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Let $T$ and $S$ be monads on categories $\mathcal{C}$ and $\mathcal{D}$, with forgetful functors $U_T,U_S$. Suppose $K:{\mathcal{D}}^S\to {\mathcal{C}}^T$ and $U:\mathcal{D}\to \mathcal{C}$ satisfy $U_{T}K=U{U}_S$. If $F⊣U$ and ${\mathcal{D}}^S$ has coequalizers of reflexive pairs, then $K$ has a left adjoint $L$. Thus the square must specify an actual lifting to Eilenberg-Moore categories; the required coequalizers are taken in the category of algebras.^[ref]

Writing $F_T,F_S$ for the free algebra functors, the construction starts withbecause the adjunctions give natural bijectionsAn algebra for a monad $(A,a)$ has the reflexive free-algebra presentation of a monad algebra $F_T(TA)\rightrightarrows F_{T}A\to (A,a)$. Transporting its two arrows through these bijections gives a reflexive pair between $F_{S}F(TA)$ and $F_{S}FA$; its coequalizer defines $L(A,a)$. The universal property supplies $L⊣K$.

This explains the theorem’s use in constructing adjoints of logical functors through power-object monadicity: the algebra presentation makes the missing adjoint explicit.

Let continuous maps $f,g:X\rightrightarrows Y$ between compact Hausdorff spaces have an underlying split coequalizer $q:Y\to Z$ with set maps $s,t$ satisfying $qs=1$, $ft=1$ and $gt=sq$. Its equivalence relation is the image of the closed subset $\{(x,x^{\prime })\in X^2:g(x)=g(x^{\prime })\}$ under $(f,f)$. One inclusion follows from $qf=qg$; for the other use $x=t(y)$ and $x^{\prime }=t(y^{\prime })$ whenever $q(y)=q(y^{\prime })$. This image is compact and hence closed in the Hausdorff product $Y^2$. The quotient is therefore compact Hausdorff, giving the preserved coequalizer required by the Beck monadicity theorem.

The dual universal family is a coend $Q$ with maps $i_C:T(C,C)\to Q$ satisfying $i_{C}T(f,1_C)=i_{D}T(1_D,f)$ on $T(D,C)$. Every compatible family to another object factors uniquely through $Q$. For small $\mathcal{C}$ in a cocomplete category, it is the coequalizer ${\coprod }_{f:C\to D}T(D,C)\rightrightarrows {\coprod }_{C}T(C,C)\to Q$.

For a reflexive pair $f,g:A\rightrightarrows B$ with common section $r$, form the pushout in a category of $f$ and $g$, with maps $j_1,j_2:B\to P$. The equation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. This common map is a coequalizer of $f,g$: every map equalizing them defines a commuting pair into the pushout and hence factors uniquely through $q$.

The cokernel of $f:A\to B$ is the universal morphism $q:B\to Q$ satisfying $qf=0$. It is the coequalizer of $f$ and the zero morphism.

For $F⊣G$ inducing $T=GF$, the Eilenberg-Moore comparison functor has a left adjoint exactly when every algebra $(A,a)$ has a coequalizer $q_a:FA\to L(A,a)$ of $Fa,{\epsilon }_{FA}:FTA\rightrightarrows FA$. Transposition identifies maps equalizing this pair with monad algebra morphisms $(A,a)\to K(D)$, proving both directions of the criterion. The unit is $\lambda =G{q}_a{\eta }_A$ and satisfies $G{q}_a=\lambda a$. In the base category, $a:TA\to A$ is a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$, split by ${\eta }_A$ and ${\eta }_{TA}$. Consequently $\lambda$ is invertible exactly when $G{q}_a$ is also a coequalizer. The unit is invertible everywhere precisely when $G$ preserves all these specified coequalizers; preservation of arbitrary coequalizers is not required.

