I am trying prove this. It is stated on the wikipedia page about discrete categories.

One way of the proof is very easy and obvious: if a category is discrete then all its subcategories are also discrete and, hence, full (they all have the identities, wich are the only arrows in the first category. So they all have all the arrows relative to the objects in the first category).

The other way is trickier. We consider some category $C$. If all its subcategories are full, then $C$ must be discrete.

I think I can see why this must be so, but I can't state a general proof. I'm thinking this way:

Consider a category $C$ with only two objects, $a$ and $b$, and a single arrow $f: a\rightarrow b$ between them (besides the identities, of course). Then it has four subcategories, but one of them is not full, i.e., the subcategory that has both objects, but not the $f$ arrow. So it seems the only way for all subcategories of $C$ to be full is $C$ being discrete. Otherwise there will always be a discrete subcategory that is not full.

Is this reasoning correct? How can I make it more general?


Consider the subcategory $D$ of $C$ which has the same set of objects and $Hom_D(X,Y)$ is empty if $X\neq Y$ and $Hom_D(X,X)=\{Id_X\}$ it is full. So $Hom_C(X,Y)=Hom_D(X,Y)$.

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    $\begingroup$ Ok, let me see if I get this. For any category $C$ we have its discrete subcategory $D$ with same objects but only identities. By our supposition, this category is full. But then this means that $C$ can't have any arrow that is not in $D$, otherwise $D$ would not be full. So $C$ is also discrete (and in fact identical to $D$). $\endgroup$ – Karlm Nov 22 '18 at 17:47

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