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?

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

This together leads to the conclusion that a category equivalent with a discrete category can only be a groupoid with the special property that every homset of it contains at most one arrow. You could say that the category is a groupoid and a preorder at the same time. Also you can recognize these categories simply as equivalence relations. In that perspective its components can be identified as equivalence classes.

For each of its components we can choose a representative object and the full subcategory based on these objects is discrete and is a skeleton.


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