Why is the commutative diagram needed in an arrow Category? Given a category $\mathcal{C}$, why is diagram commutativity necessary for the morphisms in $\mathcal{C}^{\to}$?
As far as I can tell, even if the arrows of $\mathcal{C}^{\to}$ form just plain $\mathcal{C}$ diagrams, then they still satisfy the associativity/identity definitions of a category. Instead of needing the commutativity condition for $\mathcal{C}^{\to}$ to be a category, is it instead motivated by some other intuition?
In case I am not being clear: if $f: A \rightarrow B$  and $g: C\rightarrow D$ are $\mathcal{C}$-arrows then it appears you can legitimately define an arrow in $\mathcal{C}^{\to}$ as $(e, h)$ where $e: A \rightarrow C$ and $h: B \rightarrow D$. In this case, why then also stipulate that $g\circ e=h\circ f$?

This definition of the arrow category $\text{Arr}(\mathcal C)$ (where the squares formed by $\text{Arr}(\mathcal C)$-morphisms are commutative) arises naturally in category theory and has better properties.

*

*$\text{Arr}(\mathcal C)$ is (isomorphic to) the function category $[2,\mathcal C$] where $2 = \{0\rightarrow 1\}$.

*A functor $\mathcal C\to \text{Arr}(\mathcal D)$ is the same as a natural transformation between two functors $\mathcal C\to\mathcal D$.

