Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 18 (Category Theory)

A monomorphism is a morphism $m:X\to Y$ such that $mu=mv$ implies $u=v$ for every pair $u,v:Z\to X$. Dually, an epimorphism $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$. A regular epimorphism is a coequalizer of some parallel pair. An isomorphism has a two-sided inverse.

For a product in a category, write ${\Delta }_A:A\to A\times A$ for its categorical diagonal. Since ${\pi }_1{\Delta }_A=1_A$, equality ${\Delta }_{A}u={\Delta }_{A}v$ gives $u={\pi }_1{\Delta }_{A}u={\pi }_1{\Delta }_{A}v=v$. Thus the diagonal is a monomorphism, indeed a split monomorphism.

Here use the left lifting property against monomorphisms as the definition of “strong”; epimorphicity will be established separately. In a lifting square write $g:C\to D$, $m:A\to B$, $k:C\to A$ and $l:D\to B$, with $mk=lg$ and $m$ a monomorphism. Any two lifts agree because $mt=m{t}^{\prime }=l$.

First let $g$ be the coequalizer of $r,s:E\rightrightarrows C$. Every coequalizer is an epimorphism: if $ug=vg$, the uniqueness clause for the coequalizer applied to this common composite gives $u=v$. Moreover,so $kr=ks$ by monomorphism cancellation. The coequalizer therefore supplies $t:D\to A$ with $tg=k$. Then $mtg=mk=lg$, and epimorphism cancellation gives $mt=l$. This proves that regular epimorphisms are strong epimorphisms.

If $g:C\to D$ is also a monomorphism, take $m=g$, $k=1_C$ and $l=1_D$. Its lift satisfies $tg=1_C$ and $gt=1_D$. Hence monic lifting-only strong morphisms are invertible.

Next suppose $gf:A\to C$ has the left lifting property against monomorphisms. Given a lifting square for $g:B\to C$, with $mk=lg$, precompose its top arrow with $f$. A lift for $gf$ gives $t:C\to X$ with $mt=l$ and $tgf=kf$. Crucially, one does not cancel $f$: instead $mtg=lg=mk$, and the monomorphism $m$ gives $tg=k$. Thus the right factor of a strong composite is strong, proving right-factor cancellation for lifting-only strong morphisms.

Finally, in $u=iv$ with $u$ strong and $i$ monic, the preceding result makes $i$ strong. The monic-strong argument then makes $i$ an isomorphism. None of these arguments assumed that a lifting-only strong morphism was already epic.

Let $g:C\to D$ have the left lifting property against monomorphisms, and suppose $\alpha ,\beta :D\to B$ satisfy $\alpha g=\beta g$. Use the categorical diagonal ${\Delta }_B:B\to B\times B$, which is a monomorphism by part (a). The square with top arrow $\alpha g$, bottom arrow $⟨\alpha ,\beta ⟩$, left arrow $g$ and right arrow ${\Delta }_B$ commutes, since both product components are $\alpha g$.

Its lift $t:D\to B$ satisfies ${\Delta }_{B}t=⟨\alpha ,\beta ⟩$. Applying the two product in a category projections gives $t=\alpha$ and $t=\beta$. Thus $g$ is an epimorphism. This binary-product criterion for lifting-only strong epimorphisms requires binary products, rather than any assumption about equalizers.

The covariant form of the Yoneda lemma says that for a locally small category $\mathcal{C}$, a functor $F:\mathcal{C}\to \mathbf{Set}$ and an object $A$, there is a bijection, natural in $A$ and $F$,Explicitly its two directions areFor $u:B\to B^{\prime }$, functoriality gives $F(u){\alpha }_B^x(f)=F(uf)(x)={\alpha }_{B^{\prime }}^x(uf)$, so ${\alpha }^x$ is a natural transformation. Conversely, naturality of $\alpha$ at $f:A\to B$ gives ${\alpha }_B(f)=F(f)({\alpha }_A(1_A))$. This proves that the displayed maps are inverses.

Naturality in $F$ follows because postcomposition by $\theta :F\Rightarrow F^{\prime }$ sends $x$ to ${\theta }_A(x)$. For $u:A^{\prime }\to A$, precomposition of transformations by $\mathcal{C}(u,-)$ sends ${\alpha }^x$ to ${\alpha }^{F(u)(x)}$. This also verifies naturality in the representing object.

