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Stream: learning: questions

Topic: Concrete categories where the hom-sets are subsets

Bernd Losert (Jul 13 2024 at 21:53):

Let (A, U) be a concrete category over a category X. In seems that in most real examples of concrete categories, it is always the case that A(A, B) ⊆ X(UA, UB). Is there a special name for this kind of concrete category? Why isn't this condition made part of the definition?

Ivan Di Liberti (Jul 13 2024 at 22:10):

what does concrete mean to you?

Bernd Losert (Jul 13 2024 at 22:10):

I'm using the definition from Joy of Cats: U : A → X is faithful.

Bernd Losert (Jul 13 2024 at 22:15):

Screenshot-2024-07-14-at-00.13.18.png
Joy of Cats adopts this convention. But why make it a convention?

Julius Hamilton (Jul 14 2024 at 01:20):

I am under the impression that you believe:

“in most real examples of concrete categories, $\mathbf{A}(A, B) \subseteq \mathbf{X}(UA, UB)$”

because you are thinking of the common definition of a concrete category as a category whose objects are sets “with additional structure” (such as groups). I think that this might imply your condition. For example, I think every group homomorphism is a map between the elements of two groups, subject to certain conditions. Naturally, those maps are a subset of the set of all possible maps between the elements of those groups, because some of those maps aren’t homomorphisms.

In the Joy of Cats, they use a different terminology:

concrete_vs_construct.jpg

They say that a concrete category “over $X$” is a category with a faithful functor into $X$. $X$ is any arbitrary category. This generalizes the above “common” definition of a concrete category, which would now be seen as “a concrete category over $\textbf{Set}$”. But they call that a “construct”.

So the reason why you wouldn’t assume in general $\textbf{A}(A, B) \subseteq \textbf{X}(UA, UB)$ is because these two Hom-sets are not in the same category. Especially in an abstract category, there is not a way to say the morphisms of category $\textbf{A}$ “are the same morphisms as” the morphisms in a different category $\textbf{X}$, unless you state they are equal.

Their idea of a “concrete category over $X$” can be thought of like this:

• First, you choose any arbitrary category $X$. This can be thought of as your “point of reference” or “point of view”.

• Then, you choose some other arbitrary category, $A$.

• To think of $A$ as “concrete” over $X$ (I think) can be thought of as “there exists a family of maps between the Hom-sets of both categories, all of which are injective”.

Recently, I have been thinking about how (I think) an injective mapping $f : X \to Y$ can be thought of as related to the notion of subsets: $X$ is not necessarily a subset of $Y$; they could be disjoint sets that share no elements. But $X$ is isomorphic to a subset of $Y$.

I think in category theory, which has so much focus on structure, two things being isomorphic is often a more useful property than them being actually equal or identical things.

So, this seems to me to be a “categorical” way of stating that the structure of $\textbf{A}$ can be “linked” to the structure of $X$, since though it is not the case that the morphisms in A are in X, but that the Hom-sets in A are “identifiable” (isomorphic) to subsets of the Hom-sets in X. So to me it is a way of expressing that A is sort of “derivative”, structurally, of X.

Is there a special name for this kind of concrete category?

I think that would be a sub-category, because if $\textbf{A}(A, B) \subseteq \textbf{X}(UA, UB)$, I think it follows that all the source objects and target objects of all the morphisms in $\textbf{A}$ are objects in $\mathbf{X}$, too. I think this would imply that all the “conventional” concrete categories like $\textbf{Grp}, \textbf{Top}$, etc., are sub-categories of $\textbf{Set}$.

Todd Trimble (Jul 14 2024 at 01:52):

I think Julius has summed up the matter pretty well. Ultimately, the convention $\mathbf{A}(A, B) \subseteq \mathbf{X}(UA, UB)$ seems to amount to a linguistic convenience of being able to say, without blushing, "this $\mathbf{X}$-morphism $f: UA \to UB$ is an $\mathbf{A}$-morphism", instead of the somewhat more cumbersome "this $f: UA \to UB$ is the underlying $\mathbf{X}$-morphism of a (uniquely determined) $\mathbf{A}$-morphism, i.e., a morphism $g: A \to B$ such that $f = U(g)$". I wouldn't assign any more significance to it than that convenience.

In fact they don't even have to enforce the subset condition; they could just explain how they plan to use language. But they probably felt it wasn't a big deal either way: people are used to thinking about $\subseteq$ anyway.

