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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?

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Consider the category of commutative monoids: if you could do what you wanted, then every commutative monoid would automatically be a group. – Zhen Lin May 3 '14 at 16:24
Be clearer, please! What are you trying to say, that it's impossible to write $-f$ in diagrammatic terms? Or rather that one cannot simply deduce the shape of $-f$ from the isomorphism $A\times B\cong A\amalg B$? I believe in the second statement. – Fosco Loregian May 3 '14 at 16:44
I made an edit. – Fosco Loregian May 3 '14 at 17:07
As far as I know, the answer to the edited question is no. Having inverses is a property, not a structure. – Qiaochu Yuan May 3 '14 at 17:17
Funny thing I was reading that very page (stuff, structure, property) when you commented. From this perspective, your argument is clear. But what would a formal definition of "structure" and "property" be? – Fosco Loregian May 3 '14 at 17:49

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.

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Thank you! The setting where I want to work is that of an abelian category, since I was interested in proving that the equalizer of $f,g$, defined as a two-fold pullback (cf. the "equalizer" page on the Lab) coincides with the kernel of $f-g$. But originally I thought everything happened in additive context. – Fosco Loregian May 3 '14 at 17:53
@tetrapharmakon You are welcome! – Marco Vergura May 3 '14 at 18:14

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