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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.

Let $\mathcal{Q}$ be the full subcategory of quotients of decidable objects. It is closed under quotients, by composition of epimorphisms, and under small coproducts, by part (ii). It is also closed under subobjects: pull back a decidable cover $D\twoheadrightarrow B$ along $B^{\prime }\hookrightarrow B$. The resulting cover of $B^{\prime }$ has domain a subobject of $D$, hence a decidable object.

The terminal object lies in $\mathcal{Q}$. If $D\twoheadrightarrow B$ and $D^{\prime }\twoheadrightarrow C$ are decidable covers, their product is an epimorphism $D\times D^{\prime }\twoheadrightarrow B\times C$. Part (ii) makes its domain decidable. Thus products, and then equalizers as subobjects of products, remain in $\mathcal{Q}$. The inclusion $I:\mathcal{Q}\hookrightarrow \mathcal{E}$ preserves finite limits.

We construct a coreflective subcategory rather than claim that every object has a decidable cover. For $X\in \mathcal{E}$, let $qX\hookrightarrow X$ be the union of all subobjects of $X$ which lie in $\mathcal{Q}$. The Grothendieck topos is well-powered, so these subobjects form a set. Choose a decidable cover of each and take their coproduct. Its map to $X$ has image $qX$, so $qX$ is itself a quotient of a decidable object. Any map from an object of $\mathcal{Q}$ to $X$ has image in $\mathcal{Q}$, and therefore factors uniquely through $qX$. This givesThe induced idempotent comonad $G=Iq$ on $\mathcal{E}$ preserves finite limits: $q$ is a right adjoint and $I$ preserves those limits. Its counit is the inclusion $qX\hookrightarrow X$, and $q(qX)=qX$.

The coalgebras of this comonad are exactly the objects of $\mathcal{Q}$. A coalgebra structure is a section $X\to qX$ of the monic counit, forcing the counit to be an isomorphism; conversely an object already in $\mathcal{Q}$ has the unique such structure. Thus $\mathcal{Q}\simeq {\mathcal{E}}^G$. Part (i) now givesThis argument proves the required elementary-topos conclusion without presuming a small family of decidable generators for $\mathcal{E}$.

The key fact is that the extra left adjoint $f_{!}$ sends each representable $yC$ to an indecomposable projective object. Let ${\coprod }_i{B}_i\twoheadrightarrow f_{!}yC$ be epic. The inverse image $f^{\ast }$ preserves epimorphisms and coproducts, because it is a left adjoint between toposes. Apply it and lift the unit $\eta :yC\to f^{\ast }{f}_{!}yC$ through the resulting epimorphism, using projectivity of $yC$. A map from $yC$ to a coproduct selects one component, by evaluation at $C$ and the Yoneda lemma. Thus for some $i$ we obtain $t:yC\to f^{\ast }{B}_i$ with $f^{\ast }(B_i\to f_{!}yC)t=\eta$.

Transpose $t$ across $f_{!}⊣f^{\ast }$ to $\bar{t}:f_{!}yC\to B_i$. The displayed equality says that its composite back to $f_{!}yC$ is the identity. This proves the required indecomposable-projective property.

Since idempotents split in $\mathcal{D}$, part (ii) supplies objects $FC\in \mathcal{D}$ and isomorphisms $f_{!}yC\cong y(FC)$. Full faithfulness of the Yoneda embedding transports the action of $f_{!}$ on representable arrows to a functor $F:\mathcal{C}\to \mathcal{D}$. For $P\in \hat{\mathcal{D}}$,These identifications are natural in both $C$ and $P$. HenceIts right adjoint is consequently the right Kan extension from part (i), uniquely up to natural isomorphism. Thus the entire geometric morphism is induced by $F$.

A representable functor $yC$ is an indecomposable projective object. Given an epimorphism ${\coprod }_i{B}_i\twoheadrightarrow yC$, evaluate at $C$. Epimorphisms and coproducts in a presheaf category are pointwise, so $1_C\in yC(C)$ is the image of some element of a particular $B_i(C)$. By the Yoneda lemma, that element defines $s:yC\to B_i$, and its composite into $yC$ corresponds to $1_C$, hence is the identity. The selected component is split epic. More generally, evaluation sends any epimorphism to a surjection, so a map from $yC$ lifts through any epimorphism; this also proves its ordinary projectivity.

Conversely, every presheaf $A$ has the canonical epimorphismwhose component is the natural transformation named by $x$. It is pointwise surjective, since an element at $D$ is reached from its own summand $(D,x)$ at $1_D$. If $A$ is indecomposable projective, one component $r:yC\to A$ has a section $s:A\to yC$. The endomorphism $sr$ of $yC$ is idempotent and therefore corresponds to an idempotent morphism $e:C\to C$.

If idempotents split in $\mathcal{C}$, choose $C\stackrel{p}{\to }D\stackrel{i}{\to }C$ with $ip=e$, $pi=1_D$. Then $A\cong yD$: the mutually inverse maps are $y(p)s:A\to yD$ and $ry(i):yD\to A$. ThusWithout that hypothesis the argument still proves that every such object is a retract of a representable. The initial presheaf is not indecomposable projective, since its identity is the empty-coproduct epimorphism and has no component to select.