For a finite diagram in a category $D:\mathcal{J}\to \mathcal{C}$, set $X={\coprod }_{j}D(j)$ and $Y={\coprod }_{a:i\to j}D(i)$. The two maps $s,t:Y\rightrightarrows X$ on the $a$-summand are the injection of $D(i)$ and the injection of $D(j)$ after $D(a)$. Their coequalizer is the colimit of $D$. The pair $[s,1_X],[t,1_X]:Y⨿X\rightrightarrows X$ is a reflexive pair with the same coequalizer. Thus finite coproduct in a category constructions and coequalizers of reflexive pairs suffice, including the empty coproduct for the empty diagram. Finite products cannot replace coproducts here: the poset with elements $0,a,b,u_0,u_1,\dots$, order $0<a,b<u_{n+1}<u_n$, and incomparable $a,b$ has finite meets and top $u_0$, hence finite categorical products. Every reflexive parallel pair is an equal pair and has its identity as coequalizer, but $a,b$ have no least upper bound and therefore no coproduct.

For algebras for a monad $(A,\alpha ),(B,\beta )$ and underlying binary coproducts in a category, put $\kappa =[T{\nu }_1,T{\nu }_2]:TA+TB\to T(A+B)$. If the algebra pair $F(\alpha +\beta ),{\mu }_{A+B}F\kappa :F(TA+TB)\rightrightarrows F(A+B)$ has a coequalizer, that coequalizer is their algebra coproduct. An algebra map from $F(A+B)$ transposes to $[a,b]:A+B\to C$; equalizing the pair is exactly the two algebra-morphism equations for $a,b$. The pair is reflexive through $F({\eta }_A+{\eta }_B)$.

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.

If each standard counit presentation is a coequalizer, the Eilenberg-Moore comparison functor is full and faithful. An monad algebra morphism $\alpha :GB\to GC$ makes ${\epsilon }_{C}F\alpha$ equalize $FG{\epsilon }_B$ and ${\epsilon }_{FGB}$; its unique descent is the required morphism $B\to C$. The triangle identities for an adjunction make $G{\epsilon }_B$ a split epimorphism, so descent has underlying arrow exactly $\alpha$. Epimorphic cancellation at ${\epsilon }_B$ proves faithfulness.

If $e=ir$ and $ri=1$, then $r$ is the coequalizer of $(e,1)$: an arrow $h$ with $he=h$ factors uniquely as $(hi)r$. Conversely a coequalizer $q$ makes $e$ factor as $iq=e$; epimorphic cancellation gives $qi=1$. Thus the splitting of an idempotent morphism is equivalent to this particular coequalizer, without assuming arbitrary coequalizers exist.

For the nested partial unary operation category tower, each one-step forgetful functor is a monadic adjunction right adjoint: it reflects isomorphisms and creates its split coequalizers. The Beck monadicity theorem identifies the first Eilenberg-Moore category with the next category in the tower. A long composite has that same first monad because free extension of nested partial unary operations leaves no new top fixed points. Repeating the comparison forgets one fewer operation at each step. The remaining forgetful functor is not full until no operation remains to forget, giving monadic length $m-n$.

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.

For $(A,a)$, consider the reflexive free-algebra presentation of a monad algebraBoth parallel morphisms are monad algebra morphisms: ${\mu }_A$ satisfies the monad associativity law, while $Ta$ is the image under $F^T$ of $a$. The map $a$ is a monad algebra morphism because $a{\mu }_A=aTa$, and the same equation says it equalizes the pair.

Suppose $h:F^T(A)\to (B,b)$ is a monad algebra morphism with $h{\mu }_A=hTa$. Put $k=h{\eta }_A:A\to B$. Naturality of $\eta$ and the unit law giveMoreover, since $h$ is a monad algebra morphism,Thus $k$ is a monad algebra morphism and factors $h$ through $a$. If $k^{\prime }a=h$, then $k^{\prime }=k^{\prime }a{\eta }_A=h{\eta }_A=k$, proving uniqueness. Therefore every monad algebra is the stated coequalizer of free algebras.

The underlying fork is a split coequalizer: take $q=a$, $s={\eta }_A$ and $t={\eta }_{TA}$. Then $qs=1_A$, ${\mu }_{A}t=1_{TA}$, and $(Ta)t=sa$. Its splitting maps need not be monad algebra morphisms; the preceding argument separately proves the coequalizer in ${\mathcal{C}}^T$.

