This is a question from "Categories and computer science" by R F C Walters. I've spent longer than I should of on this question. The constraints imposed on the category are:
(i) every arrow has an inverse;
(ii) for each pair of objects, theres is an arrow from one to another;
(iii) for each object $A$ there are $n$ arrows with domain $A$.
and the goal is:
Show that the total number of endomorphisms (loops) is $n$
Progress: I think it's the case that
- As every arrow is an isomorphism, for any three objects $X, Y, Z$ and unique arrows $a_1, a_2$ from $X$ to $Y$, and arrow $b$ from $Y$ to $Z$, it has to be the case that $ba_1\neq ba_2$, because if $b$ is an isomorphism, it's mono, meaning $ba_1=ba_2 \implies a_1 = a_2$. This means there must be at least as many arrows from $X$ to $Z$ as there are arrows from $X$ to $Y$.
- You can show there must be at least as many arrows from $X$ to $Z$ as there are arrows from $Y$ to $Z$ by the same method as #1, but using the epi property of isomorphisms, in place of the mono.
- Replacing $Y$ in the above #2 with $X$ shows there must be at least as many arrows from $X$ to $Z$ as there are arrows from $X$ to itself (endomorphisms on $X$).
- Similarly to #3 Replacing $Z$ in point #1 with $X$ shows there must be at least as many arrows from endomorphisms on $X$ as there are arrows from $X$ to $Y$.
- Combining #3 and #4 gives the number of arrows from one object to another must be equal to the number of endomorphisms on the domain object
- And because each isomorphic arrow between two objects must have a unique arrow in the opposite direction as it's inverse, we know the number of arrows from one object to another is the same as the arrows in the oppisite direction.
- This means every pair of objects has the same number of arrows between them as endomorphisms on one object of the pair
- So each object has $n$ arrows with it as it's domain, and from #7 and that each pair of objects has atleast an arrow between them, the number of arrows going to each of the $o$ objects (including itself) in the category is the same. Giving $m$ arrows from the object to itself, for each of the o objects, for $m*o=n$ endomorphisms
The main reason I'm not convinced is the large number of steps compared to the other problems in the textbook, I'm convinced there must be a better way to the proof.