Verify the axiom of the category of Poset

Question

Given a poset $(S,\leq)$, we can define a category $\mathcal{C}$ as follows: objects are elements in $S$, and for any $a,b\in S$ we define $Mor(a,b):=\varnothing$ if $a\leq b$ is false, and $Mor(a,b)=\{a\longrightarrow b\}$ if $a\leq b$ is true, where $\longrightarrow$ is an arrow from $a$ to $b$ indicating the relation $a\leq b$, and is uniquely determined by $a$ and $b$.

Now, I want to check that this is indeed a category. That is, to verify the axioms.

I understand that the composition law follows from the transitivity of partial order, and the identity follows from the anti-symmetric.

However, I am stuck in the first axiom, namely

If $a\neq c$ or $b\neq d$, then $Mor(a,b)\neq Mor(c,d)$.

Firstly, how do we use $a=b$? For instance, to verify this axiom, I need to show that if $a=c$ but $b\neq d$, then $Mor(a,b)\neq Mor(c,d)$. But how can we directly use $a=c$? The antisymmetry only said that if $a\leq c$ and $c\leq a$ then $a=c$, it does not say the converse. So how can I express the first axiom properly in the context of partial order?

Secondly, even if we ignore the first problem, we still have anther one: we can indeed have the follow result:

Since $\longrightarrow$ is uniquely determined by $a$ and $b$, we know that as long as $a\leq b$ and $c\leq d$ are true, then we must have $Mor(a,b)\neq Mor(c,d)$ as long as $a\neq b$ or $c\neq d$, since the arrow from $a$ to $b$ is not the same as the arrow from $c$ to $d$. If $a\leq b$ is not true but $c\leq d$ is true, then it is immediate that $Mor(a,b)\neq Mor(c,d)$, vice versa.

However, in the case of $a\leq b$ is not true and $c\leq d$ is not true, we have two empty sets, which are equal to each other.

So the first axiom is actually not satisfied?

Where did I make mistakes? Thank you!

Edit 1:

After accepting Albert's answer, I was away for a while to study other courses. Tonight I came back and then finally fully understood Albert's answer. I've forgot the definition of relation for a long time, and thanks to Albert and to this exercise, I reviewed this notion but in a totally different view.

I just answered my own post to attach a detailed proof, inspired by Albert.

I am also really grateful for all users involving in this discussion. Thank you so much!

Answer 1

As said in the comments, the first axiom must be $Mor(a,b)\cap Mor(c,d)=\emptyset$ if $(a,b)\neq (c,d)$. Now your definition of the poset as a cateogry is "right" but it's not a rigorous definition as you are not explicitly giving what is the element in the set $Mor(a,b)$. I'll make it clearer:

Let $(S,\leq)$ be a poset. Remember that $\leq$ is a relation, that is $\leq\subseteq S\times S$ and $\leq$ satisfy the poset conditions. So, when you write $a\leq b$ it is another way to write $(a,b)\in\leq $. Thus, you can formally define $$ Mor(a,b) = \{(a,b)\} $$ if $a\leq b$, and $Mor(a,b)=\emptyset$ otherwhise. With this definition, the identity element for an object $a$ is the pair $(a,a)$, and the composition is given by $$ (b,c)\circ (a,b) = (a,c). $$ Now you are in position to prove the first axiom without problems.

Answer 2

After accepting Albert's answer, I was away for a while to study other courses. Tonight I came back and then finally fully understood Albert's answer. I've forgot the definition of relation for a long time, and thanks to Albert and to this exercise, I reviewed this notion but in a totally different view.

Below is my detailed proof of this exercise, I attached here for other students who may study this subject in the future.

I also need to express my sincere appreciation for all the users participating in this discussion. Thank you guys!

---

Exericse: Given a set $S$ and preorder $\leq$, we can define a category $\mathcal{C}$ as follows: objects are elements in $S$, and for any $a,b\in S$, $Mor(a,b):=\varnothing$ if $a\leq b$ is false, and $Mor(a,b)=\{(a,b)\}$ if $a\leq b$ is true. Check that $\mathcal{C}$ is indeed a category. (This is without doubt also true if $\leq$ is partial order).

Proof:

The exercise is actually vaguely stated. Let me argue in a more clear way.

Let $(S,\leq)$ be a poset. Remember that $\leq$ is a relation. In other words, $\leq\subset S\times S$ and $\leq$ satisfies the poset condition. Therefore, when we write $a\leq b$, it is another way to write $(a,b)\in\leq.$

Thus, we can formally define $Mor(a,b):=\{(a,b)\}$ if $a\leq b$ and $Mor(a,b):=\varnothing$ otherwise.

Now, remember that $a\leq b$ means the pair $(a,b)$ belongs to the relation, so if $a\neq c$ or $b\neq d$, then we must have $Mor(a,b)\cap Mor(c,d)=\varnothing$, since the relation $(a,b)$ is not the same as the relation $(c,d)$. (If one of $Mor(a,b)$ or $Mor(c,d)$ is empty, then the intersection is empty trivially).

Now, the composition law inherits from the logic operator $\wedge$. That is, $(a,b)\circ (b,c):=a\leq b\wedge b\leq c$. This is well-defined due to the transitivity of $\leq$. Indeed, by the transitivity, $$(a,b)\circ (b,c):=a\leq b\wedge b\leq c=a\leq c=(a,c).$$

Then, the identity element for any object $a$ is $id_{a}:=(a,a)$. This exists due to the reflexive property of partial order relation, that is $a\leq a$. Then, for any $a,b\in S$, we have $$a\leq b\wedge b\leq b=a\leq b=a\leq a\wedge a\leq b,$$ and thus for any $(a,b)\in Mor(a,b)$ we have $$(a,b)\circ id_{b}=(a,b)\circ (b,b)=(a,b)=(a,a)\circ (a,b)=id_{a}\circ (a,b).$$

Finally, the composition is associative since the logic operator $\wedge$ is associative. Thus, this indeed forms a category.