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A functor is conservative when it reflects isomorphisms: if $F(f)$ is invertible then $f$ is invertible. It need not be full or faithful functor. The crude monadicity theorem uses this reflection property to lift invertibility of the comparison-adjunction counit from the base category.

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 $F$.

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 $h$.

A functor $F:\mathcal{C}\to \mathcal{D}$ is full when each function $\mathcal{C}(A,B)\to \mathcal{D}(FA,FB)$ is surjective. It is a faithful functor when each is injective. A full and faithful functor identifies its source, up to equivalence, with a full subcategory of the target. Fullness only concerns morphisms between image objects; it does not require every target object to lie in the image.

A balanced category is one in which every morphism that is both monic and epic is an isomorphism. If a faithful functor sends $f$ to an isomorphism, then $f$ is monic: equality $fg=fh$ becomes $FfFg=FfFh$, cancellation gives $Fg=Fh$, and faithfulness gives $g=h$. The same argument on the other side makes $f$ epic. Balance therefore makes $f$ an isomorphism. Thus a faithful functor out of a balanced category reflects isomorphisms.

Under the adjunction, the transpose of $Ff:FX\to FA$ is ${\eta }_{A}f:X\to GFA$. Hence if $F$ is faithful, ${\eta }_{A}f={\eta }_{A}g$ forces $Ff=Fg$ and then $f=g$, making every ${\eta }_A$ monic. Conversely, if the unit is pointwise monic and $Ff=Fg$, naturality gives ${\eta }_{B}f=GFf{\eta }_A=GFg{\eta }_A={\eta }_{B}g$, whence $f=g$. ThereforeWe also use the standard adjunction fact with its justification: $F$ is full and faithful exactly when $\eta$ is invertible. Indeed the composite hom-set map $\mathcal{C}(X,A)\to \mathcal{D}(FX,FA)\cong \mathcal{C}(X,GFA)$ is postcomposition by ${\eta }_A$. If $F$ is full and faithful these maps are bijective; taking $X=GFA$ produces a right inverse, and taking $X=A$ gives injectivity and therefore a left inverse. Conversely an invertible unit makes these hom-set maps bijective.

For the final example, view the two-element ordered set $0<1$ as a category $\mathcal{C}$, and let $\mathcal{D}$ have one object and only its identity. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to the functor $G$ selecting $1$, since every $\mathcal{C}(x,1)$ is a singleton. Every arrow in an ordered-set category is monic, so the unit arrows $x\to 1$ and the identity counit are monic. But $F$ is not full: $\mathcal{C}(1,0)$ is empty whereas $\mathcal{D}(F1,F0)$ is a singleton. The example does not contradict the upcoming equivalence because $\mathcal{C}$ is not balanced; its arrow $0\to 1$ is both monic and epic but not invertible.

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A strong monomorphism $m:A\to B$ is a monomorphism right orthogonal to all epimorphisms: whenever $e:X\to Y$ is epic and $a:X\to A$, $b:Y\to B$ satisfy $ma=be$, there is a unique $d:Y\to A$ with $de=a$ and $md=b$. A regular monomorphism is an equalizer of some parallel pair of morphisms. Its equalizer property supplies the indicated lift after epic cancellation, so every regular monomorphism is strong.

If every monomorphism is strong and $m:A\to B$ is also epic, apply strongness to the square whose left and right sides are both $m$, with horizontal arrows $1_A$ and $1_B$. The lift $d:B\to A$ satisfies $dm=1_A$ and $md=1_B$. Thus $m$ is invertible and the category is balanced.

Conversely, suppose the category is balanced and has pullbacks in a category. Given the lifting square above, form $P=A{\times }_{B}Y$ with projections $p:P\to Y$, $q:P\to A$. The pair $(a,e)$ gives $h:X\to P$ with $ph=e$, $qh=a$. The pullback of the monomorphism $m$ is monic, so $p$ is monic. It is also epic: $rp=sp$ implies $re=se$, hence $r=s$ because $e$ is epic. Balancedness makes $p$ invertible. Now $d=q{p}^{-1}$ satisfies both required equations, and monicity of $m$ proves uniqueness. Hence in a balanced category with pullbacks, every monomorphism is strong.