A monomorphism $m:A\to B$ is left cancellable: $mu=mv$ implies $u=v$ for all parallel arrows into $A$. A strong monomorphism is a monomorphism with the following lifting property: if $e:X\to Y$ is an epimorphism and $u:X\to A$, $v:Y\to B$ satisfy $mu=ve$, there is a diagonal $d:Y\to A$ with $de=u$ and $md=v$. It is unique by monicity. An extremal monomorphism is a monomorphism $m$ for which every factorization $m=he$ with $e$ an epimorphism forces $e$ to be an isomorphism. We use the strong lifting formulation for the paper's terminology; below we also justify its equivalence with the extremal formulation under the complete, well-powered hypotheses. A regular monomorphism is an equalizer of two parallel arrows.

First, a strong monomorphism is extremal. In a factorization $m=he$, apply its lifting property with upper arrow $1_A$ and lower arrow $h$. This gives $d$ with $de=1_A$. Since $e$ is an epimorphism, $ede=e$ implies $ed=1$, so $e$ is an isomorphism.

If $m:A\to B$ is an equalizer of $r,s:B\rightrightarrows C$, consider a lifting square as above. The equation $rve=smu=sve$ and the epimorphism property give $rv=sv$. The equalizer property therefore gives a unique $d$ with $md=v$. Monicity then gives $de=u$. Hence every regular monomorphism is a strong monomorphism.

For the intersection of strong subobjects, let $m_i:A_i\to B$ be strong and let $p_i:P\to A_i$ exhibit their intersection, with $m=m_1{p}_1=m_2{p}_2$. Given $mu=ve$ with $e$ an epimorphism, the two strong monomorphisms supply diagonals $d_i:Y\to A_i$ satisfying $d_{i}e=p_{i}u$ and $m_i{d}_i=v$. The pullback in a category property supplies $d:Y\to P$ with $p_{i}d=d_i$. Its two projections show $de=u$, and $md=v$. Thus the intersection is strong. The same argument works for every existing small intersection of strong monomorphisms. Also, composites of strong monomorphisms are strong: first lift through the outer one, then through the inner one.

Write $\mathcal{S}$ for the full subcategory formed by every saturated object with respect to anodyne morphisms, using the paper's anodyne morphism in a category convention. Let $m:C\to B$ be a strong monomorphism and $B$ saturated. For an anodyne morphism in a category $e:X\to Y$ and $u:X\to C$, saturation extends $mu$ to $v:Y\to B$. Since $e$ is an epimorphism, the strong lifting property supplies $d:Y\to C$ extending $u$. Therefore every strong subobject of a saturated object is saturated.

Now assume $\mathcal{C}$ is a complete category and a well-powered category, with the given saturated embeddings. Choose a monomorphism $a:A\to B$ into a saturated object. Intersect all strong subobjects of $B$ through which $a$ factors. They form a set by the well-powered category hypothesis, and the identity of $B$ is one of them. Completeness constructs their intersection, so we obtainBy the intersection of strong subobjects, $m$ is strong; hence $LA$ is saturated. Since $a$ is a monomorphism, so is ${\eta }_A$.

To prove that ${\eta }_A$ is an epimorphism, take $u,v:LA\rightrightarrows D$ with $u{\eta }_A=v{\eta }_A$. Their equalizer $j:E\to LA$ is a regular monomorphism, and ${\eta }_A$ factors through it. The composite $mj:E\to B$ is strong, so it belongs to the family defining $LA$. Minimality gives $t:LA\to E$ with $mjt=m$. Cancelling $m$ gives $jt=1_{LA}$; since $j$ is a monomorphism, also $tj=1_E$. Thus $j$ is an isomorphism and $u=v$. We have provedThis is the saturated reflection from a strong-subobject intersection.

The construction of a least strong subobject and the equalizer argument work for any morphism into any object, without requiring the source map to be monic or the target to be saturated. They give an epimorphism followed by a strong monomorphism. Applying this factorization to an extremal monomorphism makes the first factor invertible; thus the extremal and strong formulations agree under the present complete, well-powered hypotheses.

For every saturated $S$, saturation extends every $f:A\to S$ across the anodyne morphism in a category ${\eta }_A$ to $\bar{f}:LA\to S$. Its extension is unique because ${\eta }_A$ is an epimorphism. Consequentlynaturally in $S$ and $A$. In particular, for $h:A\to A^{\prime }$ define $Lh$ by $Lh{\eta }_A={\eta }_{A^{\prime }}h$; uniqueness proves the identity and composition laws. This constructs the reflector and proves that $\mathcal{S}$ is a reflective subcategory.

Finally we prove that $\mathcal{S}$ is a balanced category. If $e:U\to V$ is epic in $\mathcal{S}$ and $u,v:V\to X$ satisfy $ue=ve$, embed $X$ by a monomorphism $j:X\to Y$ into a saturated object. Fullness makes $ju,jv$ arrows of $\mathcal{S}$, so $ju=jv$ and hence $u=v$. Thus an epimorphism in $\mathcal{S}$ is also epic in $\mathcal{C}$.

A monomorphism $f:U\to V$ in $\mathcal{S}$ is also monic in $\mathcal{C}$: for $a,b:X\to U$, extend them uniquely along ${\eta }_X$ to $\bar{a},\bar{b}:LX\to U$. If $fa=fb$, cancellation of the ambient epimorphism ${\eta }_X$ gives $f\bar{a}=f\bar{b}$, so monicity in $\mathcal{S}$ gives $\bar{a}=\bar{b}$ and $a=b$. Now if $f$ is both monic and epic in $\mathcal{S}$, it is an anodyne morphism in a category in $\mathcal{C}$. Saturation of $U$ extends $1_U$ across $f$ to $r:V\to U$ with $rf=1_U$. Ambient epicity gives $fr=1_V$. Fullness puts this inverse in $\mathcal{S}$, proving