can a category,between any two of whose objects there are maps,not necessarily unique, be regarded as a preordered set?

Question

In my book on category theory it has been stated that a category in which for each pair of its objects there is "at most" one map between them can be regarded as a preordered set. I do not know the logic behind the expression "at most" here,since I think this hypothesis can be removed without losing the content. Can anybody illustrate the matter?

Answer 1

I will use the following conventions:

1). A preordered set is a pair $(X,\le_X)$, where $X$ is a set and $\le_X$ is a reflexive transitive relation on it. The category of preordered sets and monotonic functions between them is denoted by $\mathbf{Preord}$.

2). A category is a $6$-tuple $(\text{Obj}(A),\text{Mor}(A),\text{dom}_A,\text{cod}_A,\text{comp}_A,\text{id}_A)$, where $\text{Obj}(A)$ and $\text{Mor}(A)$ are sets and others are functions with additional properties. The category of categories and functors between them is denoted by $\mathbf{Cat}$.

Let $A$ be a category, $a$ and $b$ be its objects. Then denote $$ \text{hom}_A(a,b)=\{f\in Mor(A)|\quad \text{dom}(f)=a,\quad \text{cod}(f)=b\}. $$

Definition. The category $A$ is called a preorder iff for every two objects $a$ and $b$ of $A$ we have $|\text{hom}_{A}(a,b)|=1$ or $|\text{hom}_{A}(a,b)|=0$, where $|\text{hom}_{A}(a,b)|$ denotes the cardinal number of the set $\text{hom}_{A}(a,b)$. Equivalently, for every two objects $a$ and $b$ the set $\text{hom}_{A}(a,b)$ contains no more than one element/at most one element.

3). The category of all preorders (not preordered sets!) is a full subcategory of $\mathbf{Cat}$; denote it by $\mathbf{CatPreord}$.

Now the main statement is that categories $\mathbf{Preord}$ and $\mathbf{CatPreord}$ are equivalent. The equivalence $\mathcal{F}\colon \mathbf{Preord}\to \mathbf{CatPreord}$ defines in the following way: $$\mathcal{F}(X,\le_X)=(X,\le_X,\text{pr}_1,\text{pr}_2,\text{tr}_{X},\text{ref}_X),$$

where $\text{pr}_1$ and $\text{pr}_2$ denote the restrictions of the first and the second projections of the product $X\times X$, the functions $\text{tr}_{X}$ and $\text{ref}_X$ define obviously by transitivity and reflexivity of $\le_X$. The inverse equivalence $\mathcal{G}\colon \mathbf{CatPreord}\to \mathbf{Preord}$ defines in the following way: $$\mathcal{G}(\text{Obj}(A),\text{Mor}(A),\text{dom}_A,\text{cod}_A,\text{comp}_A,\text{id}_A)=(\text{Obj}(A),\le_A),$$

where $\le_A$ is such relation that $a\le_Ab$ iff $|\text{hom}_A(a,b)|=1$.

Edit 3. You ask why we can't remove a hypothesis about "at most one map". The answer is that if you add to $\mathbf{CatPreord}$ some other categories with more than one arrow between two objects, then it becomes not equivalent to $\mathbf{Preord}$. That's why we can't omit the requirement about "at most one map".

Edit 2. Sorry for my inattantiveness at the first time. I have added some actual information to this answer. Kevin Carlson gave a good explanation in comments. I want to add that a preorder (category) is not a preordered set. Such arguments are true only with regarding the categories of appropriate objects (see Edit 1).

Edit 1. Of course, this definition is motivated by the connection between such categories and preordered sets. The rigorous expression of such connection is that the category $\mathbf{Preord}$ of preordered sets is equivalent to the full subcategory of category $\mathbf{Cat}$, whose objects are preorders.