# How should I think about concrete functors and in particular about concrete isomorphism?

All the definitions that follow is taken from The Joy of Cats.

Definition 1. Let $$\bf{X}$$ be a category. A concrete category over $$\bf{X}$$ is a pair $$({\bf{A}},U)$$, where $$\bf{A}$$ is a category and $$U :{\bf{A}} \to X$$ is a faithful functor.

Definition 2. If $$({\bf{A}},U)$$ and $$({\bf{B}}, V )$$ are concrete categories over $$\bf{X}$$ , then a concrete functor from $$({\bf{A}},U)$$ and $$({\bf{B}}, V )$$ is a functor $$F : {\bf{A}}\to {\bf{B}}$$ with $$U = V \circ F$$. We denote such a functor by $$F : ({\bf{A}},U)\to ({\bf{B}}, V )$$.

Definition 3. If $$({\bf{A}},U)$$ and $$({\bf{B}}, V )$$ are concrete categories over $$\bf{X}$$ , then a concrete functor from $$F:({\bf{A}},U)\to ({\bf{B}}, V )$$ is said to be a concrete isomorphism iff $$F:{\bf{A}}\to{\bf{B}}$$ is an isomorphism.

## Question

We know that isomorphic categories can be viewed as being "essentially same". But then how should I view concretely isomorphic concrete categories? More generally how should I view a concrete functor?

In The Joy of Cats the following is written,

A concrete isomorphism $$F : (\mathbf{A},U) \to (\mathbf{B}, V )$$ between concrete categories over $$\bf{X}$$ is a concrete functor that is an isomorphism of categories. ...That such a concrete isomorphism exists means, informally, that each structure in $$\bf{A}$$, i.e., each object $$A$$ of $$\bf{A}$$, can be completely substituted by a structure in $$\bf{B}$$, namely $$F(A)$$ (keeping, of course, the same morphisms). For example, the standard descriptions of topological spaces by means of

• neighborhoods,

• open sets,

• closure operators, or

• convergent filters,

give technically different constructs, all of which are concretely isomorphic. This is why the differences between the various descriptions are regarded as inessential and we can in good conscience call each of them "$$\bf{Top}$$". The concept of concretely isomorphic concrete categories gives rise to an equivalence relation that is stronger than the relation of isomorphism of categories. For example, assuming that no measurable cardinals exist, $$\bf{Top}$$ (and, indeed, any construct) can be thought of as being isomorphic to a full subcategory of $$\bf{Rel}$$. However, $$\bf{Top}$$ is not concretely isomorphic to such a subcategory, because there are more topologies on $$\Bbb{N}$$ (namely, $$2^{2^{\aleph_0}}$$) than there are binary relations on $$\Bbb{N}$$ (namely, $$2^{\aleph_0}$$).

Here $$\bf{Top}$$ denotes the category of topologies and continuous functions and $$\bf{Rel}$$ denotes the category whose objects are pairs of the form $$(X,\rho)$$ (where $$\rho$$ is a binary relation on the set $$X$$) and whose morphisms are relation-preserving functions.

But I don't understand why the same description as given in the following, "That such a concrete isomorphism exists means, ...the same morphisms)." can't be said about isomorphic categories.

## 2 Answers

A concrete category should be thought of more or less as a category of structured sets (sets equipped with some sort of specified structure) and morphisms between them; the faithfulness to $$Set$$ means that morphisms are completely determined by their underlying functions.

For concrete isomorphism, let's take an example different from the one in The Joy of Cats: modules over a matrix algebra $$M = M_n(\mathbb{R})$$ (for simplicity, let's keep everything finite-dimensional). This algebra acts on $$V = \mathbb{R}^n$$, and in fact every module is a power $$V^k$$ of this one simple module. Moreover, $$M$$-module maps $$V^k \to V^j$$ are in natural bijection with ordinary $$j \times k$$ matrices with coefficients in $$\mathbb{R}$$. It follows that the category of $$M$$-modules of the form $$V^j$$ is abstractly isomorphic to the category of $$\mathbb{R}$$-vector spaces $$\mathbb{R}^j$$.

