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I have the following definition of finite category:

Let $\mathcal{C}$ be a category. We say that $\mathcal{C}$ is finite if $\bigcup_{x,y\in\text{Ob}(\mathcal{C})}\text{Hom}_{\mathcal{C}}(x,y)$ is finite. (So in particular, $\text{Ob}(\mathcal{C})$ is finite).

I don't understand why $\bigcup_{x,y\in\text{Ob}(\mathcal{C})}\text{Hom}_{\mathcal{C}}(x,y)$ being finite implies that $\text{Ob}(\mathcal{C})$ is finite.

For each $x\in\mathcal{C}$, we know that $\{1_x\}\subset\text{Hom}_{\mathcal{C}}(x,x)$. This implies that $$\bigcup_{x\in\text{Ob}(\mathcal{C})}\{1_x\}\subset\bigcup_{x,y\in\text{Ob}(\mathcal{C})}\text{Hom}_{\mathcal{C}}(x,y)$$ and so $\bigcup_{x\in\text{Ob}(\mathcal{C})}\{1_x\}$ is finite.

Now define a mapping $f:\text{Ob}(\mathcal{C})\rightarrow\bigcup_{x\in\text{Ob}(\mathcal{C})}\{1_x\}$ by writing $f(x):=1_x$ for all $x\in\text{Ob}(\mathcal{C})$. We're done if we can show that $f$ is an injection. But $f$ is not an injection: take $x,y\in\text{Ob}(\mathcal{C})$ such that $f(x)=f(y)$; this means that $1_x=1_y$ and how can we conclude that this implies $x=y$? I can't see how.

For example, if this latter equality were in Set, then the equality of $1_x=1_y$ would imply that $x=y$ from the definition of mapping (i.e. equal maps must have equal domains). I'm not even sure what the equality $1_x=1_y$ means in an arbitrary category...

Anyways, how can one show that $\text{Ob}(\mathcal{C})$ is finite?

user831160
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  • The big union should really be a big disjoint union. Otherwise the statement is false (consider a discrete category with a singleton as the set of morphisms). – varkor Feb 01 '22 at 18:00

1 Answers1

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There are multiple ways of axiomatizing categories, which are ultimately equivalent in any appropriate sense, and I suggest not worrying about the differences too much.

If your axiomatization of categories requires that the Hom-sets for distinct pairs of objects are disjoint, then $1_x$ and $1_y$ belong to distinct Hom-sets if $x\neq y$, whence cannot be equal since those are disjoint.

If your axiomatization of categories does not require that the Hom-sets for distinct pairs of objects are disjoint, then the definition of finite categories is inadequate, as varkor has already pointed out, and you need to require the disjoint union of all Hom-sets to be finite instead.

In either case, finite categories are precisely those with finitely many objects and only finitely many morphisms between each pair of objects. Perhaps this is the definition best to remember.

Thorgott
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  • The definition of a category typically does not require the hom-sets are disjoint. It is the definition of "finite category" that is incorrect instead. – varkor Feb 01 '22 at 18:00
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    It's a matter of convention, if anything, but the OPs context will have to assume this. Otherwise, the definition is mistaken, as you say. I guess we'll have to wait for OP to elaborate on their preferred definition of a category. – Thorgott Feb 01 '22 at 18:06
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    @varkor I think some might disagree to some degree, but at any rate it is sometimes not made explicit in the axioms. – rschwieb Feb 01 '22 at 18:09
  • If you add a disjointness condition, you don't recover small categories as categories internal to Set, so I would suggest that a definition including such a condition is not the right definition. – varkor Feb 01 '22 at 18:15
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    @varkor Really? According to the definition of internal category in the Wikipedia article you linked to, it seems Hom sets are disjoint (since they are the fibers of the map $(d_0,d_1)\colon C_1\to C_0\times C_0$). – Alex Kruckman Feb 01 '22 at 18:28
  • I've edited my answer to be a bit more flexible, because I believe the question of "what the "right" definition of a category is" should not be one of OPs concerns here. Nonetheless, I do agree with @AlexKruckman. The disjointness of hom-sets is a concrete instance of morphisms in any internal category having defined source and target morphisms, which is what's needed to generalize the composability condition into the pullback expressed using those. Either definition, though, yields equivalent 2-cateories of categories (mod size issues), so this isn't a pressing issue anyhow. – Thorgott Feb 01 '22 at 18:40
  • @Thorgott nice answer, thank you very much! – user831160 Feb 01 '22 at 19:10