= Solution
Fix an object $C$. For $x:C\to B$, use $rx:C\to A$ as its identity arrow. For $a:C\to A$, its source is $fa$ and target is $ga$. If $ga=fb$, define the composite of $a$ followed by $b$ by
$$
\boxed{b\circ a=a+b-rga,\qquad a^{-1}=rfa+rga-a.}
$$
These expressions use addition and subtraction in the <hom-set> of the <preadditive category>. The <reflexive pair> identities $fr=gr=1_B$ give
$$
f(b\circ a)=fa+fb-ga=fa,\qquad g(b\circ a)=ga+gb-ga=gb,
$$
so the composite has the required endpoints. Also $f(a^{-1})=ga$ and $g(a^{-1})=fa$.
The unit identities follow directly: $a\circ rfa=rfa+a-rfa=a$ and $rga\circ a=a+rga-rga=a$. If $ga=fb$ and $gb=fc$, either parenthesization of three arrows equals $a+b+c-rga-rgb$, so composition is associative. The inverse identities are
$$
a^{-1}\circ a=a+a^{-1}-rga=rfa,\qquad a\circ a^{-1}=a^{-1}+a-rfa=rga.
$$
Therefore these operations define a <groupoid> on the specified object and arrow <sets>. Precomposing by $k:C'\to C$ respects each expression, by additivity of composition, so the <groupoids> are natural in $C$.
\b[Every <reflexive pair> therefore has the required <groupoid> structure on generalized points.] This is the <reflexive-pair groupoid formula in a preadditive category>. If the composable-arrow <pullback in a category> $A\times_{g,B,f}A$ exists, its projections $p,q$ give the actual internal composition map $p+q-rgp$. The printed generalized-point definition does not assume that this pullback exists, and preadditivity alone would not ensure it.
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