Showing disjointness of the collection of $\mathrm{Hom}$-sets in $\mathscr{C}^2$ I’m working on 0.12 in Rotman’s Algebraic Topology text. The problem is to show that $\mathscr{C}^2$ is a category for any category $\mathscr{C}$. But Rotman apparently has an unusual definition for category, specifically in requirement 1:

Definition.
A category $\mathscr{C}$ consists of three ingredients:
a class of objects, $\operatorname{obj} \mathscr{C}$;
sets of morphisms $\mathrm{Hom}(A, B)$, one for every ordered pair $A, B \in \operatorname{obj} \mathscr{C}$;
composition $\mathrm{Hom}(A, B) \times \mathrm{Hom}(B, C) \to \mathrm{Hom}(A, C)$, denoted by $(f, g) \mapsto g \circ f$, for every $A, B, C \in \operatorname{obj} \mathscr{C}$, satisfying the following axioms:

*

*The family of $\mathrm{Hom}(A, B)$’s is pairwise disjoint;


*composition is associative when defined;


*for each $A \in \operatorname{obj} \mathscr{C}$, there exists an identity $1_A \in \mathrm{Hom}(A, A)$ satisfying $1_A \circ f = f$ for every $f \in \mathrm{Hom}(B, A)$, all $B \in \operatorname{obj} \mathscr{C}$, and $g \circ 1_A = g$ for every $g \in \mathrm{Hom}(A, C)$, all $C \in \operatorname{obj} \mathscr{C}$.
(Original scan)

1. Problem 0.12.

0.12.
Given a category $\mathscr{C}$, show that the following construction gives a category $\mathscr{M}$.
First, an object of $\mathscr{M}$ is a morphism of $\mathscr{C}$.
Next, if $f, g \in \operatorname{obj} \mathscr{M}$, say, $f \colon A \to B$ and $g \colon C \to D$, then a morphism in $\mathscr{M}$ is an ordered pair $(h, k)$ of morphisms in $\mathscr{C}$ such that the diagram
$$
  \require{AMScd}
  \begin{CD}
    A        @>{f}>>  B        \\
    @V{h}VV           @VV{k}V  \\
    C        @>>{g}>  D
  \end{CD}
$$
commutes.
Define composition coordinatewise:
$$
  (h', k') \circ (h, k) = (h' \circ h, k' \circ k) \,.
$$
(Original scan)

2. My confusion.
It's clear that the $\mathrm{Hom}$-sets in $\mathscr{C}^2$ are disjoint for any two pairs $(f, g)$ and $(f', g')$ of morphisms in $\mathscr{C}$ that don’t agree on domain and range – but what if they do?
So let $f, f' \in \mathrm{Hom}(A, B)$ such that $f \neq f'$ and let $g \in \mathrm{Hom}(C, D)$. Now suppose $(h, k) \in \mathrm{Hom}_{\mathscr{C}^2}(f, g) \cap \mathrm{Hom}_{\mathscr{C}^2}(f', g)$. In other words,
$$
  k \circ f = g \circ h \,, \quad  k \circ f' = g \circ h \,.
$$
It’s not clear to me why this is impossible, but it would violate the disjointness requirement of the collection of $\mathrm{Hom}_{\mathscr{C}^2}$-sets.
Update
The first commenter to the post, Randall, has given an excellent answer.
 A: Both definitions are fine on their own, but not compatible in the way that they are written.
For two morphisms $f \colon A \to B$ and $g \colon C \to D$ in $\mathscr{C}$, the current definition is
$$
  \operatorname{Hom}_{\mathscr{M}}(f, g)
  =
  \{
    (h, k)
    ∈
    \operatorname{Hom}_{\mathscr{C}}(A, C)
    ×
    \operatorname{Hom}_{\mathscr{C}}(B, D)
    \mid
    k ∘ f = g ∘ h
  \} \,.
$$
It can absolutely happen that for two different choices of morphisms $f$, $g$ and $f'$, $g'$ the sets $\operatorname{Hom}_{\mathscr{M}}(f, g)$ and $\operatorname{Hom}_{\mathscr{M}}(f', g')$ are not disjoint.
(However, this still requires that $f$ and $f'$ have the same domain and same codomain, and that similarly $g$ and $g'$ have the same domain and same codomain.)
In other words, there might exist an element $(h, k)$ in the intersection $\operatorname{Hom}_{\mathscr{M}}(f, g) ∩ \operatorname{Hom}_{\mathscr{M}}(f', g')$.
As mentioned by Randall in the comments, we nevertheless want to treat the morphism $(h, k) \colon f \to g$ as distinct from the morphisms $(h, k) \colon f' \to g'$.