In the Category of sets, an epimorphism is exactly a surjective function. A surjective function is right-cancellable. If $q:X\to Y$ misses $y\in Y$, the constant-zero function and the function that is one at $y$ and zero elsewhere are distinct maps $Y\to \{0,1\}$ with equal composites with $q$.

If every ${\alpha }_A$ is surjective, equality $\beta \alpha =\gamma \alpha$ for natural transformations $\beta ,\gamma :G\Rightarrow H$ gives ${\beta }_A={\gamma }_A$ for every $A$. Thus $\alpha$ is an epimorphism in the functor category.

For the converse, construct the pointwise amalgamated doublewhere exactly the two copies of each element of $\mathrm{im}{\alpha }_A$ are identified; different elements of $G(A)$ remain different. Define $H(u)[y,i]=[G(u)y,i]$. Naturality of $\alpha$ implies that $G(u)$ takes its image into the image at the target, so this formula is well-defined and gives a functor. The maps $j_{i,A}(y)=[y,i]$ form natural transformations $j_0,j_1:G\Rightarrow H$, with $j_0\alpha =j_1\alpha$.

If $\alpha$ is an epimorphism, $j_0=j_1$. Since the two copies of an element outside $\mathrm{im}{\alpha }_A$ would be distinct, every element must lie in that image. Hence $\alpha$ is epic if and only if every component is epic. This proves the pointwise epimorphism in a functor category criterion directly, including the needed existence and naturality of the separating functor.

A projective object in a category is an object $P$ such that, for every epimorphism $e:X\to Y$ and every $u:P\to Y$, there is $v:P\to X$ with $ev=u$.

Let $P={\coprod }_{i\in I}{P}_i$ be a coproduct in a category of projective objects in a category, with injections ${\iota }_i$. Given $e:X\to Y$ epic and $u:P\to Y$, projectivity supplies $v_i:P_i\to X$ with $e{v}_i=u{\iota }_i$. Choose these lifts for the set-indexed family. The coproduct in a category supplies a unique $v:P\to X$ satisfying $v{\iota }_i=v_i$. Since $ev{\iota }_i=u{\iota }_i$ for every $i$, its universal property gives $ev=u$.

Thus coproducts of projective objects are projective. For an empty family, $P$ is the initial object, and the lifting assertion follows directly from its unique maps. The family-of-lifts step uses the usual axiom of choice.

Take an epimorphism $\alpha :F\Rightarrow G$ in $[\mathcal{C},\mathbf{Set}]$ and a natural transformation $\beta :\mathcal{C}(A,-)\Rightarrow G$. By the Yoneda lemma, $\beta$ corresponds to $x={\beta }_A(1_A)\in G(A)$. By the pointwise epimorphism in a functor category criterion, choose $y\in F(A)$ with ${\alpha }_A(y)=x$.

The Yoneda lemma gives ${\gamma }_B(f)=F(f)(y)$, a natural transformation $\mathcal{C}(A,-)\Rightarrow F$. Naturality of $\alpha$ yieldsTherefore covariant representables are projective in the set-valued functor category. Local smallness ensures that $\mathcal{C}(A,-)$ is set-valued.

A functor $U:\mathcal{C}\to \mathbf{Set}$ is a representable functor if some object $R$ admits a natural isomorphism $\mathcal{C}(R,-)\cong U$. Here the representable is covariant.

For the identity functor on the Category of sets, choose the singleton $1=\{\ast \}$. The evaluation mapsare bijections with inverse $x\mapsto (\ast \mapsto x)$. For $u:X\to Y$, evaluation of $uf$ is $u(f(\ast ))$, proving naturality. Hence the identity functor on sets is represented by a singleton.

For a set $S$, let ${\iota }_s:1\to S$ select $s$. Given any family $f_s:1\to X$, define $f:S\to X$ by $f(s)=f_s(\ast )$. Then $f{\iota }_s=f_s$, and these equations determine every value of $f$, so it is unique.

This is precisely the universal property of a coproduct in a category, givingThe empty set gives the empty coproduct in a category, namely the initial object of the Category of sets.

Suppose $F:\mathbf{Set}\to \mathcal{C}$ is a left adjoint to $U$. The adjunction gives bijections, natural in $X$,The second map is evaluation at the singleton element, as in part (a). Thus $U$ is represented by $F1$. This argument uses the one-point set as a generator of the particular set-valued adjunction; it does not claim that every arbitrary right adjoint is representable.

