Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 25 / 7

A semi-additive category has commutative-monoid enrichment: each hom-set is a commutative monoid, and composition in a category preserves addition and zero in each variable. In the finite-biproduct convention, which we use here, it also has finite products in a category and coproducts in a category. A preadditive category has abelian groups as its hom-sets, with composition additive in each variable; this definition alone does not require finite products in a category or a zero object.

Let $P={\prod }_{i=1}^n{A}_i$ with projections $p_i$. Define $i_j:A_j\to P$ by $p_i{i}_j=1_{A_j}$ for $i=j$ and $p_i{i}_j=0$ otherwise. Bilinearity givesso product uniqueness gives ${\sum }_j{i}_j{p}_j=1_P$. For a family $f_j:A_j\to X$, set $f={\sum }_j{f}_j{p}_j$. Then $f{i}_j=f_j$. If $h:P\to X$ has the same restrictions, $h=h{\sum }_j{i}_j{p}_j={\sum }_j{f}_j{p}_j=f$. Thus the product in a category is also the coproduct in a category, with its canonical injections. Conversely, the analogous construction from a finite coproduct in a category gives projections and the same identities, proving that it is a product in a category.

For the empty family, a terminal object $T$ has a singleton endomorphism monoid, so $1_T=0$. Every map $T\to X$ equals its composite with $1_T=0$, hence equals the zero map; this map exists by enrichment. Therefore $T$ is also an initial object, a zero object. The dual argument applies to an initial object. Consequently finite products in a category and coproducts in a category coincide canonically as biproducts, including the nullary case.

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 $b$ byThese expressions use addition and subtraction in the hom-set of the preadditive category. The reflexive pair identities $fr=gr=1_B$ giveso 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 areTherefore these operations define a groupoid on the specified object and arrow sets. Precomposing by $k:C^{\prime }\to C$ respects each expression, by additivity of composition, so the groupoids are natural in $C$.

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.