Does that mean we simply identify $$M$$-modules with vector spaces? Not necessarily, because according to how we think of these structures, the underlying sets are quite different ($$V^j$$ in one case, $$\mathbb{R}^j$$ in the other). They are different as structured sets, even though abstractly the categories are isomorphic.

Cf. a common abuse of language, where we say a category $$C$$ is monadic over $$Set$$. Normally we don't worry about this too much because we have a specific underlying set functor $$U: C \to Set$$ in mind, but really the question should be: monadic how? We should really be speaking of functors $$U: C \to Set$$ as monadic. For another example, consider finite Boolean algebras. Usually we say that a finite Boolean algebra has $$2^n$$ elements for some $$n$$, but actually there are other monadic underlying set functors $$Bool \to Set$$ where we would be perfectly within our rights to say, for example, that the underlying sets have cardinality $$3^n$$.

In summary, concrete isomorphism is a more refined notion which captures a way in which objects are viewed as sets with structure.

• Does there exist any criteria of two categories being concretely isomorphic in terms of fibre subcategories? It seems that (and I now came to realize this after going through the example given in The Joy of Cats) that if two concrete categories are concretely isomorphic then their fibre subcategories must also be isomorphic (i.e., roughly the "number" of structured objects over the base category should be same and the "number" of relations they are in should also be same). Is this a good intuition? – user57432 Aug 1 '19 at 7:14
• I don't know of any general such criterion. A trouble is that considering the fibers abstractly on their own loses the information of how they are "woven together" via the underlying-set functor. But an enhanced criterion in this vein holds for fibrations, where you could say that two fibrations over $C$ are concretely equivalent if their associated fiber functors $C \to Cat$ are naturally equivalent. – Todd Trimble Aug 1 '19 at 11:59
• My answer has just been downvoted. I invite the downvoter to explain why. – Todd Trimble Dec 27 '19 at 15:41

Let me add little to Todd's answer, because you asked specifically about "the same morphisms". When we have a concrete category $$(\mathbf A,U)$$ over $$\mathbf X$$, we can use $$U$$ to identify morphisms $$f:a\to b$$ in $$\mathbf A$$ with their images $$U(f):U(a)\to U(b)$$ in $$\mathbf X$$. This works because $$U$$ is faithful. In other words, $$U$$ provides a monic function from the set $$\mathbf A(a,b)$$ to $$\mathbf X(U(a),U(b))$$, and we use this monomorphism of sets to identify $$\mathbf A(a,b)$$ with a subset of $$\mathbf X(U(a),U(b))$$. Note that this agrees with common terminology in "ordinary" (non-category-theoretic) mathematics: "A homomorphism of groups is a function such that $$\dots$$"; "a smooth embedding is a function such that $$\dots$$"; etc. So when the passage you quoted says "the same morphisms", it really means "morphisms in $$\mathbf A$$ and $$\mathbf B$$ that become the same when identified, via $$U$$ and $$V$$, with morphisms in $$\mathbf X$$". So this "same" really boils down to just the requirement that $$V\circ F=U$$.

• I have one more small question. What does the phrase "[t]hat such a concrete isomorphism exists means, informally, that each structure in $\bf{A}$, i.e., each object $A$ of $\bf{A}$, can be completely substituted by a structure in $\bf{B}$, namely $F(A)$ (keeping, of course, the same morphisms)" mean? Especially what exactly should be "same morphisms"? Is it "same $\bf{A}$-morphisms" or is it "same $\bf{B}$-morphisms"? What exactly is meant by "complete substitution" in this context? – user57432 Aug 1 '19 at 15:00
• @user170039 As I said in my answer, "same morphisms" means "same $\mathbf X$-morphisms" (after using $U$ and $V$ to identify morphisms from $\mathbf A$ and $\mathbf B$ with morphisms in $\mathbf X$). "Completely substituted" means that an object $A$ of $\mathbf A$ and $F(A)$ in $\mathbf B$ can be regarded as the same object of $\mathbf X$, with the extra structure that makes them elements of $\mathbf A$ and $\mathbf B$. – Andreas Blass Aug 1 '19 at 18:20