Question

First of all, I know the concepts of isomorphism and equivalence between categories, and that the latter one is the more interesting one, whereas the first is rather rare and uninteresting.

Are there isomorphisms of categories, which are not trivial and not pathological? I regard the examples on wikipedia as trivial, because these are only reformulations of the definitions of the objects in consideration. Thus perhaps the question is: Are there nontrivial reformulations?

There are lots of nontrivial equivalences of categories (affine schemes <-> rings (dual), compact hausdorff spaces <-> unital commutative C*-algebras (dual), finite abelian groups <-> finite abelian groups (dual), skeletons such as the algebraic extensions of function fields over fixed prime fields in the category of fields), but I wonder if these categories are actually isomorphic. Of course, in the examples of interest, you can't take the known equivalence as an isomorphism, but perhaps there is another one?

Answer 1

Whether this counts as trivial is a subjective matter, but here goes.

Any adjunction $$ F: C \to D,\ \ \ G: D \to C $$ (with $F$ left adjoint to $G$) gives rise canonically to a monad $T = GF$ on $C$ and a "comparison" functor $K: D \to C^T$. Here $C^T$ is the category of algebras for the monad $T$. The adjunction is said to be monadic if $K$ is an equivalence of categories.

Now in fact, for most of the obvious examples of monadic adjunctions, the comparison is actually an isomorphism. For example, if $G$ is the forgetful functor from groups to sets then it's an isomorphism. The same is true if you replace groups by any other algebraic theory (rings, Lie algebras, etc).

Indeed, if you look in Categories for the Working Mathematician, you'll see that Mac Lane calls a functor monadic if $K$ is an isomorphism. He does the whole basic theory of monads with this definition. I suspect this is because $K$ really is an isomorphism in the standard examples. CWM was published in 1971, and since then it's become clear that Mac Lane's definition was too narrow. Whether the pioneers of monad theory (such as Beck) also used this narrow definition, I don't know.

Answer 2

A different flavor of example:

Connes' cycle category Λ can be described as follows. It has one object (n) for each positive integer n which we think of as an oriented circle with n marked points. A map from (n) to (m) is an isotopy class of degree 1 maps which sends marked points to marked points. Alternatively, we can think of it as a map between the sets of marked points which preserves the cyclic orderings. (Note: I am calling (n) what is usually called something like [n-1], for reasons that aren't relevant here.)

Given a map f : (n) → (m), we can also look at what happens to the intervals of the circle between the marked points. Each interval in (m) is hit by exactly one interval of (n), and the data of, for each interval of (m), which interval of (n) hits it, determines f. So, f also determines a map from an arrangement of m arcs on a circle to an arrangement of n arcs on a circle. The conclusion: Λ is isomorphic to its opposite category Λ^op.

If you prefer working with the presentation of Λ by generators and relations, then the generator $d_i$ corresponding to inserting a new point in an interval is "dual" to the generator $s_j$ collapsing the two resulting intervals to one (and rotation is "dual" to rotation).

This fact is worth knowing when learning about Hochschild homology if only so that you don't use it accidentally! If you lose track of whether you are attaching your algebra to the marked points or the intervals of the circle, confusion will ensue.

Answer 3

One general rule that unites some of the examples above is that if you have two categories whose objects are sets endowed with some structure, and there is an equivalence between these two categories that assigns to a set with a structure the same set with a different (but equivalent) structure, than such an equivalence of categories is an isomorphism of categories. One can also have objects of some other fixed category in place of sets and some collections of morphisms in place of the structures on sets (see the very last example below).

To give a simple nontrivial example of this, the category of G-modules for a group G is isomorphic to the category of modules over the group ring Z[G], or the category of modules over a Lie algebra g is isomorphic to the category of modules over its enveloping algebra U(g), or the category of comodules over a finite-dimensional coalgebra C is isomorphic to the category of modules over the dual algebra C*.

Another series of examples of isomorphisms of categories is provided by equivalences between categories whose classes of objects are the same though morphisms are different but isomorphic. This includes equivalences between various quotient categories or localizations of a given category (which all have the same objects as the original category).

Here is another example of this kind. Let C be a category, R:C->C be a monad on C, and L:C->C be a functor left adjoint to C. Then L is a comonad. The categories of R-algebras and L-coalgebras in C can be quite different. However, one can consider the category of free R-algebras in C; this is a category whose objects are formally just the objects X of C while morphisms X->Y are the R-algebra morphisms R(X)->R(Y). Analogously one defines the category of cofree L-coalgebras in C whose objects are the objects X of C and morphisms are the L-coalgebra morphisms L(X)->L(Y). Then the categories of free R-algebras and cofree L-coalgebras are isomorphic; this is called the isomorphism of Kleisli categories. To give a concrete example of this, the categories of cofree left comodules and free left contramodules over a given coalgebra are isomorphic.

