Is there a non-discrete abelian category which has only finitely many objects?

Just out of curiosity I am wondering if such an abelian category exists, while the usual examples of abelian categories that I know contain infinitely many objects.

If the finiteness is too strong a condition, then per chance one might ask

Is there an abelian category whose objects form a set?

I have no idea whether such a small abelian category exists; I am not even sure if the set of all abelian groups exists.


Thanks to @Oskar, this question already has an answer here. But I cannot mark this question as duplicate, as that is a math overflow question.

As a brief summary, @Jeremy Rickard gave an answer there: there is an abelian category with two objects, which is any skeletal category of the quotient category of the category of countable dimensional vector-spaces by the Serre subcategory of finite-dimensional vector-spaces.

Any help is sincerely appreciated.

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    $\begingroup$ The category of all finite-dimensional $k$-vector spaces is an abelian category; so a skeleton of this category, e.g. consisting of $k^n$ for $n \in \mathbb N$, is a set-sized abelian category. $\endgroup$ – Mees de Vries Feb 12 at 13:00
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    $\begingroup$ My intuition says no, but I don't know enough to make it formal. Note that if $X$ is any object, you also need $X \oplus X \oplus \cdots \oplus X$ in the category. You could choose $X$ with $X \oplus X = X$, so that you only get finitely many objects -- like picking $X = \mathbb Z^{\oplus \omega}$. Of course such an $X$ is then quite complicated, and it seems like you should get infinitely many quotients from it.... $\endgroup$ – Mees de Vries Feb 12 at 13:12
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    $\begingroup$ For set-sized abelian category, you should read about the embedding theorem from Freyd. It will interest you and you will read examples which will help you in your question. A good reference is the Handbook of Categorical Algebra, of F. Borceux, just read the last section of the first chapter of Volume 2. For finiteness, my intuition was the same as Mees de Vries. So I believe it will be complicated, but one should try to construct an example. $\endgroup$ – Sov Feb 12 at 13:21
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    $\begingroup$ @Mees: Right. Then take the power set of that group, as an Abelian group with symmetric difference as addition. But you're right that some basic knowledge in set theory is needed. But why is that a bad thing? :) $\endgroup$ – Asaf Karagila Feb 12 at 13:45
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    $\begingroup$ See mathoverflow.net/questions/111965/… $\endgroup$ – Oskar Feb 12 at 19:44

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