Write the adjunction unit and adjunction counit of $F⊣G$ as $\eta ,\epsilon$. Its induced monad is $T=GF$, with multiplication $\mu =G{\epsilon }_F$. The Eilenberg-Moore comparison functor isThe unit law for $a_d$ is $G{\epsilon }_d{\eta }_{Gd}=1_{Gd}$, a triangle identity for an adjunction. Naturality of $\epsilon$ at ${\epsilon }_d$ gives ${\epsilon }_{d}FG{\epsilon }_d={\epsilon }_d{\epsilon }_{FGd}$; applying $G$ gives $a_{d}T{a}_d=a_d{\mu }_{Gd}$. Naturality at $h$ gives the monad algebra morphism equation for $Gh$. Thus $K$ is well-defined.

For each $d$, consider the fork in $\mathcal{D}$It commutes by the naturality equation just used. Its image under $G$ is the fork $T^2(Gd)\rightrightarrows T(Gd)\to Gd$ of the preceding question, with parallel morphisms ${\mu }_{Gd},T{a}_d$ and final map $a_d$. It is split using $s={\eta }_{Gd}$ and $t={\eta }_{T(Gd)}$. By the assumed reflection, ${\epsilon }_d$ is the coequalizer of this pair in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ for $h,k:d\to e$, naturality gives $h{\epsilon }_d={\epsilon }_{e}FGh={\epsilon }_{e}FGk=k{\epsilon }_d$. A coequalizer is an epimorphism, so $h=k$.

For fullness, let $u:K(d)\to K(e)$ be a monad algebra morphism. Thus $u:Gd\to Ge$ satisfies $u{a}_d=a_{e}Tu$. Set $v={\epsilon }_{e}Fu:FGd\to e$. We verify explicitly that $v$ equalizes the two morphisms of the displayed fork. The monad algebra morphism equation givesNaturality of $\epsilon$ first at $Fu$ and then at ${\epsilon }_e$ givesThe coequalizer therefore supplies a unique $h:d\to e$ with $h{\epsilon }_d=v$. Applying $G$ yields $Gh{a}_d=Gv=a_{e}Tu=u{a}_d$. The map $a_d$ is split epic with section ${\eta }_{Gd}$, so cancellation gives $Gh=u$. This proves fullness, and the earlier cancellation proves faithfulness. HenceNo claim that $K$ is essentially surjective is needed; that stronger conclusion needs the corresponding existence hypotheses.

For $G:\mathcal{D}\to \mathcal{C}$, a functor-split coequalizer pair is a parallel pair $f,g:Y\rightrightarrows Z$ such that its image admits a split coequalizer in $\mathcal{C}$. In one orientation this means morphisms $q:GZ\to Q$, $s:Q\to GZ$ and $t:GZ\to GY$ satisfyThese equations prove that $q$ is a coequalizer: if $hGf=hGg$, then $h=hGft=hGgt=hsq$, so $hs$ factors $h$; uniqueness follows from $qs=1$. They remain true under every functor, making a split coequalizer an absolute colimit.

To say that $G$ reflects such coequalizers means that any fork $Y\rightrightarrows Z\stackrel{p}{\to }W$ in $\mathcal{D}$ whose image is a split coequalizer is itself a coequalizer in $\mathcal{D}$. This is a reflection assertion about an existing fork; it does not by itself require lifting every split fork from $\mathcal{C}$, or creating its splitting maps in $\mathcal{D}$.

Let $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$. A categorical end is an object $E$ with morphisms $e_A:E\to T(A,A)$ satisfying, for every $f:A\to B$,It is universal among such families: any $x_A:X\to T(A,A)$ satisfying the same equations is uniquely $e_{A}h$ for one $h:X\to E$. This is a universal dinatural transformation from the constant value $E$, and is written $E={\int }_{A}T(A,A)$.

