Additive inverse of a morphism in additive category, diagrammatically

Question

I can't find anything useful through the pages of Freyd's Abelian Cats and Google is of little help.

I would like to understand how is the (additive) inverse $-f$ of $f\colon A\to B$ defined, for $f$ a morphism in an additive category: the sum operation $f+g$ is defined diagrammatically as $$ f+g = \nabla (f \oplus g) \Delta \quad\colon\quad A\to A\times A\to B\times B\cong B\amalg B\to B$$ for any $f,\,g: A \to B$, where $\Delta$ and $\nabla$ are the diagonal and codiagonal morphisms. But what about the inversion?

Freyd's proof (Thm. 2.39) that $\hom(A,B)$ is an abelian group shows that an inverse must exist because of the invertibility of $\left(\begin{smallmatrix} 1 & f \\ 0 & 1\end{smallmatrix} \right )$. It is rather implicit and I would like to see what is $-f$ made of.

Obviously, if I'm able to invert identity maps I'm able to build $-f$ for any $f$, since $-f$ must equal $(-1_B)\circ f$ and/or $f\circ (-1_A)$; so, to sum up, I have to define an involution on the set of objects (confused with their identity arrows) of $\cal C$, naturally deduced from the data of an additive category; maybe it is deduced from the isomorphism between (binary) products and coproducts?

Edit: I see there's a clash in notations; maybe it's better to rephrase the entire question is simpler terms. When $\cal C$ has biproducts, then it is automatically enriched over abelian monoids. Can the enrichment over $\bf Ab$ be seen as a consequence of another property of $\cal C$? If yes, there can be a diagrammatic interpretation of $f\mapsto -f$ in terms of categorical data?

Answer 1

According to MacLane (Categories for The Working Mathematician, exercise 4 $\S$ VIII.2), given a category $\mathscr{A}$ with a zero object, finite products and coproducts and such that $A\amalg B\simeq A\times B$, the definition of $f+g$ in $\mathscr{A}$ as $\nabla (f \oplus g) \Delta\colon A\to A\times A\to B\times B\cong B\amalg B\to B$ turns the commutative monoid $\mathscr{A}(A,B)$ into an abelian group if, for each object $A$, there is an arrow $v_{A}\colon A\to A$ such that $v_{A}+1_{A}=0\colon A\to A$. Hence it seems that the condition of having already (a priori) an additive inverse for the identities is unavoidable, at this level of generality.

On the other hand, if you are interested in recovering the inverse $-f$ diagrammatically in an abelian category, I would suggest you to take a look (again, as I am sure you know it well) at Handbook of Categorical Algebra 2, by Borceux, $\S$ 1.6. Shortly, $-f$ is the composite

$$ A\stackrel{(0,f)}{\to}B\times B\stackrel{q}{\to}Q\stackrel{r^{-1}}\to B, $$ where $q$ is the coequaliser of $\Delta_{B}$ and $r$ is the composite $q\circ (1_{B},0)$.

Hope this helps somehow.

Answer 2

I know I arrive several years after the OP posed the question, and by the way this is not a real answer since the OP was interested in the case of abelian categories, but I have been convinced until today that it was always true that a category having a zero object and biproducts (i.e. having binary products and coproducts and such that the canonical morphisms $A\coprod B\rightarrow A\times B$ are isomorphisms) must be canonically preadditive, i.e. enriched in abelian groups, while I've just realized you only have an enrichment in commutative monoids, in general. I just wanted to share the (really simple) counterexample that came to my mind.

Consider the full subcategory $\mathcal{C}$ of the category $\mathrm{CMon}$ of small commutative monoids with objects the powers $M^k$ for $k\geq0$, where $M$ is a fixed non-zero commutative monoid that is not an abelian group (for example, we can take $M=\mathbb{N}$). Then $\mathcal{C}$ clearly has biproducts $M^k\oplus M^t=M^{k+t}$ and a zero object $0=M^0$, and in fact it has the natural enrichment over $\mathrm{CMon}$, but it cannot be enriched over $\mathrm{Ab}$; in fact, this would be, in particular, an enrichment in commutative monoids, and so it would be uniquely determined by the categorical structure of $\mathcal{C}$, so it would coincide with the standard $\mathrm{CMon}$-enrichment. But it is obvious that if $k>0$ and $f:M^k\rightarrow M^h$ is a monoid morphism (i.e. an additive map) and there is an $m\in M^k$ such that a component of $f(m)$ has no additive inverse (for example, $f=1_{M^k}$, or any $f\neq0$ if $M=\mathbb{N}$), then $f$ cannot have an additive inverse $-f$.