For the finite example, choose a strong monomorphism $m:A\to B$ that is not regular. It is not an isomorphism, since an isomorphism is the equalizer of an identical pair. It is not epic either, since an epic strong monomorphism is invertible by the first argument. Consequently there are distinct arrows $r,s:B\rightrightarrows D$ with $rm=sm$. Since $m$ is not the equalizer of this pair, there is a map $h:X\to B$ with $rh=sh$ that does not factor through $m$. Monicity rules out failure of uniqueness, so failure of existence is the only possible failure of its equalizer property.

Define ${\mathcal{C}}_0$ to have four formally distinct objects $a,b,x,d$ and exactly the following six nonidentity arrows:Their only nontrivial composites areThere are no longer composable strings of nonidentity arrows, so these rules define an associative category. Sending $a,b,x,d$ to $A,B,X,D$, respectively, and the six arrows to $m,h,r,s,rm,rh$, defines a functor. It is a faithful functor: every hom-set has at most one arrow except ${\mathcal{C}}_0(b,d)=\{r_0,s_0\}$, and their images are distinct. Faithfulness does not require the four image objects to be distinct.

The map $m_0$ is monic, by direct inspection of arrows into $a$. Neither $m_0$ nor $h_0$ is epic, since the distinct pair $r_0,s_0$ agrees after either. Every other nonidentity arrow has target $d$ and is epic, because there is only one arrow out of $d$. A lifting square against $m_0$ whose left side is one of these epimorphisms would require an arrow $d\to b$, which does not exist. Squares whose left side is an identity have the evident unique lift. Thus $m_0$ is a strong monomorphism.

It is not a regular monomorphism. The only distinct parallel pair out of $b$ is $r_0,s_0$, and $h_0$ equalizes it but has no factorization through $m_0$, since there is no arrow $x\to a$. An identical pair has $1_b$ as its equalizer and cannot have the noninvertible $m_0$ as its equalizer. Therefore

For the monad $(T,\eta ,\mu )$, an algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. A morphism of algebras for a monad $f:(A,a)\to (B,b)$ satisfies $fa=bTf$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad identities verify the algebra laws. For the forgetful functor $U$, the free adjunction iswith inverse $v\mapsto bTv$. The algebra law makes the latter an monad algebra morphism. The identities $h{\eta }_A\mapsto bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$ and $bTv{\eta }_A=b{\eta }_{B}v=v$ prove the bijection.

In the category of adjunctions inducing a fixed monad, objects are adjunctions $F⊣G$ with their induced monad identified with $T$. A morphism to $F^{\prime }⊣G^{\prime }$ is a functor $H$ between the right-hand categories satisfying $G^{\prime }H=G$, $HF=F^{\prime }$ and compatibility with units and counits. With these strict identifications, defineThe triangle identities give the unit algebra law; naturality of $\epsilon$ at ${\epsilon }_B$ gives the multiplication law. Naturality at $f$ makes $Gf$ an monad algebra morphism. Moreover $UK=G$, $KF=F^T$ because $G{\epsilon }_{FA}={\mu }_A$, and $K{\epsilon }_B$ is the free-adjunction counit at $K(B)$. Hence $K$ is a morphism into the Eilenberg-Moore adjunction.

For any other such $H$, its underlying object at $B$ must be $GB$. Counit compatibility forces its algebra action to be $G{\epsilon }_B$, and the forgetful functor, which is a faithful functor, forces $H(f)=Gf$. Thus $H=K$. This proves terminality of the Eilenberg-Moore adjunction. If adjunctions are specified only up to coherent isomorphisms, the same argument gives uniqueness up to the corresponding compatible natural isomorphism.