David Michael Roberts (Jul 14 2024 at 03:22):

At the cost of passing to an isomorphic category with the same objects, a faithful functor can be replaced by one where the domain's hom sets are literal subsets of the target's.

Bernd Losert (Jul 14 2024 at 07:37):

Thank you for the responses.

So the reason why you wouldn’t assume in general A(A,B)⊆X(UA,UB) is because these two Hom-sets are not in the same category.

I'm not sure I understand this statement. Surely you can have two different categories with different objects, but the homsets happen to be e.g. sets of natural numbers.

I wouldn't assign any more significance to it than that convenience.

But if you have to adopt strange linguistic conventions to work with a definition, then I think that says something about the deficiency of the definition, no? I honestly can't think of any good examples where the extra generality of the definition of "concrete category" is used. I can only think of some artificials examples.

John Baez (Jul 14 2024 at 07:52):

In category theory we generally want properties and structures to be invariant under equivalence of categories. So, if $U: A \to \mathsf{Set}$ is a concrete category and we have an equivalence $f: B \to A$ we want $U \circ F: B \to \mathsf{Set}$ to be concrete as well. This forces the greater generality of the definition of concrete category.

Todd Trimble (Jul 14 2024 at 08:00):

But if you have to adopt strange linguistic conventions to work with a definition

Part of my point (see the second paragraph) was that you don't.

Bernd Losert (Jul 14 2024 at 08:41):

John Baez said:

... we have an equivalence $f: B \to A$ we want $U \circ F: B \to \mathsf{Set}$ to be concrete as well. This forces the greater generality of the definition of concrete category.

Thanks for pointing this out. That does indeed make a lot of sense.

Ralph Sarkis (Jul 14 2024 at 08:49):

The first example that came to my mind is the category $\mathbf{Mat}(k)$ of matrices over a field $k$. Its objects are natural numbers and morphisms are matrices, but it can be made concrete over $\mathbf{Set}$ or $\mathbf{Vect}_k$ by sending a number to a vector space of that dimension and a matrix to the corresponding linear transformations.

Bernd Losert (Jul 14 2024 at 09:02):

Thanks for the Mat(k) example - it is a good example where the subset condition makes no sense.

Vincent Moreau (Jul 14 2024 at 14:43):

To me it makes complete sense to consider that the hom-set $\mathbf{Mat}(k)(m, n)$ is a subset of $\mathbf{Set}(k^m, k^n)$. It is the subset made of functions which are linear!

Vincent Moreau (Jul 14 2024 at 14:50):

While I'm at it, I have the feeling that somehow, equipping a category $\mathbf{C}$ with a faithful functor into $\mathbf{Set}$ amounts to some kind of determinization process. In the case of $\mathbf{FinRel}$ for instance, the category of finite sets and relations between them, the usual functor into $\mathbf{Set}$ is the powerset functor, and relations $R \subseteq X \times Y$ correspond exactly to join-preserving function $\mathcal{P}(X) \to \mathcal{P}(Y)$ sending $\{x\}$ on $\{y \mid (x, y) \in R\}$. This is closely related to determinization in automata theory. This example and the one of $\mathbf{Mat}(k)$ are instances of the general case of matrices over semirings.

Ralph Sarkis (Jul 14 2024 at 16:14):

Vincent Moreau said:

To me it makes complete sense to consider that the hom-set $\mathbf{Mat}(k)(m, n)$ is a subset of $\mathbf{Set}(k^m, k^n)$. It is the subset made of functions which are linear!

Sure, but then you are giving another definition of the category of matrices. The set of matrices is not a subset of the set of functions. However, there is an injection between them, so we can still consider my definition of $\mathbf{Mat}(k)$ to be concrete over $\mathbf{Set}$, and that effectively identifies $\mathbf{Mat}(k)(m, n)$ with the subset of $\mathbf{Set}(k^m, k^n)$.

David Michael Roberts (Jul 15 2024 at 01:04):

@Bernd Losert I think the discussion above highlights the difference between a concretisable category, and a concrete category. The latter is equipped with a specific faithful functor, and with that fixed, one can gloss over the slight difference between a subset inclusion and a specified injection (technically speaking, the hom-sets of the domain are representatives of subobjects of the codomain). The former has ... some faithful functor, but we don't really know how to identify the domain hom-sets with subsets of the codomain hom-sets, because we aren't given any particular injection.