Let $U\cong \mathcal{C}(R,-)$ be a representable functor, and let $(L,p_j)$ be the categorical limit of a small diagram $D:J\to \mathcal{C}$. A morphism $R\to L$ is uniquely equivalent to a family $f_j:R\to D(j)$ satisfying $D(u)f_j=f_k$ for every $u:j\to k$.

Such compatible families are exactly the elements of the categorical limit of the set-valued diagram $\mathcal{C}(R,D(-))$. Consequently the canonical comparisonis a bijection. Transporting it through the representing natural isomorphism proves that $U$ preserves every small limit that exists in $\mathcal{C}$. This proves that covariant representables preserve limits. For an empty diagram this says that maps into a terminal object form a singleton.

Choose a representing object $R$ and a natural isomorphism ${\theta }_X:\mathcal{C}(R,X)\cong UX$. Using the assumed small coproducts in a category, defineFor a function $v:S\to S^{\prime }$, define $F(v)$ by $F(v){\iota }_s={\iota }_{v(s)}$. The coproduct in a category uniqueness clause proves preservation of identities and composition, so this is a functor.

Restriction to the coproduct summands, followed by $\theta$, givesThese bijections are natural in $S$ by the definition of $F(v)$, and natural in $X$ by naturality of $\theta$. They establish $F⊣U$, the left adjoint to a covariant representable functor. The empty set is sent to the empty coproduct.

Write $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$. The characterization concerns an adjunction with the specified unit and counit of an adjunction $\eta$ and $\epsilon$. Its triangle identities for an adjunction areAssume these identities. Define the hom-set mapsNaturality of $\eta$ and $\epsilon$ makes these maps natural in both objects. Naturality and the triangle identities for an adjunction giveThus they are inverse bijections and define $F⊣G$.

Conversely, from the natural hom-set bijections of an adjunction, define ${\eta }_A=\Phi (1_{FA})$ and ${\epsilon }_D={\Phi }^{-1}(1_{GD})$. Naturality gives the same formulas for $\Phi$ and $\Psi$ above. Applying $\Psi \Phi$ to $1_{FA}$ and $\Phi \Psi$ to $1_{GD}$ gives the two triangle identities for an adjunction. Therefore these identities are exactly the compatibility conditions on the specified unit and counit. If the printed equivalence were read as mere existence of some adjunction, independently of the supplied transformations, its only-if direction would be false: on the category of abelian groups, $F=G=1$ are adjoint, but choosing both transformations to be zero does not satisfy either triangle on a nonzero object.

Put $e_A={\epsilon }_{FA}F{\eta }_A$. This is a natural transformation $F\Rightarrow F$. Only the $G$-triangle $G{\epsilon }_D{\eta }_{GD}=1_{GD}$ is assumed. Naturality gives two useful absorption identities:For the second equality in the last line use naturality of $\epsilon$ at ${\epsilon }_D$, and then the assumed $G$-triangle. Naturality of $\epsilon$ at $F{\eta }_A$ and of $\eta$ at ${\eta }_A$ now givesHence $e$ is an idempotent in the functor category: this is the one-triangle adjunction idempotent.

For the splitting of an idempotent morphism, suppose this idempotent morphism splits as natural transformations $r:F\Rightarrow H$ and $s:H\Rightarrow F$, with $sr=e$ and $rs=1_H$. DefineThe first absorption identity givesFor the other triangle, naturality of $s$ at ${\eta }_A^{\prime }$ and of $\epsilon$ at $r_A$ givesThus the triangle identities for an adjunction prove $H⊣G$.

Conversely, suppose $H⊣G$, with unit $\psi :1_{\mathcal{C}}\Rightarrow GH$ and counit $\phi :HG\Rightarrow 1_{\mathcal{D}}$. The unit has target $GH$, as its type requires. DefineThese are natural transformations. Transposition under $H⊣G$ gives $G{s}_A{\psi }_A={\eta }_A$. Independently, naturality of $\eta$ at ${\psi }_A$ and the assumed $G$-triangle giveThe transpose of $r_A{s}_A$ is therefore $G{r}_{A}G{s}_A{\psi }_A={\psi }_A$, the transpose of $1_{HA}$. Injectivity of the hom-set bijection implies $r_A{s}_A=1_{HA}$. Naturality of $\epsilon$ at $s_A$ givesConsequently $G$ has a left adjoint if and only if $e$ splits. This is the criterion for splitting a one-triangle adjunction idempotent. The argument gives both the explicit splitting and the new unit and counit, without assuming the other triangle for $F$.