We prove full faithfulness from counit coequalizers. Let $B,C\in \mathcal{D}$ and let $\alpha :K(B)\to K(C)$ be an monad algebra morphism. ThusSet $u={\epsilon }_{C}F\alpha :FGB\to C$. Its composites with the two arrows of the printed presentation are equal: naturality of $\epsilon$ givesIn the last equality we used naturality at $F\alpha$. The coequalizer property therefore gives a unique $v:B\to C$ satisfying $v{\epsilon }_B=u$.

Apply $G$ to that identity. The algebra-morphism equation gives $GvG{\epsilon }_B=\alpha G{\epsilon }_B$. The triangle identity makes $G{\epsilon }_B$ a split epimorphism with section ${\eta }_{GB}$, so $Gv=\alpha$. This proves fullness of $K$.

If $v,w:B\to C$ have $Gv=Gw$, naturality gives $v{\epsilon }_B={\epsilon }_{C}FGv={\epsilon }_{C}FGw=w{\epsilon }_B$. The counit is epic since it is a coequalizer, so $v=w$. Thus $K$ is a faithful functor. Both parallel arrows matter: the converted TeX loses the second one, ${\epsilon }_{FGB}$, which is visible in the original PDF.

An equivalence of categories consists of functors $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ together with invertible natural transformations $GF\cong 1_{\mathcal{C}}$ and $FG\cong 1_{\mathcal{D}}$. A strict isomorphism of categories instead requires a functor with an inverse whose composites are literally identity functors.

First suppose $F$ belongs to an equivalence of categories. The isomorphisms $FGD\cong D$ give essential surjectivity. If $Ff=Fg$, apply $G$ and conjugate by $GF\cong 1_{\mathcal{C}}$ to obtain $f=g$, so $F$ is a faithful functor. Similarly $G$ is a faithful functor. Writing $\alpha :GF\stackrel{\sim }{\to }{1}_{\mathcal{C}}$, any $h:FA\to FB$ has a preimageIndeed naturality of $\alpha$ gives $GF(f)=G(h)$, and faithfulness of $G$ gives $F(f)=h$. Thus $F$ is a full and faithful functor.

Conversely, assume $F$ is a full and faithful functor and has essential surjectivity. Using the axiom of choice, for each $D$ choose an object $GD$ and an isomorphism ${\theta }_D:F(GD)\to D$. Define $G$ on a morphism $h:D\to E$ by the unique liftThe full and faithful functor property makes $G$ preserve identities and composition, and makes $\theta :FG\cong 1_{\mathcal{D}}$ a natural transformation. For $A\in \mathcal{C}$, lift ${\theta }_{FA}:FGFA\to FA$ uniquely to ${\alpha }_A:GFA\to A$. Lifting its inverse shows that ${\alpha }_A$ is invertible. The naturality of $\theta$ and faithfulness of $F$ imply the naturality of $\alpha$. This provesFor large categories, this choice argument is understood in a fixed universe, or with the corresponding class-choice convention; ordinary set-sized axiom of choice suffices for small categories.

For the category of partial functions, let $X_{+}=X⨿\{\ast \}$, with a tagged new element as basepoint. Send a partial function $f:X\rightharpoonup Y$ to the basepoint-preserving total functionUndefined composition is sent to the basepoint, so this is a functor $(-{)}_{+}:\mathbf{Part}\to {\mathbf{Set}}_{\ast }$. Restriction away from the basepoint recovers each partial function uniquely; hence it is a full and faithful functor. Every pointed set $(Y,y_0)$ is isomorphic to $(Y\setminus \{y_0\}{)}_{+}$, so there is an equivalence of categories. This particular equivalence can also be constructed explicitly, without choice, by deleting and adjoining the basepoint.

