# The Unapologetic Mathematician

## Equivalence of categories

It should be pretty clear from all the examples in groups, rings, and modules that two categories $\mathcal{C}$ and $\mathcal{D}$ are isomorphic if there are functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$ so that $F\circ G$ is the identity functor on $\mathcal{D}$ and $G\circ F$ is the identity functor on $\mathcal{C}$.

Unfortunately, that’s not really as useful as it is in more mundane structures. The problem is that isomorphism is really strict, and categories are more naturally sort of loose and free-flowing. We have to weaken the notion of isomorphism to get something really useful. I’ll try to motivate this with one example. It’s sort of simple, but it gets to the heart of the problem pretty quickly.

Remember the category $\mathbf{1}$ with one object and its identity morphism. We draw it as $\bullet$, taking the identity as given. Let’s also consider a category with two objects and two non-identity morphisms, one going in either direction: $\bullet\rightleftarrows\bullet$. Call the objects $1$ and $2$, and the four arrows $f_{1,1}$, $f_{1,2}$, $f_{2,1}$, $f_{2,2}$, where $f_{i,j}$ goes from $i$ to $j$. Then we have $f_{1,2}\circ f_{2,1}=f_{2,2}$ and $f_{2,1}\circ f_{1,2}=f_{1,1}$. That is, the two non-identity arrows are inverses to each other, and the two objects are thus isomorphic.

Now since the two objects in our toy category are isomorphic, they are “essentially the same”. The category should (and does) behave pretty much the same as the category $\mathbf{1}$, but they’re clearly not isomorphic as categories.

Here’s another way they’re “the same” category. Since each hom-set in our toy category has at most one object, we can consider it as a preorder. We have the set $\{1,2\}$ with $1\leq2$ and $2\leq1$ in the order. If we (as usual) move to a partial order, we identify these two points and just get the unique partial order on one point — the category $\mathbf{1}$!

Okay, so if they’re not isomorphic, what are they? The key here is that the two objects in our toy category are isomorphic, but not identical. The definition of isomorphism of categories requires that (for example) $G(F(C))$ is exactly $C$ again. But really should it matter if we don’t get back $C$ itself, but just an object isomorphic to it? Not really. And we also shouldn’t care if we get different results for every object in $\mathcal{C}$ as long as they’re all isomorphic to the object we stick into $G\circ F$, and as long as these isomorphisms all fit together nicely with morphisms between the objects.

The upshot of all this is that we do know how to weaken the concept of isomorphism of categories: we just change the functor equalities in the definition to natural isomorphisms of functors! That is, we have natural isomorphisms $\eta:1_\mathcal{C}\rightarrow G\circ F$ and $\epsilon:F\circ G\rightarrow1_\mathcal{D}$. Let’s unpack this all for our toy category.

The functor $F$ from $\bullet\rightleftarrows\bullet$ to $\bullet$ is the only one possible: send both objects to the single object in $\mathbf{1}$ and all the arrows to its identity. For a functor $G$ the other way we have two choices. Let’s send the singlt object of $\mathbf{1}$ to the object $1$. The identity arrow must then go to $f_{1,1}$. Now we need natural isomorphisms.

To find these, let’s write down the compositions explicitly: $F\circ G$ sends the single object of $\mathbf{1}$ to the object $1$ in our toy category, which then gets sent back to the single object in $\mathbf{1}$. Its identity arrow just comes along for the ride. Thus $F\circ G$ is exactly the identity functor on $\mathbf{1}$, so we can just use its identity natural transformation as our $\epsilon$.

On the other hand, $G\circ F$ sends both objects of our toy category to the object $1$, and all four arrows to the identity arrow $f_{1,1}$. This is clearly not the identity functor on our category, so we need to find a natural isomorphism. That is, we need to find isomorphisms $\eta_1:1\rightarrow G(F(1))=1$ and $\eta_2:2\rightarrow G(F(2))=1$ so that the two nontrivial naturality squares commute. Clearly we can only pick $\eta_1=f_{1,1}$ and $\eta_2=f_{2,1}$. Now if we write out the naturality squares for $f_{1,2}$ and $f_{2,1}$ (do it!) we see that they do commute, so $\eta$ is a natural isomorphism.

When we have two functors running back and forth between two categories so that their composites are naturally isomorphic to the respective identities like this, we say that they furnish an “equivalence”, and that the categories are “equivalent”. Every isomorphism of categories is clearly an equivalence, but there are many examples of categories which are equivalent but not isomorphic. This is a much more natural notion than isomorphism for saying categories are “essentially the same”.

May 30, 2007 - Posted by | Category theory

## 9 Comments »

1. Minor typo in the paragraph that begins “The upshot of all this …” You’re missing a backslash on a mathcal. Great stuff, keep it coming!

Comment by Cotton Seed | May 31, 2007 | Reply

2. [...] on a category constitute a full subcategory of all contravariant -valued functors on , which is equivalent to the category [...]

Pingback by What does Yoneda’s Lemma mean? « The Unapologetic Mathematician | June 7, 2007 | Reply

3. [...] The functor that we described from to is an equivalence. [...]

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4. Can you show that there is an equivalence of categories between PreSet( the category of preordered sets) and PoSet (the category of partialy ordered sets) ? :)

Comment by Lucia | February 1, 2009 | Reply

5. Yes, I can. I even mentioned the important bit once before.

Comment by John Armstrong | February 1, 2009 | Reply

6. I read the post about Divisibility, but it still isn’t very clear to me why every preordered set is equivalent to a partially ordered set…when you have time, please detail, it’s important for me to understand.

Comment by lucia | February 1, 2009 | Reply

7. Every partial order is also a preorder, so to go from partial orders to preorders you just forget that it’s a partial order.

In that post I tell how to go from a preorder to a partial order.

All that’s left is to show that these constructions are functors, and check whether they furnish an equivalence of categories.

Comment by John Armstrong | February 1, 2009 | Reply

8. Thanks :)

Comment by Lucia | February 1, 2009 | Reply

9. The equivalence of PreSet and PoSet that you have indicated may be in error. If F is the forgetful functor from PoSet to PreSet, and G is the functor that you indicated, then G is not faithful. Moreover, while GF = Id, there cannot be any isomorphism between a finite proper PreSet X and FG(X), since these sets have different cardinalities.

Is there some other equivalence that I am missing? It seems to me that these categories are not equivalent.

Comment by Chris | December 5, 2010 | Reply