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An absolute monomorphism is a morphism sent to a monomorphism by every functor out of its category. The absolute monomorphisms are exactly the split monomorphisms.

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.

For $F⊣G$, the left adjoint $F$ is faithful exactly when every unit component ${\eta }_A:A\to GFA$ is a monomorphism. Under the adjunction, equality after ${\eta }_A$ corresponds precisely to equality after applying $F$.

In a regular category, the image of $f:A\to B$ is the monomorphism in its essentially unique factorizationas a regular epimorphism followed by a monomorphism. Images are stable under pullback.

A normal monomorphism is a monomorphism that is the kernel in a category of some morphism. In a pointed category with kernels and cokernels, a monomorphism is normal exactly when it is the kernel of its own cokernel.

For a locally small category $\mathcal{C}$, an object $A$, and a functor $F:\mathcal{C}\to \mathbf{Set}$, the covariant Yoneda lemma is the natural bijectionIts inverse sends $x\in F(A)$ to the natural transformation whose component at $B$ maps $f:A\to B$ to $F(f)(x)$.

Suppose now that $\mathcal{C}$ is a small category. For $F:\mathcal{C}\to \mathbf{Set}$, form the coproduct in a categoryThe Yoneda lemma associates to every summand the natural transformation determined by $x$, and these transformations combine to a map $p:P\to F$. At an object $B$, the element $y\in F(B)$ is the image of $1_B$ in the summand indexed by $(B,y)$, so $p$ is a pointwise epimorphism in a functor category. Each representable functor $\mathcal{C}(A,-)$ is a projective object in a category, sinceand evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so $P$ is the required projective object. This is the projective cover of a set-valued functor by representables.

We next prove the three equivalent conditions. If every morphism of $\mathcal{C}$ is a monomorphism, then for $f:B\to C$ and every $A$, postcompositionis injective. Thus every covariant representable functor is a monofunctor. Conversely, taking $A=B$ shows that injectivity for every representable implies that $fg=fh$ forces $g=h$, so every $f$ is monic.

If all representables are monofunctors, the object $P$ above is a monofunctor because a coproduct of injective functions is injective. Hence every $F$ is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable $R$. Choose an epimorphism $q:M\twoheadrightarrow R$ with $M$ a monofunctor. Since $R$ is projective, $1_R$ lifts to $s:R\to M$ with $qs=1_R$. Thus $R$ is a retract in a category of $M$. Every retract of a monofunctor is a monofunctor: if $R(f)x=R(f)y$, then injectivity of $M(f)$ applied to $s(x)$ and $s(y)$ gives $s(x)=s(y)$, and applying $q$ gives $x=y$. This completes the equivalence.

Finally, every functor $\mathcal{C}\to \mathbf{Set}$ is a monofunctor exactly when every morphism of $\mathcal{C}$ is a split monomorphism. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix $f:A\to B$ and form a quotient of $\mathcal{C}(A,-)\bigsqcup 1$ by identifying the distinguished point with every arrow of the form $kf:A\to X$, where $k:B\to X$. In the resulting functor $Q$, the two elements $[1_A]$ and $\ast$ of $Q(A)$ have equal images under $Q(f)$ because $[f]=\ast$. If every functor is a monofunctor, $Q(f)$ is injective, so $[1_A]=\ast$. By construction this means $1_A=rf$ for some $r:B\to A$. Thus $f$ is split monic. Equivalently, every morphism of $\mathcal{C}$ must be an absolute monomorphism.

For an adjoint functor pair $F:\mathcal{D}\rightleftarrows \mathcal{C}:G$, the unit and counit of an adjunction areUnder the adjunction bijection, ${\eta }_A$ corresponds to $1_{FA}$ and ${\epsilon }_B$ corresponds to $1_{GB}$. They satisfy the triangle identitiesConversely, natural transformations with these identities recover the adjunction through the mutually inverse maps

The fully faithful adjoint criterion gives the first equivalence directly. If $F$ is full and faithful, there is a unique $r_A:GFA\to A$ with $F(r_A)={\epsilon }_{FA}$; the triangle identity and faithfulness show that $r_A$ and ${\eta }_A$ are inverse, so the unit is an isomorphism. If the unit is an isomorphism, the displayed adjunction bijection shows thatis bijective, so $F$ is full and faithful. This also proves that either condition gives a natural isomorphism $GF\cong 1_{\mathcal{D}}$.

