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!

  • 1
    $\begingroup$ You mean you don't see the equivalence $\text{Arr}(A)\simeq \text{Fun}(D,A)$? $\endgroup$ Jul 24 '20 at 14:02
  • 2
    $\begingroup$ maybe i'm being silly here but yes. Thank you ! $\endgroup$
    – gerrard
    Jul 24 '20 at 14:05
  • $\begingroup$ 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? $\endgroup$
    – drhab
    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 $F, G : I \to \mathscr C$). enter image description here

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 $\mathrm{Arr}(\mathscr C)$.


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