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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.

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.

The assertion that filtered colimits commute with finite limits in sets means that for every filtered category $\mathcal{J}$, every finite category $\mathcal{I}$, and every functor $D:\mathcal{I}\times \mathcal{J}\to \mathbf{Set}$, the canonical comparisonis a bijection. Surjectivity follows because an element on the right uses only finitely many representatives and compatibility equations, so filteredness moves all of them to one common stage. For injectivity, equality likewise consists of finitely many equalities, which become true at one later common stage.

The dual assertion fails in ${\mathbf{Set}}^{\mathrm{op}}$. Equivalently, cofiltered limits need not commute with finite colimits in sets. Index inverse systems by $n\in \mathbb{N}$, with arrows $n+1\to n$, and putusing the inclusions $A_{n+1}\hookrightarrow A_n$ and the unique maps $A_n\to B_n,C_n$. Since $A_n$ is nonempty, the pushout $B_n{\bigsqcup }_{A_n}{C}_n$ is a singleton at every stage, so its inverse limit is a singleton. Butand the pushout of the inverse limits is $1{\bigsqcup }_{\varnothing }1$, which has two elements. Thus a cofiltered limit fails to preserve this finite pushout.

Let $F$ be a presheaf of sets on a topological space $X$. The neighborhoods of $x$, ordered by reverse inclusion, form a filtered category, andThe first result therefore says that the stalk functor for presheaves of sets preserves finite limits.

For a set $S$, define a presheaf $R_{x}S$ bywith identity restrictions between neighborhoods of $x$ and the unique maps to $1$ otherwise. A natural transformation $F\to R_{x}S$ is exactly a compatible family of maps $F(U)\to S$ over neighborhoods of $x$, hence exactly a map $F_x\to S$. Thusso $R_x$ is right adjoint to the stalk functor.

Now restrict to sheaves. Suppose two morphisms $\alpha ,\beta :F\rightrightarrows G$ induce the same map on every stalk. For $s\in F(U)$ and every $x\in U$, equality of the two germs gives a neighborhood $V_x\subseteq U$ on which $\alpha (s)$ and $\beta (s)$ agree. The $V_x$ cover $U$, so the uniqueness axiom for $G$ gives $\alpha (s)=\beta (s)$. Therefore the joint stalk functor on sheaves of setsis faithful.

The presheaf $R_{x}S$ above is already a sheaf: an open containing $x$ has a cover member containing $x$, and compatibility forces one common element of $S$. Hence a right adjoint to $P$ sends a family $(S_x)$ to the product ${\prod }_x{R}_x{S}_x$, whose existence follows from the assumed closure of $\mathbf{Sh}(X)$ under limits.

Each stalk preserves finite limits, so $P$ preserves equalizers. If $P(f)$ is an isomorphism, its monicity and epicity are reflected by the faithful functor $P$; since $\mathbf{Sh}(X)$ is assumed balanced, $f$ is an isomorphism. Thus $P$ reflects isomorphisms. The Beck comonadicity theorem now applies: $P$ has a right adjoint, reflects isomorphisms, and preserves the required equalizers. Consequently $P$ is comonadic.

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.