T. W. Hungerford Graduate Text in Mathematics. Algebra. Definition 7.1

Question

At the very beginning of Chapter 1.7 the definition of a category is given as:

A category is a class $\mathfrak{C}$ of objects (denoted $A,B,C,...$) together with
(i) a class of disjoint sets, denoted $hom(A,B)$, one for each pair of objects in $\mathfrak{C}$;(an element $f$ of $hom(A,B)$ is called morphism from $A$ to $B$ and is denoted $f: A \rightarrow B $);
(ii) ...

Definition continues with the composition property of arrows.
My problem is:
According to the part (i), $hom(A,B)$ is a class of disjoint sets. Members of this class are morphisms that is if $f,g$ are two morphisms from $A$ to $B$ then $f,g \in hom(A,B)$ and since $hom(A,B)$ consists of disjoint sets $f$ and $g$ must be disjoint sets. My question is what is the meaning of "two morphisms being disjoint".

Please do not give any other definitions of a category. If you understand my problem and can see my mistake here and if you know this book well you are more than welcome. Thanks in advance

Answer 1

It seems to me that you misunderstood something.

My problem is:
According to the part (i), $hom(A,B)$ is a class of disjoint sets. Members of this class are morphisms that is if $f,g$ are two morphisms from $A$ to $B$ then $f,g \in hom(A,B)$ and since $hom(A,B)$ consists of disjoint sets $f$ and $g$ must be disjoint sets. My question is what is the meaning of "two morphisms being disjoint".

Hungerford did not say that $hom(A,B)$ is a class of disjoint sets. He said that

1. For each pair $(A,B)$ of objects $A, B$ of $\mathfrak C$ one has a set $hom(A,B)$ whose elements are called morphisms from $A$ to $B$.

2. These sets are disjoint for distinct pairs $(A,B)$ and $(A',B')$. That is, if $(A,B) \ne (A',B')$ (which means $A \ne A'$ or $B \ne B')$, then $hom(A,B) \cap hom(A',B') = \emptyset$.

He also did not say that the morphisms of $\mathfrak C$ are sets. However, in some axiomatic approaches to set theory one can regard each morphism of $\mathfrak C$ as a set because every $x$ satisfying a relation $x \in X$ is defined to be a set. But that is irrelevant here. Whether or not elements $f, g \in hom(A,B)$ are sets, we can only say that they are distinct ($f \ne g)$ or not ($f = g$). Note in particular that if $f,g$ are sets, we may have $f \ne g$ and $f \cap g \ne \emptyset$.

Answer 2

It does not refer to any particular $\text{Hom}(A, B)$ or elements of such, but the Hom's are disjoint: Assume $(A, B)\neq(A', B')$. Then $\text{Hom}(A, B)\cap\text{Hom}(A', B')=\emptyset$. One for each pair $(A, B)$ of objects, disjoint from one another.

For example, a consequence of this: given any non-surjective function $f\colon A\to B$, we have a surjection $\overline{f}\colon A\to\text{im}(f), a\mapsto f(a)$ and the inclusion $i\colon\text{im}(f)\colon B$. While $f(a)=\overline{f}(a)$ for each $a\in A$, these are not the same function because $(A, \text{im}(f))\neq(A, B)$, as the Hom-sets are disjoint.

Answer 3

Please do not give any other definitions of a category. If you understand my problem and can see my mistake here and if you know this book well you are more than welcome.

Doing it, however, makes things easy.

If a category is presented with two classes $C_0,C_1$ of "objects" and "morphisms", the source and target functions $s,t : C_1\to C_0$ define a function $\langle s,t\rangle : C_1 \to C_0\times C_0$ sending a morphism to the pair of its domain and codomain.

Definition. $\hom(a,b) := \langle s,t\rangle^{-1}(a,b)$.

In other words, $\hom(a,b)$ is the fiber, or inverse image, of $(a,b)\in C_0\times C_0$ along said function $\langle s,t\rangle$.

It is now a property of functions that fibers over different points $(a,b), (a',b')$ are disjoint. And there you have that if $(a,b)\ne (a',b')$, then $\hom(a,b)\cap \hom(a',b')=\varnothing$.

What I implicitly assumed is that class functions behave as expected as opposed to set functions; but this is almost exactly how classes were engineered.

Now, your definition is equivalent to this if you accept that $C_1$ is the class formed as $\sum_{ab}\hom_C(a,b)$, or more evocatively the class of triples $(a,b,f:\hom_C(a,b))$: again by definition of what is means forming this class, there are source and target maps $s,t : \sum_{ab}\hom_C(a,b) \to C_0$ and you can repeat the previous line of reasoning.