= Solution
A <preadditive category> has an <abelian group> structure on every hom-set, with composition additive in each variable. Neither a <zero object> nor <biproducts> are part of this definition.
Fix $C$ and use the <reflexive pair> $f,g:A\rightrightarrows B$, $r:B\to A$, with $fr=gr=1_B$. Take objects $x:C\to B$ and arrows $a:C\to A$, with source $fa$, target $ga$, and identity at $x$ equal to $rx$. For composable $a,b$, so $ga=fb$, define
$$
\boxed{b\circ a=a+b-rga.}
$$
The <preadditive category> axioms give
$$
f(b\circ a)=fa+fb-ga=fa,\qquad
g(b\circ a)=ga+gb-ga=gb.
$$
Thus the formula has the required endpoints. The identities satisfy $(rga)\circ a=a$ and $a\circ(rfa)=a$, using $gr=fr=1_B$.
For $ga=fb$ and $gb=fc$, both ways of composing three arrows equal
$$
a+b+c-rga-rgb.
$$
Indeed $g(b\circ a)=gb$, while expanding $c\circ b$ and then composing with $a$ gives the same expression. Hence composition is associative.
Every arrow has inverse
$$
\boxed{a^{-1}=rfa+rga-a.}
$$
Its source is $ga$ and its target is $fa$. Substituting in the composition formula gives $a^{-1}\circ a=rfa$ and $a\circ a^{-1}=rga$. Therefore this is a <groupoid>.
For $u:C'\to C$, precomposition by $u$ preserves sources, targets, identities, composition and inverses by bilinearity. Thus the construction is natural in $C$, giving the requested <internal groupoid> structure in its hom-set formulation. If the composable-arrow <pullback in a category> exists, the same formula defines its internal composition morphism. The <reflexive-pair groupoid formula in a preadditive category> requires no extra additive-category hypotheses.
Back to article page
