# What is characterization of categories that are equivalent to discrete?

If we define a category to be discrete iff every morphism is identity, than how can we characterize the categories that are equivalent to discrete categories? I'm trying to look at an equivalent category through the fully faithful essentially surjective functor. Let $F: C \to D$ be an equivalence and $G: D \to C$ it's inverse, where $C$ is discrete. Then I've got that every morphism in $D$ is an isomorphism, but it seems that I can't get anything else about $D$, for example is it's morphisms could be other than identify?

• If it's equivalente to its $\pi_0$?! – Ivan Di Liberti Feb 8 '18 at 11:25
• And what is $\pi_0$? – Vladislav Romanovskiy Feb 8 '18 at 11:31
• There is a term for such a category: setoid. – user14972 Feb 8 '18 at 14:32

Two categories are equivalent if and only if they have isomorphic skeletons. If in a category every morphism is an identity then the category only has one skeleton which is the category itself. So a category $\mathcal D$ will be equivalent to a discrete category if its skeletons are discrete. The skeletons are isomorphic so it is enough to focus on one skeleton. We could say that $\mathcal D$ must have a full subcategory $\mathcal C$ that is discrete and such that for every object $d$ of $\mathcal D$ there is a unique object $c$ in $\mathcal C$ such that there is an isomorphism $f:d\to c$. Now if there is also some morphism $g:d\to c'$ where $c'$ is again an object of $\mathcal C$ then there is a morphism $g\circ f^{-1}:c\to c'$ and the fact that $\mathcal C$ is discrete then tells us that $c=c'$ and $g\circ f^{-1}=\text{id}_c$ or equivalently $g=f$. We conclude that for every object $d$ in $\mathcal D$ there is only one arrow $f$ that start is $d$ and ends at some object $c$ in $\mathcal C$ and that moreover it is an isomorphism. Now let it be that $f:d\to c$ and $f':d'\to c'$ are such morphisms. Then the existence of a morphism $h:d\to d'$ lead to the conclusion that $c=c'$ because $f'\circ h\circ f^{-1}:c\to c'$ must be an identity. Then $h=f'^{-1}\circ f$ showing that at most one arrow $d\to d'$ can exist, and that this arrow must be an isomorphism again.