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  • $\newcommand{\mc}{\mathcal}$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 $\forall B \in \mc D,\,\exists A \in \mc C,\,FA \simeq B$.


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    You write both the functors and the nats have the same domains and codomains.2014-11-02

2 Answers 2

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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"

Please see also

http://en.wikipedia.org/wiki/Essentially_surjective_functor

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    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 $\alpha$.2012-12-04
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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.

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    You are right, indeed. It's just the onl$y$ reformulation I know which contains the word "fully faithful subcategory" inside.2012-12-04