Does an Ab-enriched category have a unique Ab-enrichment?

Question

I know that the group structure on Hom sets can be recovered from biproducts if they exit. Indeed, if $f, g : A \to B$ are two maps then there is a uniquely defined map $f \oplus g : A \oplus A \to B \oplus B$ and then there are diagonal and codiagonal maps giving $$ A \to A \oplus A \to B \oplus B \to B $$ so we get a composition law on Hom sets.

However, if products/coproduct/biproducts don't exist in my category, then I don't see why the Ab-enrichment should be a "property" and not a "structure." Therefore, I am wondering if it possible for some wacky pre-additive category without biproducts to have multiple Ab-enriched structures?

Answer 1

As Martin Brandenburg and Maxime Ramzi suggest, it is easy to construct examples on small categories.

For example, a one object Ab-enriched category is exactly a ring. The category corresponds to the monoid (multiplication) of the ring and the Ab-enrichment to the addition law. There are monoids which have multiple abelian group structures that make them a ring.

Even more simply, consider the category with two objects $x,y$ and $n$ parallel morphisms $x \to y$. We need two morphisms $0, \mathrm{id} \in \mathrm{Hom}(x,x)$ so we may set $\mathrm{End}(x,x) \cong \mathbb{Z}$ as a ring generated by $\mathrm{id}$. Then we set $\mathrm{Hom}(y,x) = 0$ to be the trivial group. The only nontrivial compositions are $\mathrm{Hom}(y, y) \times \mathrm{Hom}(x, y) \to \mathrm{Hom}(x,y)$ and $\mathrm{Hom}(x,y) \times \mathrm{Hom}(x,x) \to \mathrm{Hom}(x,y)$ which must be the unique $\mathrm{Z}$-module structure on any abelian group because the generator $\mathrm{id}$ acts as a unit under composition.

Therefore, an Ab-enrichment is exactly an abelian group structure on $\mathrm{Hom}(x,y)$ of which there are many.