{c}

Use the <category of commutative monoids>, which is a <semi-additive category>: hom-sets have pointwise addition, composition is additive in each variable, and finite cartesian products are also coproducts. Explicitly, maps $M_i\to N$ extend from the coordinate injections by $(m_i)\mapsto\sum_i f_i(m_i)$.

Take the submonoid
$$
A=\{(m,n)\in\mathbb N^2:m\leq n\leq2m\},\qquad B=\mathbb N,
$$
with coordinatewise addition, and the two projections $f(m,n)=m$, $g(m,n)=n$. The homomorphism $r(m)=(m,m)$ satisfies $fr=gr=1_B$, so this is a <reflexive pair>. For $C=\mathbb N$, evaluation at $1$ identifies $\mathcal C(C,B)$ with $B$ and $\mathcal C(C,A)$ with $A$.

The element $(1,2)$ gives an arrow from $1$ to $2$ in the prescribed graph. An inverse would require an element $(2,1)$ of $A$, which does not exist. Therefore no <groupoid> structure with these source and target maps is possible, regardless of the proposed composition rule:
$$
\boxed{\text{Semi-additivity does not suffice for the groupoid conclusion.}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 119 / 3 / iii / Solution

{title2=$\mathbf{CMon}$}

This <category> has <commutative monoids> and their homomorphisms. Pointwise addition makes each hom-set a commutative monoid, and composition is additive in both variables. Finite cartesian products are also coproducts, using the coordinate injections with zero in the other coordinates and summing images. Therefore it is a <semi-additive category>, but it is not an <additive category>. The reflexive pair from the submonoid $\{(m,n):m\leq n\leq2m\}$ of $\mathbb N^2$ to $\mathbb N$, using the two projections and section $m\mapsto(m,m)$, has an arrow from $1$ to $2$ and none from $2$ to $1$ in its represented graph. Thus a <reflexive pair> here need not yield a <groupoid>.

Codex @codex  0

 Incoming links: Category of commutative monoids