Mathematics Asked on December 29, 2021
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!
Let $0$ and $1$ denote the objects of $D$ and write $a:0rightarrow 1$ for the only non-identity arrow of $D$. To every functor $F:Dto A$ associate the morphism $F(a):F(0)to F(1)$ in $A$. Conversely, to every morphism $f:Xto Y$ in $A$ associate the functor $hat f:Dto A$ given by $hat f(0)=X$, $hat f(1)=Y$ and $hat f(a)=f$. Can you continue from here?
Answered by Jackozee Hakkiuz on December 29, 2021
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 $mathrm{Arr}(mathscr C)$.
Answered by varkor on December 29, 2021
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