I define a one-morphism category as a category with exactly one arrow between any two objects(I didn't find a name for this type of categories). As Zev says in this answer, such a category must be a groupoid, and its objects must be isomorphic to one another, because every arrow has an inverse.
I would like to know when two one-morphism categories are isomorphic. Clearly, if they have different amount of objects(different cardinality), they cannot be isomorphic, as no functor between them can have an inverse.
So, my question is as follows:
What necessary and sufficient condition for two one-morphism categories to be isomorphic?
Do having the same cardinality of objects suffice?
My try:
If $C_1$ and $C_2$ are one-morphism categories with same amount of objects, then the object class of each category are isomorphic as classes. Let $h$ be the isomorphism. Clearly $h^{-1}$ exists. Then, we can construct a functor $H:C_1\to C_2$ such that $H(a_1)=h(a_1)$ and the unique arrow $a_1\to b_1$ in $C_1$ gets mapped to the unique arrow $H(a_1)\to H(b_1)=h(a_1)\to h(b_1)$ in $C_2$. In other words, $H(a_1\to b_1):H(a_1)\to H(b_1)$. We can use the same method to contruct a functor $G:C_2\to C_1$ with the mapping $h^{-1}$ between objects, so $G(a_2)=h^{-1}(a_2)$. This means that $GH(a_1\to b_1)=G(H(a_1)\to H(a_2))=G(h(a_1)\to h(b_1))=G(h(a_1))\to G(h(b_1))=h^{-1}(h(a_1))\to h^{-1}(h(b_1))=a_1\to b_1$ and $GH(a_1)=G(h(a_1))=h^{-1}(h(a_1))=a_1$ meaning $C_1\cong C_2$.
Is this correct? Thanks.
Best Answer
A category with exactly one morphism between any (ordered) pair of objects is called an indiscrete category. If it has at least one object, it is equivalent to the terminal category. Any two indiscrete categories are therefore equivalent and, as you suggest, are isomorphic when they have isomorphic collections of objects, meaning their collections of objects have the same cardinality. This is intuitively because you can forget the information of the morphisms and simply treat the categories as sets or classes.