arrow category and functor category

Let A be an abelian category and D the category having two objects and only one nonidentity morphism between them.

The functor category A$$^D$$ is also abelian and it is called an arrow category with objects morphisms in A and morphisms commutative squares.

I cannot see the equivalence between the functor category and the arrow category. I understand arrow category but how it is equivalent to the functor category? Any help would be appreciated!

• You mean you don't see the equivalence $\text{Arr}(A)\simeq \text{Fun}(D,A)$? Jul 24 '20 at 14:02
• maybe i'm being silly here but yes. Thank you ! Jul 24 '20 at 14:05
• Observe that the unique category having two objects and only one non-identity arrow has the looks of an arrow. Functors $D\to A$, that is the objects of $A^D$, correspond with arrows in $A$, don't they? Jul 24 '20 at 14:08

Let $$0$$ and $$1$$ denote the objects of $$D$$ and write $$a:0\rightarrow 1$$ for the only non-identity arrow of $$D$$. To every functor $$F:D\to A$$ associate the morphism $$F(a):F(0)\to F(1)$$ in $$A$$. Conversely, to every morphism $$f:X\to Y$$ in $$A$$ associate the functor $$\hat f:D\to A$$ given by $$\hat f(0)=X$$, $$\hat f(1)=Y$$ and $$\hat f(a)=f$$. Can you continue from here?
This is true more generally for any category. Let $$I = \{ X \xrightarrow{f} Y \}$$ be the interval category (which you call $$D$$). The objects of $$\mathscr C^I$$ are functors $$F : I \to \mathscr C$$, i.e. assignments $$F(X) \in \mathscr C$$, $$F(Y) \in \mathscr C$$ and morphisms $$F(f) : I(X) \to I(Y)$$, which are the same as arrows in $$\mathscr C$$. The morphisms of $$\mathscr C^I$$ are natural transformations $$\eta$$: the naturality condition in this case amounts to the following square commuting (for $$F, G : I \to \mathscr C$$). Note that, because $$F(f)$$ and $$G(f)$$ are arbitrary arrows in $$\mathscr C$$, this simply amounts to two arrows $$\eta_X$$ and $$\eta_Y$$ in $$\mathscr C$$ such that the square commutes. That is, the data of $$\mathscr C^I$$ is exactly the same as $$\mathrm{Arr}(\mathscr C)$$.