Dually a categorical coend is $Q$ with morphisms $j_A:T(A,A)\to Q$ satisfyingAny family $x_A:T(A,A)\to X$ satisfying these equations is uniquely $h{j}_A$ for one $h:Q\to X$. Write $Q={\int }^{A}T(A,A)$. For small $\mathcal{C}$, the categorical end can be constructed as the equalizer of ${\prod }_{A}T(A,A)\rightrightarrows {\prod }_{f:A\to B}T(A,B)$ when these objects exist. The categorical coend is the coequalizer of ${\coprod }_{f:A\to B}T(B,A)\rightrightarrows {\coprod }_{A}T(A,A)$ when those objects exist. The two morphisms impose exactly the displayed equations for a dinatural transformation.

A regular category has finite limits, factorizations of every arrow as a regular epimorphism followed by a monomorphism, and pullback-stable regular epimorphisms. Equivalently one may require coequalizers of kernel pairs and pullback stability of the resulting quotients. A regular epimorphism is a coequalizer of some parallel pair; an extremal epimorphism is an epimorphism $e$ such that $e=mh$, with $m$ monic, forces $m$ to be invertible.

First, every regular epimorphism is extremal without assuming regularity of the whole category. Suppose $e:A\to B$ coequalizes $r,s$ and $e=mh$. Since $m$ is monic, $hr=hs$, so the universal property gives $k:B\to \mathrm{dom}(m)$ with $ke=h$. Then $mke=e$; cancellation of the epimorphism $e$ gives $mk=1_B$. Monicity of $m$ now gives $km=1$, so $m$ is an isomorphism. Conversely, in a regular category, factor an extremal $e$ as $mq$ with $q$ regular epic and $m$ monic. Extremality makes $m$ invertible, so $e$ is regular epic. Thus covers and regular epimorphisms coincide.

For a counterexample, work in the category of small categories, which has finite limits constructed on objects and arrows. Let $C$ be two disjoint walking arrows $a\stackrel{u}{\to }b$ and $c\stackrel{v}{\to }d$. Let $D$ be the walking isomorphism, with two objects $0,1$ and inverse arrows $p:0\to 1$, $q:1\to 0$. Define $F:C\to D$ by $a,d\mapsto 0$, $b,c\mapsto 1$, $u\mapsto p$, $v\mapsto q$. These images generate all of $D$, so two functors agreeing after $F$ agree everywhere; thus $F$ is epic.

A monic functor is injective on objects and faithful, as tested by functors from the terminal category and from the walking arrow. If $F$ factors through such a functor $m:D^{\prime }\to D$, its image contains both objects and both arrows $p,q$. Faithfulness forces their lifted composites to be the identities, so the image contains every arrow of $D$ and $m$ is an isomorphism. Hence $F$ is extremal.

It is not regular. Its kernel pair identifies the endpoint pairs $a,d$ and $b,c$ but has no relation forcing the composites of the images of $u,v$ to be identities. Concretely, send all four objects to the one-object category with endomorphism monoid $\{1,e\}$, $e^2=e\ne 1$, and send both $u,v$ to $e$. This functor equalizes the kernel pair: corresponding identities agree, and the only nonidentity arrow pairs in the kernel pair are an arrow paired with itself. It cannot factor through $D$, because images of $p,q$ would have to be inverse, whereas $e^2\ne 1$. In a category with kernel pairs, every regular epimorphism is the coequalizer of its kernel pair: its original coequalized pair factors through that kernel pair, giving the universal property. Therefore $F$ is an extremal but nonregular epimorphism, the promised nonregular extremal epimorphism in the category of categories.

Put $P(A)={\Omega }^A$, the power object, with $P(f)$ acting by inverse image of predicates. Transposing a predicate on $A\times B$ gives natural bijectionsThus $P^{\mathrm{op}}:\mathcal{E}\to {\mathcal{E}}^{\mathrm{op}}$ is left adjoint to $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$. Its unit ${\eta }_A:A\to PPA$ sends $a$ to evaluation at $a$, namely the predicate $S\mapsto (a\in S)$. This unit is monic: two generalized elements having the same evaluation on all predicates agree by testing the singleton predicate of the first, or equivalently its graph in the relevant parameter product.

