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An abelian category has enough projectives when every object admits an epimorphism from a projective object in a category. Repeating this on the kernel in a category constructs a projective resolution of every object. Finite direct sums of projective objects remain projective and give the degreewise split horseshoe construction.

Fix a prime number $p$. Take the full subcategory of finitely generated abelian groups with no element of order $p^2$. It is closed under finite direct sums and subgroups. If $t_p(G)$ is the $p$-primary torsion subgroup, the map $G\to R(G)=G/(p{t}_p(G))$ is a reflection into this subcategory: every map to an allowed group kills $p{t}_p(G)$. Therefore ordinary kernels in a category and reflected cokernels in a category supply all finite categorical limits and colimits. No nonzero finitely generated group has zero reflection, so the epimorphisms are precisely surjections and are all conormal epimorphisms. Multiplication by $p^2$ on $\mathbb{Z}$ is a monomorphism but not a normal monomorphism: its cokernel here is $\mathbb{Z}/p\mathbb{Z}$, whose kernel is $p\mathbb{Z}$, not $p^2\mathbb{Z}$. Multiplication by $p$ is normal; composing it with itself gives this nonnormal monomorphism.

In the categorical usage where no model structure is specified, an anodyne morphism may mean a morphism that is both a monomorphism and an epimorphism. It need not be an isomorphism unless the category is balanced.

In a balanced category with pullbacks in a category, pull a monomorphism back across the bottom map of a lifting square whose left map is an epimorphism. The resulting monic projection is epic because its composite with the lifted top map is epic. Balancedness makes that projection invertible, producing the unique lift. Conversely, if all monomorphisms are strong monomorphisms, applying the lifting property of a bimorphism to its own square gives its inverse, so the category is balanced. Pullback stability of epimorphisms is not needed.

In a category with binary products in a category, a map with the left lifting property against monomorphisms is an epimorphism. If $ue=ve$, lift the square with right side $\Delta$ and bottom side $⟨u,v⟩$. The two projections of its filler give $u=v$. This argument does not require equalizers.

The category of unital rings has identity-preserving ring homomorphisms as morphisms. The zero ring, allowing $0=1$, is terminal, and $\mathbb{Z}$ is initial. Nonzero rings admit no maps from the zero ring. Monomorphisms are injective: maps from $\mathbb{Z}[x]$ probe individual elements. It is a regular category, with surjective homomorphisms as regular epimorphisms; these are distinct from arbitrary categorical epimorphisms.

In an additive category with finite categorical limits and colimits in which every epimorphism is a conormal epimorphism, closure of normal monomorphisms under composition forces every monomorphism to be normal. To prove this, write a monomorphism $m$ as $kg$, where $k=\ker \mathrm{coker}m$, and put $l=\ker \mathrm{coker}g$. If $kl=\ker t$ were normal, $tm=0$ would force $tk=0$, hence $l$ would be an isomorphism. This makes $\mathrm{coker}g=0$, so $g$ is epic as well as monic. An epic monomorphism which is a cokernel is invertible, forcing $m$ to be normal. Contrapositively, any failure of normality provides two normal monomorphisms whose composite is not normal.

A conormal epimorphism is an epimorphism that is the cokernel in a category of some morphism. It is the concept dual to a normal monomorphism.

A set-indexed coproduct in a category of projective objects in a category is projective. Restrict a map out of the coproduct to each summand, lift each restriction through the given epimorphism, and assemble the lifts by the coproduct universal property. Choosing an arbitrary family of lifts uses the usual axiom of choice. The empty coproduct, an initial object, is projective as well.

Every epimorphism in an abelian category is the coequalizer of its kernel pair. Its kernel-pair square is a pullback in a category along an epimorphism and therefore also a pushout. A map equalizing the two projections supplies the same map on both pushout legs and so factors uniquely through the original epimorphism. In particular, all such epimorphisms are regular epimorphisms.

An epimorphism $e$ is extremal when a factorization $e=mh$ with $m$ a monomorphism forces $m$ to be an isomorphism. Every regular epimorphism is extremal: its coequalizer property factors $h$ back through $e$, giving an inverse to $m$. In a regular category, the converse follows by applying extremality to the regular-epi/mono image factorization.