For the remaining direction, suppose merely that $GF$ is naturally isomorphic to the identity. Transport the monad induced by an adjunction along this isomorphism. Its underlying endofunctor is then the identity, its unit is a natural endomorphism $u:1\Rightarrow 1$, and its multiplication is a natural endomorphism $m:1\Rightarrow 1$ with $mu=1$. Naturality makes $u_A$ commute with $m_A$, so also $um=1$. Hence the transported unit, and therefore $\eta$, is an isomorphism. The three conditions are equivalent.

Now let $F⊣G⊣H$. If $F$ is full and faithful, then $GF\cong 1_{\mathcal{D}}$. For $A,B\in \mathcal{D}$, the two adjunctions give natural bijectionsThe Yoneda lemma therefore gives $GHA\cong A$, naturally in $A$. The fully faithful adjoint criterion applied to $G⊣H$ shows that $H$ is full and faithful. Conversely, if $H$ is full and faithful, then $GH\cong 1_{\mathcal{D}}$ andAnother application of the Yoneda lemma gives $GFA\cong A$, so $F$ is full and faithful. Thus $F$ is full and faithful exactly when $H$ is.

Assume henceforth that $G$ is full and faithful. Then the counit $\epsilon :FG\to 1_{\mathcal{C}}$ and unit $\alpha :1_{\mathcal{C}}\to HG$ are natural isomorphisms. ConsiderApplying the faithful functor $G$, then using naturality and the four triangle identities, turns both composites intoTherefore ${\theta }_1={\theta }_2$; denote their common value by $\theta$, the double-adjoint comparison transformation.

The pointwise monicity criterion is clearest from the following natural square, in which both vertical maps are bijections:The left vertical map is the adjunction $G⊣H$, while the right one precomposes with the isomorphism ${\epsilon }_B:FGB\to B$. Thus every ${\theta }_A$ is a monomorphism exactly when $F$ is faithful on all morphisms $GB\to A$, namely morphisms whose domains lie in the image of $G$.

Dually, the natural squarehas bijective vertical maps, using $F⊣G$ on the left and the isomorphism ${\alpha }_B:B\to HGB$ on the right. Hence every ${\theta }_A$ is an epimorphism exactly when $H$ is faithful on all morphisms $A\to GB$, namely morphisms whose codomains lie in the image of $G$.

A regular category has finite limits, every morphism factors through its image of a morphism in a regular category as a regular epimorphism followed by a monomorphism, and regular epimorphisms are stable under every pullback in a category. A cover is a strong epimorphism. Every regular epimorphism is strong: if $e$ is the coequalizer of $r,s$ and a square has $e$ on the left and a monomorphism $m$ on the right, monicity shows that the upper arrow coequalizes $r,s$. It therefore factors through $e$, and the epimorphism property of $e$ shows that this factor is the required diagonal. Conversely, factor a strong epimorphism $f$ as $f=me$ with $e$ regular epic and $m$ monic. The lifting property gives a two-sided inverse to $m$, so $m$ is an isomorphism and $f$ is regular epic. Thus regular epimorphisms and covers coincide.

Let $L:\mathcal{C}\to \mathcal{D}$ be the left-exact reflector and let $m:A^{\prime }\hookrightarrow A$. Since $L$ preserves finite limits, $Lm$ is monic. Define $c_A(A^{\prime })$ as the pullbackNaturality of the unit supplies a map $A^{\prime }\to c_A(A^{\prime })$ over $A$, proving $A^{\prime }\le c_A(A^{\prime })$. A factorization $A^{\prime }\le A^{\prime \prime }$ induces $L{A}^{\prime }\le L{A}^{\prime \prime }$ and therefore $c_A(A^{\prime })\le c_A(A^{\prime \prime })$, so $c_A$ is order-preserving.

Apply $L$ to the defining pullback. Left exactness and the fact that $L{\eta }_A:LA\to LLA$ is an isomorphism identify $L(c_A(A^{\prime }))$ with $L{A}^{\prime }$. Pulling back once more therefore givesFor a map $f:B\to A$, left exactness identifies $L(f^{\ast }{A}^{\prime })$ with $(Lf{)}^{\ast }(L{A}^{\prime })$. Pasting the two pullback squares then yieldsso this closure operation induced by a left-exact reflector commutes with pullback.

