A Category with Finite Number of Sets as Objects is never an Equivalence

Question

Consider the following statement:

Let $ \mathscr{S} $ be the class of all sets; for $ A,B \in \mathscr{S} $, $ hom(A,B) $ is the set of all functions $ f: A \to B $. Then $ \mathscr{S} $ is a category. And, a morphism $ f $ of $ \mathscr{S} $ is an equivalence iff $ f $ is bijective.

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I was wondering, while digging into conclusions, what would happened if the objects (ie. the collection of sets) constituting $ \mathscr{S} $ were of a finite cardinal.

I found my answer with following reasoning:

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Let $ \mathscr{S} $ be a class of sets $ (A_1,A_2,...,A_m) $ with finite cardinality of $ m \in \mathbb{N} $.

Let $ hom(A_i,A_j) $ be the class of pairwise disjoint sets, for any $ (A_i,A_j) \in \mathscr{S}^2 $ with $ i,j \le m $ and pairwise disjoint. Consequently we have: $ card(hom(A_i,B_j)) \le m $. Thus;

1. for all $ i \le m $, $ A_i \subset \mathscr{S} $, and,
2. for all $ i,j \le m $ and pairwise distinct, $ hom(A_i,A_j) \subset \mathscr{S} $.

But we know by Set Theory that:

• $ card(hom(A_i,A_j)) \le 2^m-m $ at least. And,
• we assumed having $ m $ objects in $ \mathscr{S} $,

which gives: $ card(\mathscr{S}) \ge 2^m $. CONTRADICTION.

CONCLUSION: So is it that: if $ \mathscr{S} $ were constructed upon a Finite Number of Sets as Objects, $ f $ is not a bijection, thus $ \mathscr{S} $ is never an equivalence?

Answer 1

Let $C$ be a subcategory of the category of sets. If $C$ is a small category, that is if the class of objects of $C$ is a set, for every object $A,B$ of $C$, $Hom_C(A,B)$ DOES NOT have to be an element of $C$, therefore there exists categories which has a finite number of objects and whose objects are finite sets.

For example you can define the category whose objects are $\{0,...,i\}, i\leq n$ and the morphisms are morphisms sets.

Answer 2

A finite class of sets can, indeed, be a category, depending on what you choose as your morphisms. Indeed, take $\operatorname{hom}(A_i,A_j)=\emptyset$ for $A_i\neq A_j,$ and take $\operatorname{hom}(A_i,A_i)=\{A_i\}$ for all $i.$ That is, we let each set be its own identity morphism, and have no other morphisms, so we have a discrete $m$-element category.

As for where your reasoning went awry, it's hard for me to say. I'm not at all sure what you mean in your first paragraph (specifically the "class of disjoint sets" and "pairwise disjoint" parts), so I can't really begin to speculate about the rest.