A monomorphism $m$ is extremal if every factorization $m=he$ with $e$ an epimorphism forces $e$ to be an isomorphism. A strong monomorphism has this property by its lifting property. In a complete category that is a well-powered category, intersecting all strong subobjects containing a given morphism yields an epimorphism–strong-monomorphism factorization; applying it to an extremal monomorphism proves that the extremal monomorphism is strong. The factorization definition and lifting definition should be kept distinct until the needed ambient hypotheses are supplied.

Take four distinct objects $a,b,x,d$ and six nonidentity arrows $m:a\to b$, $h:x\to b$, $r,s:b\rightrightarrows d$, $k:a\to d$, $l:x\to d$, with $rm=sm=k$ and $rh=sh=l$. These and identities are all morphisms. The only nonidentity epimorphisms have target $d$, which admits no arrow to $b$, so every lifting square against $m$ either has an identity left side or does not exist. Thus $m$ is a strong monomorphism. It is not a regular monomorphism: $h$ equalizes $r,s$ but cannot factor through $m$, and an identical pair has $1_b$ as equalizer. Every strong nonregular monomorphism in any category occurs as the image of $m$ under a faithful functor from this category. Indeed, such a monomorphism is not epic, so choose distinct $r,s$ agreeing on it; failure of its equalizer property then supplies $h$.

For a morphism in an abelian category, its coimage $\mathrm{coker}(\ker f)$ is canonically isomorphic to its image $\ker (\mathrm{coker}f)$. This yields an epimorphism followed by a monomorphism, unique up to unique compatible isomorphism. Every epi-mono factorization has this same middle object.

For a small category $\mathcal{C}$, an object $P$ of $[\mathcal{C},\mathbf{Set}]$ is an irreducible projective when $\mathrm{Nat}(P,-)$ preserves coproduct in a category objects and epimorphisms. Preservation of coproducts means every map from $P$ into a coproduct factors uniquely through one summand. Such objects are precisely retracts of representable functors; when the indexing category $\mathcal{C}$ is a Cauchy-complete category, they are representable. To see the retract, split the canonical epimorphism from a coproduct of representables using projectivity, and use coproduct preservation to factor the splitting through one summand.

A map $e:X\to Y$ has this property if every commutative square with $e$ as its left side and a monomorphism $m:A\to B$ as its right side has a diagonal filler $t:Y\to A$. The filler is unique because $m$ is monic. This definition does not itself stipulate that $e$ is an epimorphism; binary products imply that conclusion by testing against the categorical diagonal.

Equivalently every open cover of $X$ contains $X$ as a member, so the top element of its open-set lattice is supercompact in the frame sense. Necessity follows from preservation of coproducts and epimorphisms by global sections. Conversely the union of all proper opens misses a point $x$, and the stalk functor there equals global sections. Its right adjoint is a skyscraper sheaf of sets. Empty spaces fail; a T1 space satisfies the condition exactly when it is a singleton.

A form of Beck's monadicity theorem compatible with equivalence-based monadic adjunctions is this. A functor $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, is a conservative functor, and $\mathcal{D}$ has coequalizers of all pairs whose images under $U$ admit a split coequalizer, with $U$ preserving these coequalizers. In the strict lifting formulation, one states creation of these coequalizers. Such a pair is a functor-split coequalizer pair; the splitting is required in the base category, not necessarily upstairs.

A split coequalizer fork $A\begin{array}{c}\stackrel{r}{\to }\\ \underset{s}{\to }\end{array}B\stackrel{q}{\to }Q$ has maps $u:B\to A$ and $t:Q\to B$ such thatEvery functor preserves this coequalizer: if $hr=hs$, then $h=hru=hsu=htq$, and $q$ is split epic, so the factor through $q$ is unique. The same equations remain true after applying any functor.

For necessity, let $U^T:{\mathcal{C}}^T\to \mathcal{C}$ be the monad algebra forgetful functor. It has the free-algebra left adjoint. It reflects isomorphisms, because if an algebra map $f$ is invertible in $\mathcal{C}$, the equation $fa=bTf$ rearranges to $f^{-1}b=aT(f^{-1})$.

