What does a Category with One Morphism between Each Pair of Objects Look like?
Recall that for any two categories $\mathcal{C}$ and $\mathcal{D}$, they are naturally equivalent if and only if their skeletons are isomorphic. That is, $\mathcal{C}$ and $\mathcal{D}$ are equivalent if and only if they are the same after modding isomorphic objects.
Now suppose $\mathcal{C}$ is a category such that for any two objects $A,B$ in $\mathcal{C}$, there exists exactly one morphism $f\in \text{Mor}(A,B)$. Then geometrically, we can represent this category in the following way (suppose in addition that $\mathcal{C}$ has 4 objects):
Since there are only one morphism between each object, every morphism is an isomorphism (the composition of the morphism in $\text{Mor}(A,B)$ and the morphism in $\text{Mor}(B,A)$ is necessarily equal to the identity morphism). It is then clear from the graph that we are only left with one object after we mod out isomorphic objects. Therefore, $\mathcal{C}$ is equivalent to the category $1$ with a single object and a single morphism.
This argument definitely also works for not only 4 but also every natural numbers. In fact, as long as $\mathcal{C}$ has at least one object, then it is equivalent to $1$. It is possible for $\mathcal{C}$ to have no objects; in this case, $\mathcal{C}$ is the empty category, which trivially satisfies the property that there exists exactly one morphism between any two objects.
Can we concretely construct $\mathcal{C}$ instead of drawing graphs? The answer is Yes. Let $X$ be a set. We can define a category $f(X)$ as follows:
- The objects set of $f(X)$ is $X$.
- item The morphisms set is $X\times X$.
- An ordered pair $(x,y)\in X\times X$ is a morphism from $x$ to $y$.
- The composition of $(x,y)$ and $(y,z)$ is $(x,z)$
$f(X)$ is then a category with exactly one morphism between each pair of objects. Also, $f(\emptyset)$ is the empty category.