Category Theory - Basic question concerning the intersection of arbitrary sets and their morphisms

Question

Take an arbitrary category $\mathscr{C}$ and let $A, B \in \text{ob} (\mathscr{C})$. Also, let $\mathscr{C}(A, B)$ the collection of all morphisms between $A$ and $B$. In his book "Basic Category Theory" Tom Leinster writes:

"If $f \in \mathscr{C}(A,B)$, we call $A$ the domain and $B$ the codomain of $f$. Every map in every category has a definite domain and a definite codomain. (If you believe it makes sense to form the intersection of an arbitrary pair of abstract sets, you should add to the definition of category the condition that $\mathscr{C}(A,B) \cap \mathscr{C}(A',B')= \emptyset$ unless $A=A'$ and $B=B'$.)"

My problem is with the statement in between brackets.

The point is that if we work in $\mathbf{Set}$ it seems to me this condition is too strong. Indeed, let a function be defined in such a way that $x \mapsto x^2$ for every $x \in \mathbb{R}$. Then we have actually two morphisms in the category, namely $f \in \mathbf{Set}(\mathbb{R}, \mathbb{R})$ and $f' \in \mathbf{Set}(\mathbb{R}, \mathbb{R}^+)$: the codomains are different but their intersection is nonempty, but $f$ and $f'$ are somewhat similar, i.e., there is some 'overlap' in between them, so we should have something like $\mathbf{Set}(\mathbb{R},\mathbb{R}) \cap \mathbf{Set}(\mathbb{R},\mathbb{R}^+) \neq \emptyset$.

Any help to clarify this point will be greatly appreciated.
Thank you for your time.

Answer 1

In a category the domain and codomain of a morphism are an intrinsic property of a morphism: the same morphism can't have two different domains or two different codomains. In the category of sets, you have to represent a morphism as something like a triple $(X, f, Y)$, where $X$ and $Y$ are sets and $f$ is a set of pairs comprising a function from $X$ to $Y$. In your example you have $(\Bbb{R}, f, \Bbb{R}) \in {\cal C}(\Bbb{R}, \Bbb{R})$ which is not the same as $(\Bbb{R}, f, \Bbb{R}^{+}) \in {\cal C}(\Bbb{R}, \Bbb{R}^{+})$.

In the example immediately following the passage that you quote, Leinster states that "a map from $A$ to $B$ in the category $\mathbf{Set}$ is exactly what is ordinarily called a map (or mapping or function)". This is rather misleading to anyone brought up on the usual set-theoretic definition of a function as a set of pairs.

Answer 2

It is a similar story as with a disjoint union of sets. If you have two sets $A,B$ that are not necessarily disjoint, you can still form the set known as $A \sqcup B$. One possible implementation is $A\times \{1\} \cup B\times \{2\}$.

Also in your case you can label each morphism with its domain and codomain, to make sure that different copies of "the same map" are distinguished when they belong to different hom-sets. And then of course you mentally immediately forget that you did this and treat the hom-sets as disjoint.