# nontrivial isomorphisms of categories

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?

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I'd conjecture that every pair of isomorphic 'real life' categories is a trivial example, but I'd surely love to see an example! –  Mariano Suárez-Alvarez Feb 3 '10 at 16:59
is this question appropriate for community wiki? –  Martin Brandenburg Feb 3 '10 at 18:28
I am puzzled/confused by something. The wiki page claims that a functor F:C \to D is an isomorphism of categories iff it is bijective on objects and on morphism sets. Does this not apply to the case of unital abelian C*-algebras and CHff spaces? –  Yemon Choi Feb 3 '10 at 18:52
@Yemon: I think you are confusing "equal to" with "is isomorphic to". "Every X is isomorphic to the space of functions on some Y" is not the same as "Every X is, as a set, exactly equal to the set of functions on some Y (and so in particular every element of X is a set which happens to be a function)" –  Kevin Buzzard Feb 3 '10 at 18:57
Maybe this is not what you are looking for, but anyway. Consider for example the category of complexes of abelian groups. Then a non trivial isomorphism to itself is given by degree shifting. –  Jan Weidner Feb 3 '10 at 21:14

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.

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this isomorphism for algebraic theories is trivial (actually, $K$ is then the identity, using the standard definitions!), but I found this very interesting. therefore I'll vote it. –  Martin Brandenburg Feb 3 '10 at 17:49
Glad you found it interesting; thanks for saying so. I'm puzzled as to why you say K is the identity, though - I think you and I must be using different definitions. E.g. if G is the forgetful functor from D = Grp to C = Set then - according to what I think of as the standard the definitions - an object of D is a group (formally, a triple (X, m, e) where X is a set, m is multiplication, and e is an identity element) whereas an object of C^T is a pair (X, h) where X is a set and h is a function from (free group on X) to X satisfying some axioms. So K isn't then the identity. –  Tom Leinster Feb 4 '10 at 2:31
Sorry I confused something. –  Martin Brandenburg Feb 4 '10 at 2:56

Here are two of my favorite examples, both taught regularly to undergraduates:

Galois extensions: if $L$ is a Galois extension of $K$ with Galois group $G$, then the opposite of the category of orbits $G/H$ and $G$-maps is isomorphic to the category of intermediate fields, via $G/H\mapsto L^H$.

The categories of finite $T_0$-spaces and finite posets are isomorphic; the categories of Alexandroff $T_0$-spaces and all posets are isomorphic.

As Tom says, trivial is subjective, but these are certainly both elementary and illuminating. The first subsumes a bunch of things usually taught as separate propositions. The second is a bridge between algebraic topology and combinatorics.

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I see Martin added the second example, but I think it wasn't there when I started answering :) –  Peter May Jun 28 '12 at 17:55

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 $\mathbb{Z}[G]$, or the category of modules over a Lie algebra $\mathfrak{g}$ is isomorphic to the category of modules over its enveloping algebra $U(\mathfrak{g})$, or the category of comodules over a finite-dimensional coalgebra $C$ is isomorphic to the category of modules over the dual algebra $C^\ast$.

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, $R:C\rightarrow C$ be a monad on $C$, and $L:C\rightarrow C$ 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 $X\rightarrow Y$ are the $R$-algebra morphisms $R(X)\rightarrow R(Y)$. 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 $L(X)\rightarrow L(Y)$. 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 $L:C\rightarrow C$ is a monad and $R:C\rightarrow C$ is right adjoint to $L$, then $R$ is a comonad and the whole categories of $L$-algebras and $R$-coalgebras in $C$ are isomorphic.

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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 Λop.

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.

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I'm afraid that in my case confusion always ensues when I start messing around with cyclic cohomology ;) –  Yemon Choi Feb 4 '10 at 4:11

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.

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What you are saying isn't true for an arbitrary Quillen model category C, e.g., this isn't true for the model category of topological spaces with weak homotopy equivalences as weak equivalences. To make your assertion correct in general, one has to take C to be the full subcategory of fibrant-cofibrant objects in a model category, rather than the whole model category. –  Leonid Positselski Feb 3 '10 at 18:57
No. I'm afraid you are mistaking what I'm saying. The class S is not a class of "weak equivalences", but of "homotopy equivalences". Maybe this is my fault because of my notation "HoC": notice that I said that this is the localized category respect_to_S, not respect any class of "weak equivalences" that I didn't need at all. You can find the proof in "A Cartan-Eilenberg approach to homotopical algebra", JPAA 214, 140-164 (2010), proposition 1.3.3 and example 1.3.4. –  a.r. Feb 3 '10 at 19:30
Oh, so I misunderstood you, sorry. Thanks for the reference. –  Leonid Positselski Feb 3 '10 at 19:42

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.