Assume $A$ lies in $\mathcal{D}$, so ${\eta }_A$ is an isomorphism. If $A^{\prime }$ also lies in $\mathcal{D}$, its unit is an isomorphism and the defining square gives $c_A(A^{\prime })\cong A^{\prime }$. Conversely, if $A^{\prime }$ is closed, that square expresses $A^{\prime }$ as a finite limit of $A$, $L{A}^{\prime }$, and $LA$, all fixed by $L$. Fixed objects of a left-exact reflective subcategory are closed under finite limits, so $A^{\prime }$ belongs to $\mathcal{D}$.

Finally suppose $\mathcal{C}$ is regular. The fixed objects have finite limits. For $f:A\to B$ in $\mathcal{D}$, factor it in $\mathcal{C}$ asApplying $L$ gives $A\stackrel{Le}{\to }LI\stackrel{Lm}{\to }B$. The map $Le$ is regular epic because a left adjoint preserves the coequalizer presenting $e$, and $Lm$ is monic because $L$ is left exact. Thus $\mathcal{D}$ has image factorizations. Their image subobject is the closure $c_B(I)$. A map in $\mathcal{D}$ is regular epic exactly when this closure is all of its codomain. Images in $\mathcal{C}$ commute with pullback, and the closure operation also commutes with pullback, so this condition is pullback-stable. Hence $\mathcal{D}$ is regular, as stated by the left-exact reflective subcategory of a regular category theorem.

A monomorphism is a left-cancellable morphism. A strong monomorphism $m:S\to A$ has the right lifting property against every epimorphism: from a commutative squarewith $e$ epic, one obtains $d:Y\to S$ satisfying $de=u$ and $md=v$. A regular monomorphism is an equalizer of a parallel pair.

Suppose $m:S\to A$ equalizes $r,s:A\rightrightarrows C$. In the square above,Since $e$ is epic, $rv=sv$, so the universal property of the equalizer gives the required $d$. Thus every regular monomorphism is strong.

Let $m_i:S_i\to A$ be strong for $i=1,2$, and let $P=S_1{\times }_A{S}_2$. Given a lifting square against $P\to A$, compose its top map with each projection. Strength of $m_i$ produces maps $d_i:Y\to S_i$ with $m_i{d}_i=v$. The pair $(d_1,d_2)$ has equal composites to $A$, hence induces $d:Y\to P$. Therefore intersections of strong subobjects are strong.

Call a morphism anodyne when it is both monic and epic, and call an object saturated when it is injective with respect to every such morphism. Let $m:S\to B$ be a strong subobject of a saturated object. Given an anodyne $e:X\to Y$ and $u:X\to S$, saturation of $B$ extends $mu$ to some $v:Y\to B$. Strength of $m$ applied to $mu=ve$ lifts $v$ to $d:Y\to S$ with $de=u$. Hence $S$ is saturated.

Now embed $A$ as a subobject of a saturated object $B$. Since the category is well-powered, the strong subobjects of $B$ through which $A\to B$ factors form a set; completeness supplies their intersection $j:R\to B$. The same coordinatewise lifting argument used for two factors shows that $j$ is strong, so $R$ is saturated.

The induced map $\eta :A\to R$ is monic. To prove it epic, let $f,g:R\rightrightarrows C$ satisfy $f\eta =g\eta$. Their equalizer $q:E\to R$ is regular and hence strong. Composites of strong monomorphisms are strong, so $jq:E\to B$ is a strong subobject containing $A$. Minimality of the intersection forces $R\to B$ to factor through $E$, which implies $f=g$. Thus $\eta$ is epic and therefore anodyne.

For every saturated $S$, each map $a:A\to S$ extends across $\eta$ to a map $\bar{a}:R\to S$. This extension is unique because $\eta$ is epic. Consequently $A\mapsto R$ is left adjoint to the inclusion of saturated objects: the full subcategory $\mathcal{S}$ is reflective. This is the saturated reflection from a strong-subobject intersection.

It remains to prove that $\mathcal{S}$ is balanced. First let $e:X\to Y$ be epic in $\mathcal{S}$. If $f,g:Y\rightrightarrows Z$ are morphisms in the ambient category, embed $Z$ into a saturated object $T$. Equality $fe=ge$ then implies equality after composing with $Z\to T$; epicity in the full subcategory gives equality there, and monicity of $Z\to T$ gives $f=g$. Thus $e$ is epic in the ambient category.