For $A\in \mathcal{C}$, the comma category $(A\downarrow G)$ has objects $(D,u)$ with $D\in \mathcal{D}$ and $u:A\to GD$. A morphism $(D,u)\to (D^{\prime },u^{\prime })$ is $h:D\to D^{\prime }$ satisfying $Ghu=u^{\prime }$. Identities and composition are those of $\mathcal{D}$, and functoriality of $G$ verifies the condition under composition.

The Freyd general adjoint functor theorem states: if $\mathcal{D}$ is a locally small category with all small categorical limits, and $\mathcal{C}$ is locally small, then $G:\mathcal{D}\to \mathcal{C}$ has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The latter means that for each $A$ there is a set-indexed family $u_i:A\to G{D}_i$ such that every $u:A\to GD$ equals $Gh{u}_i$ for some $i$ and some $h:D_i\to D$.

For necessity, use the standard result that a right adjoint preserves categorical limits. If $F⊣G$, the singleton family containing the unit of an adjunction ${\eta }_A:A\to GFA$ is a solution set, since transposition gives $u=Gh{\eta }_A$ for a unique $h:FA\to D$.

For sufficiency, use the following standard limit fact: if $\mathcal{D}$ is complete and $G$ preserves limits, the projection $(A\downarrow G)\to \mathcal{D}$ creates small limits. Indeed, a compatible family $A\to G{D}_j$ induces a unique arrow into $G(\lim D_j)$, and this makes the underlying limit a limit in the comma category. The comma category is locally small because each of its hom-sets is a subset of a hom-set of $\mathcal{D}$. Its solution family is a weakly initial set.

We prove the remaining initial-object lemma for complete categories with a weakly initial set. In any locally small category $\mathcal{K}$ with all small limits and a weakly initial set $(K_i)$, form $W={\prod }_i{K}_i$. It is weakly initial: for any $X$, some $K_i\to X$ exists and may be composed with the projection $W\to K_i$. The empty family cannot be weakly initial in a nonempty complete category, which has a terminal object.

The set $\mathrm{End}(W)$ is small. Form a simultaneous equalizer $e:E\to W$ of every endomorphism of $W$ and $1_W$; thusThis equalizer exists by completeness, for example as the equalizer of two maps $W\rightrightarrows W^{\mathrm{End}(W)}$. The object $E$ is still weakly initial, since it maps to $W$.

Given $a,b:E\to X$, take their equalizer $j:Y\to E$. Weak initiality of $W$ gives $t:W\to Y$. Since $ejt$ is an endomorphism of $W$, we have $ejte=e$, and cancellation of the monomorphism $e$ gives $jte=1_E$. Thus $j$ is a split epimorphism as well as a monomorphism, so it is an isomorphism. From $aj=bj$ follows $a=b$. There is at least one map $E\to X$ by weak initiality, so $E$ is initial.

Apply this lemma to every $(A\downarrow G)$ and choose its initial object $(FA,{\eta }_A)$. For $v:A\to A^{\prime }$, initiality gives the unique $Fv:FA\to F{A}^{\prime }$ satisfying $GFv{\eta }_A={\eta }_{A^{\prime }}v$. Uniqueness proves the functor laws. The same initiality gives natural bijectionsHence $F⊣G$, completing the theorem without invoking another adjoint functor theorem.

A monad consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1\Rightarrow T$ and $\mu :T^2\Rightarrow T$, the unit and multiplication of a monad, satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aTa$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ satisfies $ha=bTh$. These objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$; its composition works because $T$ is a functor.

For the list monad, $TX={\coprod }_{n\ge 0}{X}^n$ is the set of finite ordered lists, including the empty list. The map $Tf$ applies $f$ to each entry; ${\eta }_X(x)=[x]$ and ${\mu }_X$ concatenates a list of lists. The unit laws say that adding singleton brackets and then flattening changes nothing. Associativity says that flattening a list of lists of lists in either order produces the same ordered sequence. These descriptions also prove naturality.

If $a:TX\to X$ is an algebra for a monad, defineThe singleton law gives $a([x])=x$. Apply the algebra associativity law to $[[],[x]]$ and $[[x],[]]$ to obtain $e\ast x=x=x\ast e$. Applying it to $[[x,y],[z]]$ and $[[x],[y,z]]$ showsThus $(X,\ast ,e)$ is a monoid. Applying the same law to $[[x_1,\dots ,x_{n-1}],[x_n]]$ shows inductively that $a$ is necessarily ordered multiplication of its entries, with the empty product $e$.

