Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 18 / 2

A monomorphism $m:A\to B$ satisfies $mu=mv\Rightarrow u=v$, and an epimorphism $e:A\to B$ satisfies $ue=ve\Rightarrow u=v$. A strong monomorphism is a monomorphism with the right lifting property against all epimorphisms: every square $mu=ve$ has a diagonal $d$ satisfying $de=u$ and $md=v$. The diagonal is unique by monicity. A regular monomorphism is an equalizer of some pair $h,k:B\rightrightarrows Z$. It is monic because two equalizer factorizations of the same arrow must agree.

The converted TeX omits the remainder of this subpart. For the printed strict monomorphism condition, an arrow $g:C\to B$ is admissible when, for every pair $h,k$ out of $B$, the implication $hm=km\Rightarrow hg=kg$ holds; strictness says each such $g$ factors uniquely through $m$. If $m$ equalizes $h_0,k_0$, every admissible $g$ satisfies $h_{0}g=k_{0}g$, and the equalizer property supplies its unique factorization. Hence every regular monomorphism is strict.

Strictness itself implies monicity: whenever $mu=mv$, the common composite is admissible, so uniqueness of its factor through $m$ gives $u=v$. Now take a square $mu=ve$ with $e$ epic. Whenever $hm=km$, we have $hve=hmu=kmu=kve$, hence $hv=kv$. Thus $v$ is admissible and has a unique factor $d$ through $m$. Monicity gives $de=u$. We have proved the chain

A split coequalizer consists of $f,g:A\rightrightarrows B$, $q:B\to Q$, $s:Q\to B$ and $t:B\to A$ with $qf=qg$, $qs=1_Q$, $ft=1_B$ and $gt=sq$. If $hf=hg$, then $h=hft=hgt=hsq$. Thus $hs$ factors $h$ through $q$, and the factor is unique because $q$ has the right inverse $s$. This proves the coequalizer property directly. Every functor preserves the diagram, since all these equations are preserved.

For idempotent splitting through a coequalizer, first suppose $e=ir$ with $ri=1_Q$. Then $re=r$. Any $h:E\to Z$ satisfying $he=h$ factors as $h=(hi)r$, uniquely since $r$ is a split epimorphism. Hence $r$ coequalizes $(e,1_E)$.

Conversely, let $q:E\to Q$ coequalize $(e,1_E)$. Since $e^2=e$, the arrow $e:E\to E$ itself equalizes the pair, so there is a unique $i:Q\to E$ with $iq=e$. Then $qiq=qe=q$, and every coequalizer is an epimorphism, so $qi=1_Q$. Thus $e=iq$ is a splitting of an idempotent morphism.