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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.

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Isn't it the case that, if $C$ and $D$ are equivalent categories and if, in both of these categories, each object is isomorphic to a proper class of other objects, then $C$ and $D$ are isomorphic (assuming global choice)? So, for example, the category of non-trivial commutative rings and the dual of the category of nonempty affine schemes are isomorphic. (I had to exclude the empty scheme, and therefore the trivial ring, because there's only one empty scheme but lots of trivial rings, which would mess up any attempt at an isomorphism.) More generally, if $F:C\to D$ is an equivalence of categories and if, for each object $a$ in $C$, the number of isomorphic copies of $a$ in $C$ equals the number of isomorphic copies of $F(a)$ in $D$, then there should (again with a generous use of choice) be an isomorphism from $C$ to $D$ (that is, furthermore, naturally isomorphic to the given $F$).

EDIT: Martin asked in a comment for a proof; I'll put a proof (or at least a sketch, which I hope will suffice) into the answer because it won't fit into a comment. Suppose $F:C\to D$ is an equivalence and, for each object $a$ of $C$, the isomorphism classes of $a$ and $F(a)$ are the same size. In $C$, choose one representative object from each isomorphism class of objects; write $a^*$ for the representative of the isomorphism class of $a$. Also choose, for each object $a$, an isomorphism $i_a:a\to a^*$, subject to the convention that $i_{a^*}$ is the identity morphism of $a^*$. Do the same in $D$, but, instead of arbitrarily choosing the representative objects, use the objects $F(a^*)$; there's exactly one of these in each isomorphism class, because $F$ is an equivalence. But the isomorphisms $i_b$, from objects $b$ of $D$ to the representatives, are still chosen arbitrarily except that, as before, for the representatives themselves we use identity morphisms. Now define a new functor $F':C\to D$ as follows. On the representative objects $a^*$, it agrees with $F$. On other objects, it acts in such a way that the isomorphism class of any $a^*$ is mapped bijectively to the isomorphism class of $F(a^*)$; this is possible because I assumed that these isomorphism classes have the same size. Finally, if $f:a\to b$ is a morphism in $C$, then $F'$ should send it to the following mess: $$i_{F'(b)}^{-1}F(i_bf{i_a}^{-1})i_{F'(a)}.$$ In perhaps more understandable language: Use $i_a$ and $i_b$ to transport $f$ to a morphism from $a^*$ to $b^*$, apply $F$ to that, and then transport the result to map $F'(a)\to F'(b)$ via the chosen isomorphisms in $D$. It should be routine to check that this $F'$ is an isomorphism.

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Since a scheme consists of a topological space and a sheaf of rings, it is not true that there is only one empty scheme. In fact, there is a canonical bijection between the set of empty schemes in a given universe and the set of trivial commutative rings (i.e., rings of cardinality 1) in the same universe. –  Fred Rohrer Jun 28 '12 at 17:47
Thanks, Fred. So my answer can be improved by deleting "non-trivial" and "nonempty", so as to exactly match one of the equivalent pairs mentioned in the question. –  Andreas Blass Jun 28 '12 at 18:00
Andreas, what is the proof of your claim? –  Martin Brandenburg Jun 28 '12 at 19:33
Martin: This is part of Exercise A in Chapter 3 of Freyd's "Abelian Categories" (page 74). I was tempted to say that MO is not for homework, so I'd omit the proof. But that seems like cheating, so I edited a sketch of the proof into my answer. –  Andreas Blass Jun 28 '12 at 20:18
@Martin: I was assuming global choice, which makes all proper classes have the same size; every proper class is equinumerous with the class of all ordinals. By the way, the exercise in Freyd's book is about the special case where all the isomorphism classes are proper classes. But the proof is the same if some (or all) of them are sets, as long as corresponding ones have the same size. –  Andreas Blass Jun 28 '12 at 22:40

The category of inverse semigroups is isomorphic to the category of etale groupoids whose unit and arrow spaces are Alexandrov spaces and the poset associated to the unit space is required to be a meet semiattice. Morphisms are required to preserve these meets.

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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.

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In the same spirit, you could mention logically-equivalent definitions of topological space: as a set equipped with open subsets, or closed subsets, or a closure operator, or an interior operator, or neighbourhoods, or... All give isomorphic categories. –  Tom Leinster Feb 3 '10 at 17:26

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.

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But it is not true that that every Boolean algebra is a field of sets, so one of Stone's functor is not surjective. –  Mariano Suárez-Alvarez Feb 3 '10 at 20:10
You're one level higher up than I am -- I'm talking about viewing a particular Boolean algebra as a category (ie, viewing a poset as a category). Then the isomorphism between it and the corresponding field of sets Stone's theorem gives you is an isomorphism of categories. –  Neel Krishnaswami Feb 3 '10 at 20:43

This is really a comment, not an answer. But since it is a not-so-short comment to many answers together, it had to become an answer.

It has been observed that (first-order) definitional equivalences give categorical isomorphisms, at least for categories of first order structures with isomorphisms as their only morphisms. In my opinion the fact that two equivalent definitions of a mathematical structure give the same isomorphisms but possibly different morphisms (which maps between [complete] lattices should one consider: isotone? meet-semilattice morphisms? join-semilattice morphisms? lattice morphisms? [complete join, or meet, or both, morphisms?]) is a big virtue: it means that two definitions give really different points of view on the same kind of structure (in a way, they formalize a kind of non-triviality of the equivalence). This also happens for second-order structures (complete lattices, uniform and topological spaces); the definitional equivalences are expressed in the natural language of Bourbaki's "scale of sets" (or natural model of type theory) above the base sets of the (multisorted) structure (detractors of Bourbaki and/or lowers of category theory would instead speak of the topos "somewhat freely" generated by the (sorts for the) base stets; when the equivalence of definitions is completely constructive one can really take a free topos, but depending of the principles of classical logic which are needed to prove the equivalence of definitions, one considers the topos freely generated in more restricted classes).

So in summary: syntactically defined equivalences induce isomorphisms between categories of structures. As Hodges notes (for example in his book "model theory"), pratically everything which in mathematics can "really" be considered a "construction" is formalizable as a interpretation or at least a "word-construction" (and moreover it is the syntactical form itself which shows what kind of morphisms more general than isomorphisms are "preserved" by the construction. I understand that few lovers of category theory would approve such a extreme syntactical view, but note that even the "categories, allegories" book by Freyd and Scedrov insists on the "Galois correspondence" between syntactical and semantical aspects; I simply happen to prefer the syntactical side). From this point of view, Hodges'remarks about (cases slightly more general than) adjunctions among quasivarieties (and universal Horn classes) induced by forgetful functors are related to the already given remark about monadic adjunctions.

Besides, the book "abstract and concrete categories" by Adameck, Herrlich, Strecker conains many examples of "concrete isomorphisms"; some of them shoulb be interesting (and all of them, if I remember correctly, can be seen as syntactically defined as above).

Incidentally, the three authors say that non reasonable concept of "concrete equivalence" can be given; I disagree since cases exist where two categories can be concretely reflected on full subcateories of objects "in normal form", and the subcategories are concretely isomorphic [for example, take affine geometry of dimension at least three: form affine spaces algebraically defined by points, group of translations, sfield of scalars one "normalizes" to the particular case where translations are a subgroup of the group of permutations of the points and scalars are a subring of the ring of endomorphisms of the group of translations. For affine spaces geometrically defined in Hilbert's Grundlagen style, the general case can be reflected onto the "normal" case with the same set of points where lines and planes are sets of points and incidence is the set-theoretic one]

It has already been observed that, in presence of choice, "isomorphic categories" means "equivalent categories where corresponding isomorphism classes of objects have the same cardinality". Freyd and Scedrov observe that, even in absence of choice, the "correct" notion of equivalence is: to have isomorphic inflations. This means that all usual examples of equivalence of categories induce examples of isomorphisms (without the trick with arbitrary choices to consider skeletons, but instead using canonical "inflations" of the isomorphism classes)

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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

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

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.

Added 28/6/12: The category of A-spaces (topological spaces in which every intersection of open subsets is open) is isomorphic to the category of preorders. Under this isomorphism $T_0$ A spaces correspond to partial orders.

