Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 119 / 6

In a semi-additive category, hom-sets are commutative monoids and composition is additive in both variables. Write $S={\coprod }_i{A}_i$, $P={\prod }_i{A}_i$, with injections ${\nu }_i$ and projections ${\pi }_i$. The canonical $c:S\to P$ has ${\pi }_{i}c{\nu }_j=1_{A_i}$ when $i=j$ and zero otherwise. DefineBilinearity givesThe universal properties of the product in a category and coproduct in a category imply $cd=1_P$ and $dc=1_S$. For the empty family, $S$ is initial and $P$ terminal. Their endomorphisms are necessarily the identity and the zero morphism; the zero map $P\to S$ therefore also gives inverse composites. Thus the comparison is an isomorphism for every finite family, giving a biproduct:

Use countably many copies of $A$, and let $c:S={\coprod }_{i\ge 1}A\to P={\prod }_{i\ge 1}A$ be the assumed isomorphism. Write $p_i={\pi }_{i}c:S\to A$, let $\delta :A\to P$ be the diagonal with every component $1_A$, and put $q=c^{-1}\delta$. Thus $p_{i}q=1_A$ for every $i$. Let $s:S\to S$ shift the injections, $s{\nu }_i={\nu }_{i+1}$, and let $\sigma :S\to A$ be the fold map, $\sigma {\nu }_i=1_A$.

The coordinate relations give $p_{1}s=0$ and $p_{i+1}s=p_i$, while $p_1{\nu }_1=1_A$ and $p_{i+1}{\nu }_1=0$. Therefore $q$ and ${\nu }_1+sq$ have identical $p_i$-coordinates. Since $c$ is invertible, these coordinates jointly distinguish arrows into $S$, soThe coproduct property gives $\sigma s=\sigma$, because both have value $1_A$ on each injection. Consequently $z=\sigma q$ satisfiesThis proves the absorbing-endomorphism assertion without assuming an infinite addition operation on hom-sets.

If the category is an additive category, each hom-set is an abelian group, so cancellation gives $1_A=0$. Any arrow $f:A\to B$ is then $f{1}_A=f0=0$. Every hom-set has exactly one arrow, and every object is isomorphic to the zero object. This is countable biproducts in an additive category force triviality:

In the category of complete join-semilattices, define addition of arrows pointwise byand let the zero arrow be the constant-bottom map. These form commutative monoids. Pointwise join of join-preserving maps again preserves all joins, including the empty join, and composition is additive in each variable. Thus the category has the required semi-additive enrichment.

For an arbitrary set-indexed family $(A_i)$, the product poset $P={\prod }_i{A}_i$, with coordinatewise joins, is the product in a category. Define ${\nu }_i:A_i\to P$ by putting an element in coordinate $i$ and bottom in all others. Given join-preserving $f_i:A_i\to B$, defineThis preserves arbitrary joins by interchanging the two joins, and $h{\nu }_i=f_i$. It is unique because every tuple is ${\bigvee }_i{\nu }_i(a_i)$. Therefore $P$ is also the coproduct in a category with these injections, and its canonical coproduct-to-product map is the identity. This proves the biproducts of complete join-semilattices property, even for arbitrary set-indexed families, and in particular makes this a semi-additive category with the hypotheses of part (ii).

The two-element chain has distinct identity and constant-bottom endomorphisms, so the category is not equivalent to $1$. Its addition is idempotent rather than cancellative; for example $1_A+1_A=1_A$ itself supplies an absorbing endomorphism. Hence