by the constant map to the basepoint). Freyd considers, instead of faithful functors , faithful pointed functors
to the category of pointed sets. A pointed functor is one preserving the zero object (and hence the zero morphisms). The following observation becomes useful:
Observation: If a pointed category is concrete, then there is a faithful pointed functor .
The proof is to start with a functor (not necessarily pointed), apply the free abelian group functor, and quotient by the image of the zero object.
The advantage of working in purely a pointed setting is that one can talk about the kernel of a map in a pointed category: it is the equalizer of with the zero morphism.
Next, we come to the set-theoretic business. It is somewhat awkward to deal with subobjects in the homotopy category, so Freyd considers something a bit more general: given an object , he considers an equivalence relation on all morphisms (rather than simply monomorphisms).
Definition 2 Let be a pointed category. We define an equivalence relation on the class of maps for a fixed object . We say that and are equivalent if, for any map , then either and are zero or neither is zero.
This use of equivalence relations on proper classes seems a bit theological, but essentially all we’re going to be interested in the following is whether the set of equivalence classes forms a set.
Definition 3 An object is good if there is a set of equivalence classes of morphisms .
For instance, every object in is good.
One can, of course, phrase this definition in a manner that makes no appeal to proper classes: is good if one has a set of morphisms such that any morphism is equivalent (in the above sense) to one of them.
Here is the main concretizability criterion in the paper.
Theorem 4 (Freyd) Let be a pointed category which admits a pointed functor sending nonzero morphisms to nonzero morphisms (e.g. a concrete category). Then every object of is good.
In fact, Freyd proves the converse as well, but we don’t need it.
The proof of this result is now straightforward. Let be an object of , and let be two morphisms. Then if is not equivalent to , it follows that is not equivalent to . This is because sends nonzero morphisms to nonzero morphisms. This means that there can’t be a huge amount of equivalence classes of morphisms because there aren’t that many equivalence classes of morphisms of pointed sets .
Alternatively, we could phrase this as follows:
Observation: A pointed functor sending nonzero morphisms to nonzero morphisms reflects goodness.
So now, since every object of is good, it follows that every object of must be good. This completes the proof.
To prove that the homotopy category is not concretizable, we’ll now need to produce un-good objects of . In fact, Freyd shows that the suspension of a Moore space is such an example. In order to get this, we’ll need to describe a general method of showing that an object is un-good.
2. Producing un-good objects
A simple example of producing morphisms for a fixed object (a fixed pointed category) is to take maps out of and then to take their kernels. This isn’t very easy to do in the homotopy category; however, we do have something slightly weaker that turns out to be useful.
Definition 5 Let be a morphism in . A weak kernel of is a map
such that for any object , we have an exact sequence of pointed sets
This means that is zero, and conversely, is “versal” with respect to this property. That is, any map which followed by is zero factors non-uniquely through . An example is provided by the theory of fiber sequences in homotopy theory.
We note the following:
Proposition 6 Two maps which are weak kernels (possibly of different maps) are equivalent if and only if factors through and factors through .
The proof follows from the definitions.
In the context of Freyd’s theorem, it is useful, however, to dualize everything: that is, to work not in but in the opposite category. It is easy to check that a category is concrete if and only if its opposite category is concrete. Instead of using the language of the opposite category, we might as well just dualize everything said before. That is, in a pointed category , for an object , there is a natural equivalence relation on maps . We can thus define an object to be co-good if The earlier theorem still applies and shows that if admits a pointed, zero-reflecting functor to , then must possess only co-good objects.
So, how might we produce co-bad (I’m going to resist the temptation to say co-un-good) objects of ? As in the previous section, a natural thing to do is to consider maps
which are weak cokernels (defined dually as in weak kernels) to some map into . Two weak cokernels are equivalent if and only if they factor through each other.
However, there is a very large supply of weak cokernels in homotopy theory: by the Barratt-Puppe sequence, the suspension of any map is a weak cokernel. Freyd’s strategy is to produce a space (the Moore space) together with a lot of maps
for other -torsion groups (where is an arbitrary ordinal).
The suspensions of these maps will turn out not to factor through one another, and consequently one gets a class’s worth of non-equivalent maps out of . This is thus a co-bad object, which proves the non-concreteness of the homotopy category.
4. A little group theory
Surprisingly, the construction will make use of some theory of abelian groups. Let us work with -torsion abelian groups here, for some fixed prime. We will find useful the notion of the height of such a group (actually, that will be sort of implicit).
Definition 7 For an ordinal , we define the functor on the category of -torsion groups as follows by transfinite induction.
- If , then is just the usual definition: the subgroup of all multiples of in .
- If is defined, we set to be . This defines the functor for non-limit ordinals.
- If is a limit ordinal, then we define
These can be nonzero for ordinals much bigger than the countable one, as we’ll see.
For an ordinal , we can give an example of a -torsion group such that consists of a . This means that is zero when .
Construction: We let be defined by the following generators: we take all finite ascending sequences of ordinals with ; the corresponding generator is denoted . The empty sequence corresponds to zero.
We impose the relation that
This guarantees that every element is -torsion. The claim is that this group does it. In fact, one can see by transfinite induction that is the subgroup generated by sequences with . So, is a on the generator .
Notation: We write for this particular generator.
Proposition 8 For , any homomorphism sends to zero.
In fact, that’s because , but . So this is just functoriality.
5. Freyd’s construction
Now we will be able to give Freyd’s construction. Let be an integer. Recall that for every ordinal , we constructed a group together with a map
coming from the generator . We saw that there was no way to find a commutative diagram
if . In other words, the ‘s can’t be factored through each other.
This suggests that if we realize the ‘s in terms of homology (or homotopy), then we’ll be able to solve the problem we desired in . So in fact, let’s consider the Moore spaces and ; we can define maps
inducing the required maps on homology. Then we get an infinite sequence of maps out of which cannot be factored through each other, because this factorization can’t even be done at the level of homology.
So we would be able to conclude that the maps were all non-equivalent to each other for different ‘s if they were weak cokernels. They might not be, but we can fix that: suspend. We can conclude that the maps are all pairwise non-equivalent. Thus is a co-bad space in , and the category cannot be concrete.
Incidentally, it’s possible to rewrite this proof using not homology but homotopy, and replacing the Moore spaces by Eilenberg-MacLane spaces; then one would have to take the loop space of a map rather than the suspension, and use the dual Barratt-Puppe sequence to argue about badness. However, Freyd wants to restrict attention to categories of finite-dimensional CW complexes and taking suspensions (unlike taking loops) preserves that.