Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 18 / 8

In a pointed category, a zero object defines zero morphisms between all objects. A categorical cokernel of $f:A\to B$ is a map $g:B\to C$ with $gf=0$ such that every $u:B\to X$ with $uf=0$ factors uniquely as $u=\bar{u}g$. Equivalently it is the coequalizer of $f$ and the zero map, so $g$ is an epimorphism.

Write $a:A\to A^{\prime }$, $b:B\to B^{\prime }$, and $c:C\to C^{\prime }$ for the vertical arrows, with $g^{\prime }=\mathrm{coker}{f}^{\prime }$ and $cg=g^{\prime }b$. The left pushout in a category applied to the compatible pair $g:B\to C$ and $0:A^{\prime }\to C$ gives $h:B^{\prime }\to C$ satisfying $hb=g$ and $h{f}^{\prime }=0$. The categorical cokernel property of $g^{\prime }$ then gives $d:C^{\prime }\to C$ with $d{g}^{\prime }=h$.

Now $dcg=d{g}^{\prime }b=hb=g$. Cancel the epimorphism $g$ to obtain $dc=1_C$. To prove the other identity, the two arrows $cd{g}^{\prime },g^{\prime }:B^{\prime }\to C^{\prime }$ agree after $b$, because $cd{g}^{\prime }b=cg=g^{\prime }b$, and after $f^{\prime }$, because both composites are zero. The pushout in a category uniqueness clause gives $cd{g}^{\prime }=g^{\prime }$. Cancelling the epimorphism $g^{\prime }$ gives $cd=1_{C^{\prime }}$.

Thus $c$ is an isomorphism: cokernel invariance under pushout holds already in pointed categories with the indicated cokernels.

Let $q:B\to Q$ be the categorical cokernel of $f$ in an abelian category. If $f$ is an epimorphism, the equality $qf=0f$ implies $q=0$. Since $q$ is also an epimorphism, $1_{Q}q=0q$ implies $1_Q=0$. An object with zero identity is a zero object: every morphism to or from it is zero. Hence the cokernel object is zero.

Conversely, suppose the cokernel object is zero. If $uf=vf$, additivity gives $(u-v)f=0$. The categorical cokernel property makes $u-v$ factor through the zero object, so $u-v=0$ and $u=v$. ThusThis is the zero-cokernel criterion for epimorphisms. In a pushout in a category, the two horizontal morphisms have isomorphic cokernels by part (a). Therefore the lower morphism is epic if and only if the upper morphism is epic. In particular pushouts reflect epimorphisms in an abelian category; the same argument also proves preservation.

Use the displayed square's notation $f:A\to B$, $h:A\to C$, $g:B\to D$, $k:C\to D$, with $gf=kh$ and $g$ epic. Form the biproduct and the morphismsThe map $q$ is an epimorphism, since its restriction to $B$ is $g$: equality after $q$ implies equality after $g$. The pullback in a category property says exactly that $j$ is a categorical kernel of $q$. Indeed $qj=gf-kh=0$, and a map into $B\oplus C$ killed by $q$ is a pair $(x,y)$ with $gx=ky$, which factors uniquely through $(f,h)$.

Use the standard abelian category property that every epimorphism is the categorical cokernel of its categorical kernel. If $u:B\to X$ and $v:C\to X$ satisfy $uf=vh$, then $[u,-v]j=0$. Hence there is a unique $w:D\to X$ with $wq=[u,-v]$. Restriction to the two summands gives $wg=u$ and $wk=v$. This is the pushout in a category universal property. Thus a pullback of an epimorphism is a pushout in an abelian category.

In this pushout, $g$ is the pushout of $h$ along $f$. The reflection result of part (b) therefore makes $h$ epic. This proves pullback stability of epimorphisms in an abelian category.

Finally, let $e:E\to B$ be an epimorphism and let $p_1,p_2:R\rightrightarrows E$ be its kernel pair. Their pullback square is a pushout by the result just proved. If $u:E\to X$ satisfies $u{p}_1=u{p}_2$, the two copies of $u$ form a pushout cocone. There is a unique $v:B\to X$ with $ve=u$. Hence epimorphisms in an abelian category are coequalizers of their kernel pairs, so they are regular epimorphisms.