The additive Yoneda lemma says that if you have an $Ab$-enriched category $\mathscr A$ and an additive functor $F:A\rightarrow Ab$, then there is a group isomorphism $$Nat(\mathscr A(A,-),F)\cong F\ A$$
- Why is $Nat(\mathscr A(A,-),F)$ an abelian group?
Consider the map $\tau := a,X,f \mapsto F\ f\ a : F\ A \rightarrow Nat(\mathscr A(A,-),F)$.
To prove the additive Yoneda lema I understand that it is enough to prove that $\tau$ (or its inverse) is a group homomorphism, since it's a bijection because of the Yoneda lemma over $Set$.
However, Borceux in his Handbook of Categorical Algebra Vol. 2 proves instead that $(\tau\ a)_X : \mathscr A(A,X)\rightarrow F\ X$ is a group homomorphism.
- Why is this enough?
$\newcommand{\nat}{\mathsf{Nat}}\newcommand{\A}{\mathscr{A}}$I would assume the group structure on $\nat(\A(A,-),F)$ is given by the following; $\alpha+\beta$ is defined to be the natural transformation whose $j$th components are $\alpha_j+\beta_j:\A(A,j)\to F(j)$. This is well-defined and clearly yields an Abelian structure.
I really don't understand this notation $\tau:=a,X,f\mapsto F f a$, I suspect there is a typo.
Let's just agree to let $\tau:FA\to\nat(\A(A,-),F)$ be the obvious map. $\tau(0)$ is the natural transformation that maps arrows $f$ to their values $F(f)(0)$, which are always zero because $F(f)$ is a homomorphism. Thus, $\tau(0)$ is the zero element of $\nat(\A(A,-),F)$. It remains to show $\tau$ is additive. Say $x,y\in F(A)$. Then $\tau(x+y)$ is the transformation that maps every $f:A\to B$ to $F(f)(x+y)=F(f)(x)+F(f)(y)=\tau(x)(f)+\tau(y)(f)$. Thus, $\tau(x+y)=\tau(x)+\tau(y)$. Thus $\tau$ is a group homomorphism. We need to also show that each $\tau(x)_X$ is a homomorphism $\A(A,X)\to F(X)$ to ensure the arrows are in $\mathsf{Ab}$, as is required.