I'm working on a question in an intro algebraic topology book:
Verify that the "arrow category" is a category. That is, if $\mathcal{C}$ is a category, then show that the following construction gives a category $\mathcal{M}$: Let $\operatorname{obj} \mathcal{M}$ be the collection of arrows of $\mathcal{C}$. For $f,g\in \operatorname{obj} \mathcal{M}$, say $f: A \to B$ and $g: C \to D$, let an arrow in $\mathcal{M}$ from $f$ to $g$ be an ordered pair of arrows $(\alpha,\beta)$ (with $\alpha:A \to C$ and $\beta: B \to D$) in $\mathcal{C}$ such that $g \circ \alpha = \beta \circ f$.
Composition of arrows of $\mathcal{M}$ is given by the rule (\alpha',\beta') \circ ( \alpha,\beta) = ( \alpha' \circ \alpha,\beta'\circ\beta).
My trouble is showing that the hom-sets are pairwise disjoint. As I understand it, $ \operatorname{hom} (f,g) \cap \operatorname{hom} (h,k)$ is empty if $(f,g)\neq (h,k)$.
I tried assuming the contrary just to get a feel for why this must be true: Suppose $(\alpha,\beta) \in \operatorname{hom} (f,g) \cap \operatorname{hom} (h,k)$ and $(f,g) \neq (h,k)$. If $f,g,\alpha,\beta$ have the domains and codomains given above, then we must have $h: A \to B$ and $k: C\to D$ for the composites to make sense.
The diagrams commute and hence $g \circ \alpha=\beta \circ f$ and $k \circ \alpha = \beta \circ h$. Since $(f,g) \neq (h,k)$, we have $f \neq h$ or $g \neq k$.
Then...nothing. I'm not sure that this line of attack is going to work. It seems like the assumption that $(f,g) \neq ( h,k)$ gives me nothing usable.
Perhaps I have a fundamental misunderstanding of categories or the construction of $\mathcal{M}$. If somebody could give me a nudge in the right direction I would greatly appreciate it.