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A first tentative guess would be "two categories $A$ and $B$ are equivalent iff each connected component of $A$ is equivalent to a (full) subcategory of $B$ and vice versa", but this fails, for example, in the case of $FiniteFields$ and $(\mathbb{N}, \leq)$ [wrong]

Trying to strengthen it to "[...] iff each connected component of $A$ is ('strictly') isomorphic to some subcategory of $B$, and vice versa" then fails (in the other direction) for 'smaller' skeletons

A somewhat compromising possibility, involving a 'global' condition, (which I'm not sure actually holds; please tell if it's false, or well-known) would be a Dedekind-Cantor-Schröder-Bernstein-type proposition: "two categories are equivalent iff for one of them, every connected component is equivalent to some subcategory of the other, and, on the opposite direction, there is a fully faithful (not necessarily essentially surjective) functor"

ac15
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    I don’t understand the first paragraph. FiniteFields and $(\mathbb{N},\leq)$ are not equivalent, but neither is it true that every connected component of FiniteFields is equivalent to a subcategory of $(\mathbb{N},\leq)$. – Jeremy Rickard Mar 23 '24 at 14:19
  • Is it supposed to be $(\mathbb N, \mid)$ where "$\mid$" means "divides" (not sure the notation is completely standard)? – krm2233 Mar 23 '24 at 14:35
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    @krm2233 Even then, it doesn’t work. The field of four elements has two automorphisms, but no element of a poset, considered as a category, does. – Jeremy Rickard Mar 23 '24 at 14:40
  • @JeremyRickard you're right – ac15 Mar 23 '24 at 14:54
  • @krm2233 i was really the usual order – ac15 Mar 23 '24 at 14:55
  • possibly relevant: https://math.stackexchange.com/q/3372047/746936 – ac15 Mar 23 '24 at 15:12

1 Answers1

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Let $A$ be the disjoint union of countably many copies of the category $\bullet\to\bullet$, and let $B$ be the disjoint union of $A$ with the category $\bullet$. Then $A$ and $B$ are not equivalent, but there are fully faithful functors in both directions, and each connected component of either is equivalent to a subcategory of the other.

Jeremy Rickard
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