These actual categories are equivalent but not isomorphic. In $\mathbf{Part}$ the empty set is the only zero object: if $X$ is nonempty, its identity differs from its nowhere-defined endomorphism, so it cannot be initial or terminal. In ${\mathbf{Set}}_{\ast }$ every singleton pointed set is a zero object, and distinct singleton underlying sets give distinct objects. An isomorphism of categories is a bijection on objects preserving zero objects; it cannot take one such object onto several. This uses the categories of all actual sets, as in the paper, rather than chosen skeletal categories of representatives.

A skeletal category has no distinct isomorphic objects. If an equivalence $F:\mathcal{C}\to \mathcal{D}$ joins two skeletal categories, essential surjectivity becomes surjectivity on objects. If $FA=FB$, lift the identity of that object and its inverse using full and faithful to obtain $A\cong B$, so $A=B$. Thus $F$ is bijective on objects and on every hom-set. Its inverse on objects and morphisms is a strictly inverse functor, proving that it is an isomorphism of categories.

Under the axiom of choice, choose one object from each isomorphism class of a small category. The full subcategory on those objects is a skeleton of a category, and its inclusion is a full and faithful functor with essential surjectivity, hence an equivalence of categories.

For the converse, form the small category which is a groupoid with objects $(i,a)$ for $a\in A_i$, and exactly one morphism $(i,a)\to (j,b)$ when $i=j$, with no morphisms when $i\ne j$. Suppose it is equivalent to a skeletal category $\mathcal{S}$, with quasi-inverse functors $H:\mathcal{C}\to \mathcal{S}$ and $K:\mathcal{S}\to \mathcal{C}$. For each $i$, the objects $H(i,a)$ for $a\in A_i$ are isomorphic, hence all equal to a uniquely determined $s_i$. The isomorphism $KH(i,a)\cong (i,a)$ ensures that $K(s_i)=(i,a_i)$ for some $a_i\in A_i$. The rule $i\mapsto a_i$ is a choice function. In particular, no representative in $A_i$ had to be chosen to define $s_i$, since it is unique. Consequently

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A faithful functor reflects both of these cancellation properties: for instance, $fu=fv$ implies $FfFu=FfFv$, so monicity of $Ff$ and faithfulness imply $u=v$; the dual argument applies to epimorphisms. If $Ff$ is invertible, it is both monic and epic. Thus a faithful functor from a balanced category reflects isomorphisms.

For the adjunction $F⊣G$, the unit and counit of an adjunction satisfyIf $F$ is a faithful functor and ${\eta }_{A}u={\eta }_{A}v$, applying $F$ and composing with ${\epsilon }_{FA}$ gives $Fu=Fv$, hence $u=v$. Thus each ${\eta }_A$ is a monomorphism. Conversely, if every ${\eta }_A$ is monic and $Fu=Fv$ for $u,v:X\to A$, naturality givesso $u=v$. This proves the faithful left adjoint criterion.

Now assume both $\eta$ and $\epsilon$ are pointwise monic. The first triangle identity makes ${\epsilon }_{FA}$ a split epimorphism; since it is also a monomorphism, it is invertible and $F{\eta }_A$ is its inverse. The faithful left adjoint criterion says that $F$ is faithful, and the balanced category argument above then reflects the invertibility of $F{\eta }_A$ to that of ${\eta }_A$.

For completeness, an invertible unit makes $F$ a full and faithful functor. For $h:FA\to F{A}^{\prime }$, set $k={\eta }_{A^{\prime }}^{-1}G(h){\eta }_A$. The triangle identity gives $F{\eta }_A={\epsilon }_{FA}^{-1}$, so naturality of $\epsilon$ yieldsFaithfulness was already proved.

Let $q:FA\to B$ be a strong epimorphism, and putThen naturality gives ${\epsilon }_{B}t=q$. The square with left edge $q$, right edge the monomorphism ${\epsilon }_B$, top edge $t$, and bottom edge $1_B$ has a diagonal $s:B\to FGB$ by the defining lifting property of a strong epimorphism. Thus ${\epsilon }_{B}s=1_B$. A monic split epimorphism is invertible, so $B\cong FGB$. The essential image of $F$ is therefore closed under strong quotients.

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.