If $e:X\to Y$ is also monic in $\mathcal{S}$, it is monic in the ambient category as well. Indeed, for $u,v:W\rightrightarrows X$ with $eu=ev$, reflect $W$ by an anodyne map $W\to RW$. Saturation extends $u$ and $v$ to $\bar{u},\bar{v}:RW\rightrightarrows X$; ambient epicity of $W\to RW$ and monicity of $e$ inside $\mathcal{S}$ give $\bar{u}=\bar{v}$, hence $u=v$. Therefore $e$ is anodyne in the ambient category. Saturation of $X$ extends $1_X$ across $e$ to a retraction $r:Y\to X$. Since $e$ is epic, $ere=e$ implies $er=1_Y$, so $e$ is an isomorphism. Hence $\mathcal{S}$ is balanced.

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A faithful functor reflects monomorphisms and epimorphisms: cancellation after applying the functor can be pulled back by injectivity on hom-sets. Therefore, if $F:\mathcal{C}\to \mathcal{D}$ is faithful, $\mathcal{C}$ is balanced, and $F(f)$ is an isomorphism, then $f$ is both monic and epic and hence is an isomorphism. Thus $F$ reflects isomorphisms.

Now let $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ be an adjunction with unit $\eta$ and counit $\epsilon$. Under the adjunction bijectionthe morphism $Fu$ corresponds to ${\eta }_{A}u$. If $F$ is faithful and ${\eta }_{A}u={\eta }_{A}v$, then $Fu=Fv$ and hence $u=v$, so every ${\eta }_A$ is monic. Conversely, if every ${\eta }_B$ is monic and $Fu=Fv$ for $u,v:A\to B$, naturality givesand monicity gives $u=v$. This proves the faithful left adjoint criterion.

Assume next that $\mathcal{C}$ is balanced, every arrow in $\mathcal{D}$ factors as a strong epimorphism followed by a monomorphism, and both $\eta$ and $\epsilon$ are pointwise monic. The unit criterion makes $F$ faithful. The triangle identitymakes the monomorphism ${\epsilon }_{FA}$ a split epimorphism, hence an isomorphism. Thus $F({\eta }_A)$ is an isomorphism. The first paragraph shows that $F$ reflects isomorphisms, so ${\eta }_A$ is an isomorphism for every $A$. By the fully faithful adjoint criterion, $F$ is full and faithful.

To prove closure under strong quotients, let $q:FA\twoheadrightarrow B$ be a strong epimorphism. Naturality givesThe right side is a strong epimorphism, while ${\epsilon }_B$ is monic. The lifting property supplies $s:B\to FGB$ withSince ${\epsilon }_B$ is also monic, it is an isomorphism. Hence $B\cong F(GB)$ lies in the essential image of $F$.

Conversely, assume $F$ is full and faithful and its image is closed under strong quotients. Then $\eta$ is an isomorphism and in particular pointwise monic. Factor a counit component aswith $e$ strong epic and $m$ monic. Closure under strong quotients gives $C\cong F{A}^{\prime }$ for some $A^{\prime }$. After choosing this isomorphism, fullness writes $e=Fh$ for a map $h:GB\to A^{\prime }$. Since $e$ is epic and $F$ is faithful, $h$ is epic. The transpose ofis $1_{GB}$, soThus $h$ is also monic. Balancedness makes $h$ an isomorphism, hence $e$ is an isomorphism and ${\epsilon }_B=me$ is monic. This proves the pointwise-monic unit-and-counit criterion.

Balancedness is necessary. Let $\mathcal{C}$ be the two-element poset $0<1$, viewed as a category, and let $\mathcal{D}$ be the terminal category. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to the functor $G$ selecting $1$. Every morphism in a poset is monic, so the unit and counit are pointwise monic. But $F$ is not full: the unique arrow $F1\to F0$ has no preimage $1\to 0$. This is the pointwise-monic adjunction over a non-balanced poset.

Suppose $\mathcal{C}$ is balanced and every morphism of $\mathcal{D}$ factors as a strong epimorphism followed by a monomorphism. For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, both unit and counit are pointwise monic exactly when $F$ is full and faithful and its essential image is closed under strong quotients.

The coequalizer universal property supplies the diagonal filler against any monomorphism, so every regular epimorphism is strong. In a regular category, every strong epimorphism is regular: factor it as a regular epimorphism followed by a monomorphism and use the lifting property to make the monomorphism an isomorphism.

A strong epimorphism has the left lifting property with respect to every monomorphism. Thus every commutative square with the strong epimorphism on the left and a monomorphism on the right has a diagonal filler.

An object $S$ is subterminal when its unique morphism $S\to 1$ to the terminal object is a monomorphism. Equivalently, every object has at most one morphism to $S$. Subterminal objects form the preorder $\mathrm{Sub}(1)$.