To compare, when L:C->C is a monad and R:C->C is right adjoint to L, then R is a comonad and the whole categories of L-algebras and R-coalgebras in C are isomorphic.

Answer 4

For any locally profinite group $G$, the category of smooth representations of $G$ is on the nose isomorphic to the category of smooth modules over its Hecke algebra.

Answer 5

Hi Martin.

Is the following example non-trivial? There are (at least) two possible definitions of an uniform space over a set $X$:

1. A uniformity can be defined as a non-empty set $\Sigma$ of covers of $X$ such that $\Sigma$ is closed wrt "upward" refinements (i.e. $\alpha\in\Sigma \wedge \alpha\preceq\beta \implies \beta\in\Sigma$), every $\alpha\in\Sigma$ has a star-refinement in $\Sigma$.

2. A uniformity can be defined as a filter $\mathcal{R}$ on $X\times X$ such that for all $R\in\mathcal{R}$ we have $\Delta_X\subseteq R$, $R^{-1}\in\mathcal{R}$ and $\exists S\in\mathcal{S}: S\circ S\subseteq R$.

Both definitions give rise to a category of uniform spaces. Both categories are isomorphic.

Answer 6

I don't know if the following example may be considered as trivial, but it's quite useful.

Let $\cal{C}$ be a category, $\cal{S}$ a class of morphism in $\cal{C}$.

Assume that, for instance, $\cal{S}$ is a class of homotopy equivalences. By which I mean that you have a cylinder (or path object) for every object in $\cal{C}$ -for example, because it is a Quillen model category-, and $\cal{S}$ is the class of morphism which are invertible up to the homotopy relation $\sim$ generated by these path or cylinder objects.

Then, on one hand, you can consider the quotient category $\cal{C}/\sim$, whose objects are those of $\cal{C}$ and whose morphisms are the homotopy classes of morphisms.

On the other hand, you can consider the localized category $\mathrm{Ho}\cal{C}$, with the same objects, but inverting the morphisms of $\cal{S}$.

Well, at least when your homotopy relation $\sim$ is generated by a cylinder or path object, these two categories are canonically isomorphic.

Remark. Do not confuse my statement with Quillen's equivalence of categories. I'm sorry for the notation $\mathrm{Ho}\cal{C}$, but I don't know how to write square brackets here.

Answer 7

Stone's representation theorem gives you an isomorphism between every Boolean algebra and a field of sets. Viewed categorically, this is an isomorphism of categories, since isomorphism and equivalence coincide for partial orders viewed as categories.

Answer 8

The categories of Boolean algebras and of Boolean rings (rings in which $a^2=a$ for all $a$) are isomorphic. The reason is that given a Boolean ring $(R,+,\cdot,0,1)$ one can define a Boolean algebra structure on its underlying set one can define $a \wedge b:= a \cdot b$ and $a \vee b:= x+y+x\cdot y$ and $\neg a:=1+a$ and gets a Boolean algebra.

Vice versa, given a Boolean algebra $(B,\vee,\wedge,0,1)$ gives a Boolean ring via $a \cdot b:=a \wedge b$ and $a+b:=(a \vee b) \wedge \neg (a \wedge b)$

If you go back and forth you get exactly the same ring/Boolean alg. structure, the underlying set didn't change anyway. I don't know if you consider this non-trivial. But I think an isomorphism of categories should be thought of as reformulation of structure.

Answer 9

I just remember an example by myself.

Let $C$ be the following algebraic category: Objects are nonempty sets $G$ together with a binary operation $/ : G \times G \to G$, such that for all $x,y,z \in G$, we have

$x/((((x/x)/z)/(((x/x)/x)/z))=y.$

A morphism $(G,/) \to (G',/)$ is a map $f : G \to G'$ preserving $/$.

This category is isomorphic to the category of groups, i.e. groups can be described by a single equation! If $G \in C$, then the corresponding group is $G$ together with the multiplication $ab := a/((a/a)/b)$. If $G$ is a group, then define $x/y = x y^{-1}$.

Reference: Higman, Graham und Neumann, Bernhard: Groups as groupoids with one law, Publicationes Mathematicae Debrecen, 2 (1952), 215-227.