If two algebra maps $r,s:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$, then $Tq$ and $T^{2}q$ are also coequalizers. Since $qb$ equalizes $Tr,Ts$, there is a unique $c:TQ\to Q$ satisfyingAfter composing with the epimorphism $q$, the equation $c{\eta }_Q=1_Q$ reduces to the unit law for $b$. After composing with the epimorphism $T^{2}q$, the equation $cT(c)=c{\mu }_Q$ reduces to $bT(b)=b{\mu }_B$. Thus $(Q,c)$ is an algebra and $q$ an algebra map. If an algebra map $h:(B,b)\to (Z,z)$ equalizes $r,s$, its unique underlying factor $k:Q\to Z$ satisfies $kc=zTk$: this follows by precomposing with the epimorphism $Tq$. Therefore this is a coequalizer upstairs as well. This proves that the monad algebra forgetful functor creates split coequalizers. The same properties transport through the comparison equivalence of categories, with the object-lifting convention distinguished in Question 9.

For sufficiency, suppose the three conditions hold, choose $F⊣U$, and let $T=UF$ with comparison $K$. For each algebra $(X,a)$, consider the pairIts image has coequalizer $a:TX\to X$. It is split: take $r={\mu }_X$, $s=T(a)$, $q=a$, $u={\eta }_{TX}$ and $t={\eta }_X$. The monad and algebra laws giveBy hypothesis the pair has a coequalizer $q:FX\to D_X$ and $Uq$ is also a coequalizer. Identify $U{D}_X$ with $X$ using the unique coequalizer isomorphism, so $Uq=a$. Since $q$ is carried by $K$ to an algebra map from the free algebra, the transported action $b:TX\to X$ satisfiesBut $T(a)$ is a split epimorphism, with right inverse $T({\eta }_X)$, so $b=a$. Hence $(X,a)\cong K(D_X)$: the comparison is essentially surjective.

Next, for every $D\in \mathcal{D}$, put $a_D=U{\epsilon }_D$. The counit of an adjunction coequalizesThis follows from naturality of the counit. The image pair has the split coequalizer $a_D$, as above. If $q:FUD\to Q$ is the hypothesized preserved coequalizer, the induced map $h:Q\to D$ has $Uh$ invertible because both $Uq$ and $a_D$ coequalize the same pair. Conservativity makes $h$ invertible. Therefore ${\epsilon }_D$ itself is a coequalizer.

Let $f:K(D)\to K(E)$ be an algebra map, so $f{a}_D=a_{E}Tf$. Set $g={\epsilon }_{E}F(f):FUD\to E$. Under the adjunction transpose bijection, the two composites of $g$ with the displayed presentation pair correspond respectively to $f{a}_D$ and $a_{E}Tf$. They are equal, so $g$ factors uniquely through ${\epsilon }_D$, giving $h:D\to E$ with $h{\epsilon }_D=g$. Applying $U$ yieldsSince $a_D$ is split epic, $Uh=f$. Thus $K$ is full. If $Uh=Uk$, the transposes of $h{\epsilon }_D$ and $k{\epsilon }_D$ are equal, so those composites agree. The counit is epic, hence $h=k$; thus $K$ is faithful.

The comparison is full, faithful and essentially surjective, so it is an equivalence of categories. This proves monadicity. The argument also explains the role of each hypothesis: split base presentations construct all algebras, and reflection of isomorphisms turns the counit presentations into actual coequalizers upstairs.

Let $a:X\to Y$ be a natural transformation of categorical presheaves. If every $a_C:X(C)\to Y(C)$ is a surjective function, and $r,s:Y\rightrightarrows Z$ satisfy $ra=sa$, then $r_C{a}_C=s_C{a}_C$ at every $C$. Surjectivity gives $r_C=s_C$ for all $C$, hence $r=s$. Thus $a$ is an epimorphism.

Conversely suppose $y\in Y(C)$ is outside the image of some $a_C$. Construct the pushout witness for failure of pointwise surjectivity explicitly: at each $D$ takeRestrictions send $[(z,k)]$ to $[(Y(f)z,k)]$; naturality of $a$ ensures that every generating identification is respected. Thus $P$ is a categorical presheaf, and the two inclusions $j_0,j_1:Y\to P$ are natural transformations with $j_{0}a=j_{1}a$.

All identifications concern the same element $z$ in its two copies, so an element outside the image is never identified with its other copy. Consequently $j_{0,C}(y)\ne j_{1,C}(y)$, proving $a$ is not an epimorphism. ThereforeThis is the pointwise epimorphism in a functor category criterion. It requires neither a componentwise choice of preimages nor a natural section.