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