What if we simply require $\mathscr{A}$ to be pre-additive, or additive? I have seen it stated without proof that if $\mathscr{A}$ and $\mathscr{B}$ are abelian categories, then any equivalence $F:\mathscr{A}\to\mathscr{B}$ must be additive, but I have not been able to prove that. This is my work so far:


Clearly, $F(x\xrightarrow{0}y)=F(x)\xrightarrow{0}F(y)$ by factoring the first morphism through the zero object. However, the fact that $F$ is fully faithful does not seem to be sufficient to prove that $F$ induces homomorphism of Hom-sets (at least not by naïve algebraic manipulations). Is there some step I'm missing?



Let $F:\mathscr{A}\to\mathscr{B}$ be an equivalence of Abelian categories. Must $F$ be additive?
I am posting the comment above as an answer.


An equivalence of categories preserves identity morphisms, finite product, and finite coproducts. Thus, it also preserves diagonal morphisms and codiagonal morphisms. In an Abelian category, the finite product equals the finite coproduct. For every $f,g\in \text{Hom}\_{\mathcal{A}}(a,a')$, the morphism $f+g$ equals the composition of the diagonal morphism, $\Delta\_a: a\to a\times a$, the product morphism $f\times g : a\times a \to a'\times a'$, and the codiagonal morphism $a'\times a' \to a'$. Thus an equivalence of Abelian categories induces group homomorphism of Hom sets.


