# Equivalence of categories and subcategories

• $\newcommand{\mc}{\mathcal}$I heard that $\mc C \equiv \mc D$ (an equivalence of categories) holds iff $\mc C$ is a fully faithful essentially surjective subcategory of $\mc D$, but subcategory seems too strict since categories could be equivalent without the objects of one being a subset of the other. Where am I going wrong?

Let $\mc C$,$\mc D$ be categories then they are equivalent if we have functors $F: \mc C \to \mc D$, $G: \mc C \to \mc D$ and natural isomorphisms $\alpha : 1_{\mc C} \to GF$, $\beta : 1_{\mc C} \to GF$.

• A category $\mc C'$ is a subcategory of $\mc C$ if its objects are a subclass of the objects of $\mc C$ and its morphisms are a subclass of the morphisms of $\mc C$.

• A functor $F : \mc C \to \mc D$ is full if the map between Hom-sets $\mc C(A,B) \to \mc D(FA,FB)$ is surjective.

• A functor $F : \mc C \to \mc D$ is faithful if the map between Hom-sets $\mc C(A,B) \to \mc D(FA,FB)$ is injective.

• A functor $F : \mc C \to \mc D$ is essentially surjective if $\forall B \in \mc D,\,\exists A \in \mc C,\,FA \simeq B$.

• I am a little confused by your question: an equivalence of categories is a functor; what do you mean when you say it's the same thing as a certain category?
– user314
Dec 3, 2012 at 19:50
• @Adeel, they imply each other so if I have one I can make the other out of it and vice versa. updated.
– user51427
Dec 3, 2012 at 19:54
• "I heard that...", Objection your Honor, that's hearsay! :-) Welcome to math stackexchange sunflower. Please look at my answer below. Dec 4, 2012 at 10:51
• $\alpha$ and $\beta$ are natural isomorphisms, not just transformations. I edited accordingly Dec 4, 2012 at 16:30
• You write both the functors and the nats have the same domains and codomains. Nov 2, 2014 at 13:34

There are different characterizations of equivalences between categories. One of them is: the skeleta of the two categories are isomorphic categories. In this sense $\mathcal C$ and $\mathcal D$ are equivalent if and only if there is a third category $\mathcal A$ together with two functors $F \colon \mathcal A \to \mathcal C$ and $G \colon \mathcal A \to \mathcal D$ which identify $\mathcal A$ with fully faithful subcategories of $\mathcal C$ and $\mathcal D$ respectively.

Does this answer your question? If you want I can provide a proof of my statement about skeleta.

• Welcome to math stackexchange Mauro. I do not think your post answers the question posed. It is just another (logically-equivalent) definition of categorical-equivalence Dec 4, 2012 at 10:40
• You are right, indeed. It's just the only reformulation I know which contains the word "fully faithful subcategory" inside. Dec 4, 2012 at 15:13

The correct statement is reported in

http://en.wikipedia.org/wiki/Equivalence_of_categories ("Equivalent characterizations" section)

In the "Joy of Cats" book

you can find, starting at page 36 a discussion/proof of these matters.

A functor which is "essentially surjective" is also called "dense" or "isomorphic dense"

• So according the real theorem would be "$\mc C \equiv \mc D$ (an equivalence of categories) holds iff there is a fully faithful essentially surjective functor $\mc C \to \mc D$"? in that case $\mc C$ is a subcategory of $\mc D$ in essence but not in terms of the definition..
• @magma: can you make more precise this last affirmation? According to my definition, an embedding is a fully faitfhul functor injective on objects. In this case, your statement is false: $\mathbf{Vect}_K$ is equivalent to $\mathbb N$, but you cannot embed it as a subcategory. The best you can do is explained in my answer below, I suppose... Dec 4, 2012 at 15:10
• Sorry for the non-standard notation. I was meaning $\mathbf{FinVect}_K$ (finite dimensional vector spaces and linear maps) and $\mathbb N$ is the (skeleton of the) category of finite sets with functions. The assignment $\mathbb N \to \mathbf{FinVect}_K$ sends $n$ to $K^n$ and the actions on maps is the obvious one. It is fully faithful because the forgetful functor $\mathbf{FinVect}_K \to \mathbf{Set}$ as a left adjoint, it is essentially surjective because every vector space has a basis. Observe that you can do the same for vector spaces of dimension $\le \alpha$, for any ordinal $\alpha$. Dec 4, 2012 at 17:01