Conversely, any monoid defines such a list-fold map $a$. The monoid unit proves $a\eta =1$, and associativity and the unit prove that multiplying flattened lists equals multiplying their individual products, including empty sublists. Hence $a\mu =aTa$. An algebra morphism preserves the empty-list value and two-entry-list values, so it is a monoid homomorphism; conversely a monoid homomorphism preserves every ordered product and is an algebra morphism. ThereforeThus list-monad algebras are monoids, with the identification also matching every morphism.

The free algebra functor isThe monad laws make ${\mu }_A$ an algebra action, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $U:{\mathcal{C}}^T\to \mathcal{C}$ forget the action. DefineThe proposed inverse is an algebra morphism becauseNaturality of $\eta$ and the algebra unit law give $(bTf){\eta }_A=f$. If $h$ is an algebra morphism, thenThe formulas are natural in both variables, so $F⊣U$. Its unit is $\eta$, and its counit at $(B,b)$ has underlying map $b:TB\to B$. Thus the monad induced by an adjunction has endofunctor $UF=T$, unit $\eta$, and multiplication $U{\epsilon }_F=\mu$. It is exactly the original monad, not merely a monad with the same endofunctor.

For an algebra for a monad $(A,a)$, consider the fork in the Eilenberg-Moore categoryHere ${\mu }_A:F(TA)\to F(A)$ is the counit at $F(A)$, and $Ta=F(a)$. The arrow $a$ is an algebra morphism by $a{\mu }_A=aTa$, which also says that it coequalizes the two arrows.

Let $h:F(A)\to (B,b)$ be an algebra morphism with $h{\mu }_A=hTa$. Define $g=h{\eta }_A:A\to B$. Naturality of $\eta$ at $a$ givesSince $h$ is an algebra morphism,Thus $g:(A,a)\to (B,b)$ is an algebra morphism with $ga=h$. Any other such factorization satisfies $g^{\prime }=g^{\prime }a{\eta }_A=h{\eta }_A=g$. This proves the full coequalizer universal property inside the algebra category.

The pair is moreover a reflexive pair: its common section is $F({\eta }_A)$, with underlying map $T{\eta }_A$, because ${\mu }_{A}T{\eta }_A=1_{TA}$ and $TaT{\eta }_A=T(a{\eta }_A)=1_{TA}$. Hence every monad algebra is a reflexive coequalizer of free algebras. This reflexive free-algebra presentation of a monad algebra needs no general existence theorem for arbitrary algebra-category colimits.

An isomorphism $f:A\to B$ is a morphism with an inverse $f^{-1}:B\to A$ satisfying $f^{-1}f=1_A$ and $f{f}^{-1}=1_B$. A groupoid is a category in which every morphism is an isomorphism.

With the given one-sided inverses,Consequently $f$ is an isomorphism with $f^{-1}=g=h$. This argument uses only the category axioms.

A preadditive category has an abelian group structure on every hom-set, with composition additive in each variable. Neither a zero object nor biproducts are part of this definition.

Fix $C$ and use the reflexive pair $f,g:A\rightrightarrows B$, $r:B\to A$, with $fr=gr=1_B$. Take objects $x:C\to B$ and arrows $a:C\to A$, with source $fa$, target $ga$, and identity at $x$ equal to $rx$. For composable $a,b$, so $ga=fb$, defineThe preadditive category axioms giveThus the formula has the required endpoints. The identities satisfy $(rga)\circ a=a$ and $a\circ (rfa)=a$, using $gr=fr=1_B$.

For $ga=fb$ and $gb=fc$, both ways of composing three arrows equalIndeed $g(b\circ a)=gb$, while expanding $c\circ b$ and then composing with $a$ gives the same expression. Hence composition is associative.

Every arrow has inverseIts source is $ga$ and its target is $fa$. Substituting in the composition formula gives $a^{-1}\circ a=rfa$ and $a\circ a^{-1}=rga$. Therefore this is a groupoid.

For $u:C^{\prime }\to C$, precomposition by $u$ preserves sources, targets, identities, composition and inverses by bilinearity. Thus the construction is natural in $C$, giving the requested internal groupoid structure in its hom-set formulation. If the composable-arrow pullback in a category exists, the same formula defines its internal composition morphism. The reflexive-pair groupoid formula in a preadditive category requires no extra additive-category hypotheses.

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.