This problem occurs not only for $\mathscr{M}$.
Suppose more generally that we have a “category” $\mathscr{C}$ that satisfies Rotman’s definition of a category except for possible part (i).
This is, there might exist for some objects $A$, $A'$, $B$ and $B'$ of $\mathscr{C}$ an element $f$ in the intersection $\operatorname{Hom}_{\mathscr{C}}(A, B) ∩ \operatorname{Hom}_{\mathscr{C}}(A', B')$.
We do then want to treat $f \colon A \to B$ as distinct from $f \colon A' \to B'$.
There is a standard way of solving this problem.
For this, we form a category $\mathscr{C}^+$ that satisfies Rotman’s definition, including part (i), as follows:

*

*The objects of $\mathscr{C}^+$ are the objects of $\mathscr{C}$.

*For every two objects $A$ and $B$ of $\mathscr{C}$, the set $\mathrm{Hom}_{\mathscr{C}^+}(A, B)$ is given by
$
  \{
    (A, f, B) \mid f ∈ \operatorname{Hom}_{\mathscr{C}}(A, B)
  \}
$.
(The elements of $\operatorname{Hom}_{\mathscr{C}^+}(A, B)$ are thus ordered triples with first entry $A$ and third entry $B$.)

*Given two such morphisms
$$
  (A, f, B) \colon A \to B \,,
  \quad
  (B, g, C) \colon B \to C
$$
in $\mathscr{C}^+$, their composite is defined by
$$
  (B, g, C) ∘ (A, f, B)
  :=
  (A, g ∘ f, C) \,,
$$
where $g ∘ f$ is the composite of $f$ and $g$ in the “category” $\mathscr{C}$.

Composition of morphisms in $\mathscr{C}^+$ is associative because the original composition of morphisms in $\mathscr{C}$ is associative.
For every object $A$ of $\mathscr{C}$, the morphism $(A, 1_{\mathscr{C}, A}, A)$ in $\mathscr{C}^+$ serves as the identity morphism of $A$ in $\mathscr{C}^+$.
This tells us that $\mathscr{C}^+$ is again a “category”.
However, $\mathscr{C}^+$ has a technical advantage over $\mathscr{C}$:
the sets $\operatorname{Hom}_{\mathscr{C}^+}(A, B)$ are pairwise disjoint, even if the original sets $\operatorname{Hom}_{\mathscr{C}}(A, B)$ were not.
Intuitively speaking, by passing from $\mathscr{C}$ to $\mathscr{C}^+$ we are forcing all morphism to remember their domain and codomain.
This then ensures that $\operatorname{Hom}$-sets are becoming pairwise disjoint.
We have for every two objects $A$ and $B$ of $\mathscr{C}$ the bijection
$$
  α_{A, B}
  \colon 
  \operatorname{Hom}_{\mathscr{C}}(A, B)
  \to
  \operatorname{Hom}_{\mathscr{C}^+}(A, B) \,,
  \quad
  f \mapsto (A, f, B) \,.
$$
These bijection are compatible with composition of morphisms and also with identity morphisms.
That is, we have
$$
  α_{B, C}(g) ∘ α_{A, B}(f) = α_{A, C}(g ∘ f)
$$
for every two composable morphisms $f \colon A \to B$ and $g \colon B \to C$ in $\mathscr{C}$, and also
$$
  α_{A, A}( 1_{\mathscr{C}, A} ) = 1_{\mathscr{C}^+, A} \,.
$$
For all practical and theoretical purposes, we can therefore replace the “category” $\mathscr{C}$ by the category $\mathscr{C}^+$.
