In what follows

$\textbf{preadditive}$: a category $\mathscr{C}$ is preadditive when $\forall\ A,B,\ \mathscr{C}(A,B)$ is an abelian group and the morphisms composition is a group homomorphism on each component (bilinear)

$\textbf{additive}$: a category $\mathscr{C}$ is additive when it is preadditive, it has a zero object and finite biproducts

$\textbf{additive functors}$: a functor between preadditive categories is additive when it is a group homomorphism on the hom-sets

I can't understand the proof of the following fact, which you can find on Borceux, Categorical algebra, vol. 2 at page 8.

$\textbf{Prop.}$ Consider an additive category $\mathscr{B}$ and a small preadditive category $\mathscr{A}$. In that case, the category $\operatorname{Add}(\mathscr{A},\mathscr{B})$ of additive functors between $\mathscr{A}$ and $\mathscr{B}$ is additive and biproducts in it are computed pointwise. Moreover, if $\mathscr{B}$ is finitely complete, so is $\operatorname{Add}(\mathscr{A},\mathscr{B})$ and finite limits are computed poitwise.

$\textbf{proof.}$ Since I have understood the first part of the proof., I will report here only the second, that concerning the existence of finite limits in the functor category.

So, since $\mathscr{B}$ has kernels, given a natural transformation $\alpha:F\Rightarrow G$, we can consider its pointwise kernel $\gamma:K\Rightarrow F$. We shall prove that $K$ is additive. Given $f,g:A\rightrightarrows A'$ in $\mathscr{A}$, we have \begin{align*} \gamma_{A'}\circ K(f-g)&=F(f-g)\circ\gamma_A\\ &=(Ff-Fg)\circ\gamma_A\\ &=(Ff\circ\gamma_A)-(Fg\circ\gamma_A)\\ &=(\gamma_{A'}\circ Kf)-(\gamma_{A'}\circ Kg)\\ &=\gamma_{A'}\circ (Kf-Kg) \end{align*} from which $K(f-g)=Kf-Kg$, since $\gamma_{A'}$ is a monomorphism. Similarly, we can prove that $K(0)=0$. Thus $\operatorname{Add}(\mathscr{A},\mathscr{B})$ has equalizers, hence it has finite limits (because products + equalizers$\Rightarrow$ limits)


1) Why does he use $K(f-g)$ and not $K(f+g)$? If he wants to prove that $K$ is additive, then $K(f+g)=Kf+Kg$ should be enough, shouldn't it?

2) Why does he prove that $K(0)=0$? If $K(f-g)=Kf-Kg$ then $K(0)=K(0-0)=K0-K0$ and the result follows by cancellation. Am I wrong?

3) (most important) Why do the additivity of $K$ imply that $\operatorname{Add}(\mathscr{A},\mathscr{B})$ has equalizers?? Here I don't have any guess unfortunately.


2 Answers 2


I'll try to answer your questions in order.

  1. I think you're right, it should be enough to prove that $K(f+g)=K(f)+K(g)$, which is exactly the definition of additivity.
  2. Again I think that your argument is sound. He's being redundant because every semigroup-homomorphism between groups preserves identities and inverses.
  3. $K$ has been defined as the pointwise kernel of $\alpha$, this implies by general results that $K$ is the equalizer of $\alpha$ and the zero morphism in $\text{Add}(\mathcal A,\mathcal B)$, i.e. it's a kernel. This proves that the category $K$ is a preadditive category with kernels, but preadditive categories have kernels if and only if they have equalizers: that's basically because an equalizer for a pair of parallel morphisms $f$ and $g$ is nothing but a kernel for the morphism $f-g$.

If I didn't do any mistake due to tiredness, I guess this should answer your question.

  1. No, because a map of abelian groups can preserve addition without preserving the identity and inverses, but this is not the case for subtraction. Thus you answer this yourself in 2: he's being redundant.

  2. A morphism of additive functors is an arbitrary natural transformation. Thus since $K$ is an equalizer with respect to natural transformations out of any functor, it's certainly an equalizer with respect to those out of additive functors, as long as it's additive itself, so that it lives in the right category.

EDIT: My response to 1 is wrong, as discussed in the comments. I don't know why he does this, but there's certainly no harm in it,

  • $\begingroup$ Why do you say that a map between abelian groups that preserve addition shouldn't preserve identities and inverse? Let $f \colon G \to H$ be a semi-group homomorphism between two groups, then if $0$ is the unit in $G$ we have that $f(0)=f(0+0)=f(0)+f(0)$. By subtracting on both sides of the equality $f(0)$ ($H$ is a group by hypothesis) we get that $0=f(0)$, where the $0$ on the left is the one in $H$. A similar argument proves that $f$ preserves inverses too. $\endgroup$ Commented Jul 16, 2016 at 22:24
  • $\begingroup$ Strange, I'm sure I have something in my head about this, and that I've seen people use this subtraction trick to simplify verifications. I'm not sure what I'm thinking. $\endgroup$ Commented Jul 16, 2016 at 22:27
  • $\begingroup$ @KevinCarlson probably the difference is useful to prove that a subset is a subgroup at once, because if a subset is closed under difference (or under multiplication by the inverse, in multiplicative notation) then it is closed under addition and it has the zero element. $\endgroup$
    – bateman
    Commented Jul 17, 2016 at 6:23
  • $\begingroup$ @bateman That's it! Thanks. $\endgroup$ Commented Jul 17, 2016 at 21:32

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