Why not just define equivalence relations on objects and morphisms for equivalent categories?

Question

My question is regarding this blog post https://unapologetic.wordpress.com/2007/05/30/equivalence-of-categories/ which I will paraphrase below:

The author gives the example of a category $\mathscr{C}$ with just one object and the identity morphism, and another category $\mathscr{D}$ consisting of two objects and 4 morphisms (2 identity and 2 non-identity). After some discussion the author reveals that all 4 of the morphisms in the category $\mathscr{D}$ must be isomorphisms, and the 2 objects in $\mathscr{D}$ are isomorphic. This example serves to motivate the notion of an equivalence of categories given that the two objects in our category $\mathscr{D}$ are "essentially the same," and that besides the number of objects and morphisms, $\mathscr{C}$ and $\mathscr{D}$ behave the same.

They then define equivalent categories as those whose composition of functors are naturally isomorphic to their respective identity functors, i.e. for functors $F:\mathscr{C}\to \mathscr{D}$ and $G:\mathscr{D}\to \mathscr{C}$ we must have that $G\circ F\cong id_{\mathscr{C}}$ and $F\circ G \cong id_{\mathscr{D}}$.

But why go through this whole process of defining this notion of equivalent categories if we can just create a single equivalence classes of objects via the equivalence relation of being isomorphic, and make morphisms defined on the same equivalence class "the same?" This way we can define an isomorphism of categories by sending the equivalence class of objects in $\mathscr{D}$ to the one object in $\mathscr{C}$ and similarly with the morphisms.

Or is there some other benefit to the notion of equivalence that I am not aware of?

Answer 1

You ask: Or is there some other benefit to the notion of equivalence that I am not aware of?. Clearly I am going to answer: Yes !

The usual first example of when equivalence of categories can be seen to be `better' than isomorphism is with the category of (real) finite dimensional vector spaces. Any such vector space, $V$, is, of course, isomorphic to $\mathbb{R}^n$ for some $n$, but the isomorphism has to be chosen, as you have to choose a basis for $V$. There are properties of vector spaces that do not depend on a choice of basis (e.g. dimension), and a basis is not an intrinsic part of a vector space. The category, $vect$, of finite dimensional vector spaces is equivalent to its full sub-category of those of the form $\mathbb{R}^n$, but not naturally so.

Isomorphism is stronger than sensible. Equivalent categories have the same categorical properties, e.g. existence of finite limits, so the natural notion of being 'essentially the same' is equivalence rather than isomorphism.

Finally, if you know some homotopy theory, there is a notion of homotopy of small categories, and `homotopy equivalence' is in that theory just categorical equivalence. So your question would be related to 'why use homotopy equivalence rather than homeomorphism in homotopy theory?' I hope this helps a little.

(Edit: in reply to the point made by Oskar in a comment. I will answer it here as that avoids character limits!

That was not how I interpreted the question as worded. Perhaps Brandon (the OP) would like to comment. The other question that you pose, O is a good one, but has a similar answer. If I understand what you mean, you interpret Brandon as saying: to prove an equivalence, pass to a skeleton of the domain category, (i.e. picking a 'nice' representing object for the ismorphism class of each object), similarly for the codomain category, then prove isomorphism of those skeleta. That could be a good attack on proving equivalence, but is a more convoluted way than that which occurs very naturally in applications. To prove $C$ and $D$ are equivalent categories I would usually look for a nice functor $F: C\to D$ and another one $G:D\to C$, that somehow undid the construction encoded in $F$. It is VERY RARE that it will give you back the same object you started with, but you just have to show the result is isomorphic (not equal)to it. With the skeleton based approach you have to prove equality but that means you have to look at the passage from the whole category to the skelton. Sometimes that will be easy to see, but not always. Also proving equality can be difficult.

It is possible to work with things like a choice of products in a category, but again that ends up being messy later on as a product is only defined up to isomorphism. End of edit.)

Answer 2

Well, I decided to include some additional information.

Definition 1. Let $\mathcal{C}$ and $D$ be categories, $T\colon\mathcal{C}\to\mathcal{D}$ and $S\colon\mathcal{D}\to\mathcal{C}$ be functors. Then the pair $(T,S)$ is called an equivalence iff $S\circ T\cong I_{\mathcal{C}}$ and $T\circ S\cong I_{\mathcal{D}}$. In this case functors $T$ and $S$ are also called equivalences, and categories $\mathcal{C}$ and $\mathcal{D}$ are called equivalent.

It is a basic definition and Tim's answer shows why we need to use equivalence even more frequently than isomorphism. Here's another important definition:

Definition 2. Let $\mathcal{C}$ be a category, $\mathcal{S}$ be a subcategory of $\mathcal{C}$. Then the category $\mathcal{S}$ is called a skeleton of $\mathcal{C}$ iff it is a full subcategory of $\mathcal{C}$ and every object of $\mathcal{C}$ is isomorphic to precisely one object of $\mathcal{S}$.

Note, that if the axiom of choice holds, then every category has a skeleton. See also nLab article. The connection between equivalences of categories and their skeletons is described in the following proposition:

Proposition 1. Let $\mathcal{C}$ and $\mathcal{D}$ be categories, $\mathcal{S}_{\mathcal{C}}$ and $\mathcal{S}_{\mathcal{D}}$ be their skeletons. Then $\mathcal{C}\simeq \mathcal{D}$ iff $\mathcal{S}_{\mathcal{C}}\cong\mathcal{S}_{\mathcal{D}}$.

The proof follows from the fact that every category is equivalent to its skeleton and if two skeletal categories are equivalent, then they are isomorphic. You can also search Mac Lane's "Categories for the working mathematician" for the details.

Thus we can use the notion of skeleton instead of the original definition of equivalence, but sometimes it is not a simplification. As it was mentioned, even an attempt to prove that a category has a skeleton may lead to the set-theoretical difficulties. Tim also gave arguments.

You write: But why go through this whole process of defining this notion of equivalent categories if we can just create a single equivalence classes of objects via the equivalence relation of being isomorphic, and make morphisms defined on the same equivalence class "the same?"

Okay, it could be a good idea if we want to define something like a skeleton. But it isn't, because the straightforward applying this idea leads to wrong definition. Let's try to do this.

Definition 3. Let $\mathcal{C}$ be a category. Then define the graph $\text{Equiv}(\mathcal{C})$ in the following way: $\text{Obj}(\text{Equiv}(C))=\text{Obj}(\mathcal{C})/\cong_{\mathcal{C}}$ and $$\text{hom}_{\text{Equiv}(\mathcal{C})}([a],[b])=(\coprod_{a'\in[a],b'\in[b]}\text{hom}_{\mathcal{C}}(a',b'))/[(f\sim g)\Leftrightarrow(\exists a,b\in \text{Iso}(\mathcal{C})|\quad g\circ a=b\circ f) ].$$

But the graph $\text{Equiv}(\mathcal{C})$ doesn't inherit the composition law from $\mathcal{C}$. The graph $\text{Equiv}(\mathcal{C})$ doesn't even coincide with graph of any skeleton of $\mathcal{C}$ in general case. For example, it may paste two morphisms with the same domain: in the category $\mathbf{Finord}$ we have $\text{end}_{\text{Equiv}(\mathbf{Finord})}([2])=\text{hom}_{\text{Equiv}(\mathbf{Finord})}([2],[2])\cong2$, but $\text{end}_{\mathbf{Finord}}(2)=2^2=4$.