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Doesn't really matter, but it would seem more natural to define a morphism by $f(x/y) = f(x)/f(y)$. –  Johannes Hahn Feb 3 '10 at 18:28
thanks, I've edited it. –  Martin Brandenburg Feb 3 '10 at 18:44
Why is this not a reformulation? –  Mariano Suárez-Alvarez Feb 3 '10 at 18:48
I'm looking for nontrivial reformulations, see also my question. –  Martin Brandenburg Feb 4 '10 at 2:54
Does your equation make any sense? It seems to imply $y_1 = y_2$ for all $y_1,y_2\in G$, shouldn't there be a $y$ somewhere on the left-hand side? –  Ketil Tveiten Jun 29 '12 at 9:10

In my first step as student, I still do not know the concepts of category theory, but I thinked in "naive" way about this:

Let $Top$ the category of topological spaces and continuous maps (such that the invese images of opens are still opens), and let $Fil_T$ the category with objects like $(X, F)$ where $X$ is a set, $F=(F_x)_{x\in X}$ with $F_x$ a filter of subset with $x\in \bigcap F_x$ and such that $\forall U\in F_x\ \exists V\in F_x: \forall y\in V: U\in F_y$. ANd with morphisms $f: (X, F)\to (Y, G)$ , the maps $f_ X\to Y$ such that $\forall x\in X: \forall V\in G_{f(x)}: \exists U\in F_x: f(U)\subset V$.

The usual representation of topology as a family of opens or as families of neighbrhoods, and the essential identification between these, is just a isomorphisms between $Top$ and $Fil_T$ .

Of course this is a elementary example of isomorphism of concrete categories, in general a concrete isomorphism is just a different (but "equivalent") representation of the structure on a set, but these representation can have very different formalization (see all the literature on concrete categories, especially topological etc.).

Another nice example (I think) is the isomorphism (that fix the objects) between the Kleisli categori of a triple $T$ inducted from an adjuntion $(F, G): \mathcal{A}\to \mathcal{C}$ (right notation) and the clone category $\mathcal{C}_F$ of $F$ that has the some objects of $\mathcal{C}$ and with hom-sets defined as: $\mathcal{C}_F(X, Y)=\mathcal{A}(F(X), F(Y))$ (all these are considered disjoint) with the natural composition and identities of $\mathcal{A}$.

The following article is about a very nice isomorphisms (the shape category as the some objects and the shape functor fix these..):

http://www.jstor.org/discover/10.2307/1996811?uid=3738296&uid=2129&uid=2&uid=70&uid=4&sid=21100881872361

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I feel one should make a distinction between categories in their so-to-speak metamathematical role, for example in using results such as the Yoneda Lemma, or a left adjoint preserves colimits, and categories used as an algebraic structure with partial operations, in which guise I cite many examples of groupoids.

One such example groupoid is the unit interval groupoid $\mathbf I$ with two objects $0,1$ and exactly one arrow between any objects. This groupoid plays a similar role in the category of groupoids to the integers in the category of groups. Also the integers are obtained from $\mathbf I$ by identifying $0,1$ in the category of groupoids: this is one explanation of why the fundamental group of the circle is the integers.

That last comment is an application of the Seifert-van Kampen Theorem for the fundamental groupoid $\pi_1(X,A)$ on a set $A$ of base points, a theorem which is about calculating homotopy $1$-types of not necessarily connected spaces. It is quite necessary in using this theorem to keep all the information about the way various components of the pieces intersect; moving to equivalence will destroy that information.

In the case of groupoids with structure, which may be a topology, or a smooth structure, or an algebraic structure (group, ring, Lie algebra,...) then the usual equivalence of a transitive groupoid to a group of course no longer applies to preservation of the extra structure. But there are simpler structures: for example one knows how to classify vector spaces with a single endomorphism, but how does one classify groupoids with a single endomorphism?

Maybe the situation is less clear with categories, rather than groupoids, but it may also be that these two roles, metamathematical, and an algebraic structure with partial operations, merge in some situations. For me, that is one of the fascinations of category (and groupoid) theory.

And my definition of Higher Dimensional Algebra is as the study of partial algebraic structures with operations whose domains are given by geometric conditions.

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why did I know that your answer will contain groupoids in every paragraph, but not a single time isomorphisms? –  Martin Brandenburg Jun 29 '12 at 16:09
@Martin: Just seen this comment. I think my paper with Higgins "The equivalence of $\omega$-groupoids and cubical $T$-complexes", Cah. Top. G\'eom. Diff. 22 (1981) 349-370 shows non trivially that two structures on an underlying cubical set are equivalent, and so gives an example of isomorphic categories. In my answer, about van Kampen situations, we need to keep groupoids up to isomorphism rather than equivalence, to model the way components of spaces intersect. Groupoids have structure in dimensions 0 and 1, and this is useful in modelling the way spaces glue together. Hope that helps. –  Ronnie Brown Oct 11 '13 at 13:43