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Take any morphism $h:A\to B$ and let $c:B\to C$ be its categorical cokernel. Let $i:I\to B$ be the categorical kernel of $c$. Since $ch=0$, kernel universality gives $e:A\to I$ with $h=ie$. The map $i$ is a monomorphism. We must establish the pseudo-epimorphism property of $e$, rather than merely name $I$ an image.

Suppose $t:I\to Z$ satisfies $te=0$. Let $k:K\to I$ be its categorical kernel. Then $e=ka$ for some $a:A\to K$. The composite $ik$ is a monomorphism, so the hypothesis that every monomorphism is normal provides a map $q:B\to D$ of which $ik$ is a categorical kernel, up to its unique compatible isomorphism. Now $qh=qika=0$. The categorical cokernel property of $c$ gives $q=\bar{q}c$, and therefore $qi=\bar{q}ci=0$.

Since $ik$ is the categorical kernel of $q$, the map $i$ factors through it: $i=ik\ell$ for some $\ell :I\to K$. Cancel the monomorphism $i$ to get $k\ell =1_I$. As $k$ is also a monomorphism, $\ell k=1_K$, so $k$ is an isomorphism. Finally $t=tk\ell =0$, proving that $e$ is a pseudo-epimorphism.

ThusThe pseudo-epimorphism factorization through the kernel of a cokernel proof uses a pointed category, existence of categorical kernels and categorical cokernels, and normality of every monic map. It does not silently replace pseudo-epic cancellation against zero by the stronger epimorphism condition.

In an abelian category, for $f:A\to B$ set $\mathrm{coim}f=\mathrm{coker}(\ker f)$ and $\mathrm{im}f=\ker (\mathrm{coker}f)$. The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.

The short five lemma states that in a commuting diagram of short exact sequences $0\to L\to M\to R\to 0$ and $0\to L^{\prime }\to M^{\prime }\to R^{\prime }\to 0$ in an abelian category, if the vertical maps $L\to L^{\prime }$ and $R\to R^{\prime }$ are isomorphisms, then so is $M\to M^{\prime }$.

For the Five lemma, consider commuting exact rowswith vertical maps $v_i:A_i\to B_i$. We prove the usual stronger version: $v_1$ epic, $v_2$ and $v_4$ invertible, and $v_5$ monic imply that $v_3$ is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.

Put $L=\mathrm{im}{d}_2$ and $R=\mathrm{im}{d}_3$, and define $L^{\prime },R^{\prime }$ similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.

First, $L\cong \mathrm{coker}{d}_1$ and $L^{\prime }\cong \mathrm{coker}{d}_1^{\prime }$: exactness at $A_2$ identifies $\mathrm{im}{d}_1$ with $\ker d_2$, and the image-coimage isomorphism for $d_2$ gives the claimed cokernel. Write $q:A_2\to L$ and $q^{\prime }:B_2\to L^{\prime }$ for these categorical cokernels. To construct an inverse to the induced map $\ell :L\to L^{\prime }$, observeCancellation of the epimorphism $v_1$ gives $q{v}_2^{-1}{d}_1^{\prime }=0$, so $q{v}_2^{-1}$ factors uniquely as $\bar{\ell }{q}^{\prime }$ with $\bar{\ell }:L^{\prime }\to L$. From $\ell q=q^{\prime }{v}_2$, composing with the epic $q,q^{\prime }$ gives $\bar{\ell }\ell =1_L$ and $\ell \bar{\ell }=1_{L^{\prime }}$.

Second, $R\cong \ker d_4$ and $R^{\prime }\cong \ker d_4^{\prime }$ by exactness at the fourth objects. Write their inclusions as $j:R\to A_4$ and $j^{\prime }:R^{\prime }\to B_4$. The equality $d_4^{\prime }{v}_4=v_5{d}_4$ gives the induced map $\rho :R\to R^{\prime }$. Sinceand $v_5$ is a monomorphism, $d_4{v}_4^{-1}{j}^{\prime }=0$. Kernel universality gives $\bar{\rho }:R^{\prime }\to R$ with $j\bar{\rho }=v_4^{-1}{j}^{\prime }$. Cancelling the monic $j,j^{\prime }$ proves $\bar{\rho }\rho =1_R$ and $\rho \bar{\rho }=1_{R^{\prime }}$.

Apply the short five lemma to the two short exact sequences: their outer maps $\ell ,\rho$ are invertible, so $v_3$ is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.

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.

In an abelian category, the image factorization in an abelian category of $f:A\to B$ iswhere the abelian-category axiom identifies coimage with image. Thus $p$ is epic and $i$ is monic. Any other epi-mono factorization $f=jq$ has $\ker q=\ker f$ and $\mathrm{coker}j=\mathrm{coker}f$. Since epimorphisms are cokernels of their kernels, its middle object is canonically isomorphic to $\mathrm{im}f$, uniquely compatibly with the two factors.

For a square $vf=f^{\prime }u$, define $I(u,v):\mathrm{im}f\to \mathrm{im}{f}^{\prime }$ byThe first arrow exists because $u\ker f$ factors through $\ker f^{\prime }$, so $p^{\prime }u$ annihilates $\ker f$. Its composite with $i^{\prime }$ equals $vi$ after the epimorphism $p$, proving the second equation. Uniqueness after $p$ proves preservation of identities and composition. This gives the functoriality of abelian image factorization as a functor from the arrow category.

For pullback stability of abelian image factorization, state the standard facts that pullbacks preserve monomorphisms, epimorphisms in an abelian category are stable under pullback, and two adjoining pullback squares have pullback outer rectangle. In the given diagram, $i^{\prime }$ is therefore monic and $p^{\prime }$ is epic, while the composite $i^{\prime }{p}^{\prime }$ is the pullback of $f$. Its epi-mono factorization is an image factorization by the uniqueness just proved. Thus the top row is the image factorization of the pulled-back arrow, with its middle object canonically the pullback of the original image subobject.

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.

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.

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.