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

A reflexive pair $f,g:A\rightrightarrows B$ has a common section $r:B\to A$ with $fr=gr=1_B$. Form the pushout in a category of $f$ and $g$, and write its two maps from $B$ as $j_1,j_2$. Its relation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. Thus $qf=qg$. Any $h:B\to C$ with $hf=hg$ gives the compatible pair $(h,h)$ for this pushout and therefore factors uniquely through $q$. This proves the coequalizer of a reflexive pair from a pushout construction.

There is a genuine transcription difference: the original PDF says finite products in the finite-colimit assertion, while the TeX says finite coproducts. The PDF assertion is false with products. Even retaining a pushout-existence assumption, the full subcategory of the Category of sets on nonempty sets has finite products, pushouts, and all coequalizers, but no initial object, hence no empty colimit. Regard the poset with elements $0,a,b,u_0,u_1,\dots$, orderand incomparable $a,b$ as a category. It has finite meets and top $u_0$, hence finite products in a category. Every reflexive pair in a poset is an equal pair and has its identity as a coequalizer. But $a,b$ have no least upper bound, so they have no coproduct in a category.

For the corrected construction using finite coproducts, let $D:\mathcal{J}\to \mathcal{C}$ be a finite diagram in a category. PutDefine $s,t:Y\rightrightarrows X$ on the summand indexed by $a:i\to j$ as ${\nu }_i$ and ${\nu }_{j}D(a)$. Maps $h:X\to Z$ with $hs=ht$ are exactly cocone under a diagram data on $D$. Hence their universal coequalizer is its colimit. Replace $s,t$ by the reflexive pairwhose common section is the second injection and whose coequalizer is unchanged. This proves construction of finite colimits from coproducts and reflexive coequalizers, including the empty diagram via the empty coproduct:

Let $r:B\to A$ split the reflexive pair $f,g$. Fix $C$, and take objects $u:C\to B$ and arrows $a:C\to A$ with source $fa$ and target $ga$. Since $\mathcal{C}$ is an additive category, its hom-sets are abelian groups, and composition is additive.

The identity at $u$ is $ru$. For $a:u\to v$ and $b:v\to w$, defineIts source is $fa-v+fb=u-v+v=u$ and its target is $ga-v+gb=v-v+w=w$. The two identity laws follow from $a-rv+rv=a$ and $ru-ru+a=a$. For $c:w\to x$, both parenthesizations of the triple composite are $a-rv+b-rw+c$, so composition is associative.

Define the inverse byIts source is $u+v-u=v$ and its target is $u+v-v=u$. Moreover $a^{-1}\star a=ru$ and $a\star a^{-1}=rv$. Thus these data form a groupoid, proving the represented form of reflexive pair in an additive category is an internal groupoid:

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 submonoidwith 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: