基于态射均为恒等的离散范畴，如何刻画其等价范畴？

Great question! You’re already halfway there by noticing all morphisms in D must be isomorphisms—let’s build on that to get the full characterization.

First, let’s recap the setup: we have a discrete category C (only identity morphisms exist, with no morphisms between distinct objects), and an equivalence (F: C \to D) with inverse equivalence (G: D \to C).

Let’s break down what the properties of (F) and (G) tell us about D:

1. All morphisms in D are isomorphisms (as you deduced):
   Since G maps D to the discrete category C, every morphism (f) in D must map to an identity morphism in C (C has no non-identity morphisms to begin with). Because G is part of an equivalence, it’s faithful and reflects isomorphisms—so (f) must be an isomorphism (its image under G is an identity, which is an isomorphism, so G’s reflectivity forces (f) to be one too).

2. Each hom-set in D has at most one morphism:

   • For any object (d) in D, consider endomorphisms (\text{Hom}_D(d, d)). G maps every endomorphism of (d) to the identity of (G(d)) (since C is discrete). Since G is faithful, the map from (\text{Hom}_D(d, d)) to (\text{Hom}_C(G(d), G(d))) (a singleton set) is injective. This means there can be only one endomorphism of (d)—the identity morphism.
   • For distinct objects (x, y) in D: if (G(x) \neq G(y)), then (\text{Hom}_C(G(x), G(y))) is empty. Since G is a functor, (\text{Hom}_D(x, y)) must also be empty (otherwise G would map a morphism to a non-existent one in C). If (G(x) = G(y)), then (x) and (y) are isomorphic (because (FG(x) \cong x), (FG(y) \cong y), and (FG(x) = FG(y)) since (G(x)=G(y))). Any morphism (f: x \to y) maps to the identity of (G(x)) via G, and G’s faithfulness implies there can be at most one such morphism (since the target hom-set in C is a singleton).

Putting these together, we get the full characterization:

A category D is equivalent to a discrete category if and only if D is a thin groupoid—meaning:

• Every morphism in D is an isomorphism (so D is a groupoid), AND
• For any pair of objects (x, y) in D, there is at most one morphism from (x) to (y) (so D is "thin").

To confirm this equivalence, consider constructing an explicit pair of functors:

• Let C be the discrete category whose objects are the isomorphism classes of D (each class is a singleton in C).
• Define (F: C \to D) by sending each isomorphism class to a representative object in that class.
• Define (G: D \to C) by sending each object to its isomorphism class.

This pair ((F, G)) forms an equivalence: F is essentially surjective (every object in D is in some isomorphism class, hence isomorphic to F’s image of that class), full (any morphism in C is an identity, and F maps the unique isomorphism between representatives to that identity), and faithful (each hom-set in D has at most one morphism, so F can’t conflate distinct morphisms).

内容的提问来源于stack exchange，提问作者Vladislav Romanovskiy