We verify a sufficient form of Beck's monadicity theorem: a right adjoint is monadic if it reflects isomorphisms and its domain has, and it preserves, coequalizers of reflexive pairs. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality of the monic unit shows that $f$ is monic. Regard it as the subobject $A\hookrightarrow B$. Its characteristic predicate and the constantly true predicate have the same restriction to $A$, so injectivity of $Pf$ makes them equal. Thus this subobject is all of $B$, and $f$ is invertible. Hence $P$ reflects isomorphisms.

A reflexive pair in ${\mathcal{E}}^{\mathrm{op}}$ is a coreflexive pair $f,g:B\rightrightarrows A$ in $\mathcal{E}$, with a common retraction $r$ such that $rf=rg=1_B$. Its equalizer $e:E\hookrightarrow B$ exists. We show that $Pe:PB\to PE$ is the coequalizer of $Pf,Pg:PA\rightrightarrows PB$. Direct image of subobjects along the mono $f$ defines ${\exists }_f:PB\to PA$, and similarly ${\exists }_e:PE\to PB$. At every parameter object the identities areFor the middle identity, an equality $f(b)=g(b^{\prime })$ forces $b=b^{\prime }$ after applying $r$; their common point must therefore lie in the equalizer. This proves the identity for parameterized predicates, not merely for global elements. If $h:PB\to Z$ satisfies $hPf=hPg$, composing with ${\exists }_f$ gives $h=h{\exists }_{e}Pe$. Thus it factors through $Pe$, and the last displayed identity gives uniqueness. This proves that power objects turn coreflexive equalizers into coequalizers. All hypotheses of the monadicity criterion have now been checked, so $P$ is monadic, with

A logical functor $L:\mathcal{E}\to \mathcal{F}$ preserves finite limits, exponentials and the classifier, hence gives compatible natural isomorphisms $LP\cong PL$ and $LPP\cong PPL$. They respect units and multiplications, because these are defined by evaluation and transposition. Under the displayed equivalences, $L^{\mathrm{op}}$ is therefore the induced functor on double-power Eilenberg-Moore categories. Their forgetful functors create finite limits, and the underlying $L$ preserves them. Consequently the induced functor preserves finite limits, so $L^{\mathrm{op}}$ does too. Reversing arrows proves every logical functor preserves finite colimits.

For the first counterexample, the constant-terminal functor $F:\mathbf{Set}\to \mathbf{Set}$ has the constant-empty functor as left adjoint: both sets of maps in the adjunction are singletons. It preserves exponentials, since $F(B^A)=1=1^1=(FB{)}^{FA}$ with the canonical comparison. It does not preserve the initial object, because $F(0)=1$. Thus preservation of exponentials alone is insufficient.

For the second, take the fixed-point functor $\Gamma :G\text{-}\mathbf{Set}\to \mathbf{Set}$ for $G=\mathbb{Z}/2$. Its left adjoint gives a set the trivial action. The classifier of $G$-sets is the two-element set with trivial action, so $\Gamma$ preserves it and its truth map. Let $X$ be the free two-element $G$-set. The coequalizer of its identity and its swapping automorphism is the singleton $G$-set. Applying $\Gamma$ gives two maps $0\rightrightarrows 0$ followed by $1$, whereas their coequalizer in sets is $0$. Thus $\Gamma$ preserves the classifier but not all finite colimits. These examples also explain why both requirements in logicalness matter.

Let $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$. An end of a functor is an object $E$ with maps $e_C:E\to T(C,C)$ satisfying $T(1_C,f)e_C=T(f,1_D)e_D$ for every $f:C\to D$, universal among all such families. Thus every compatible family with vertex $A$ factors uniquely through $E$. Write it as ${\int }_{C}T(C,C)$. These are the wedge equations for a dinatural transformation from a constant functor to $T$.

Dually, a coend of a functor is an object $Q$ with maps $i_C:T(C,C)\to Q$ satisfying $i_{C}T(f,1_C)=i_{D}T(1_D,f)$ as maps from $T(D,C)$, universal among compatible families with a common target. Write it as ${\int }^{C}T(C,C)$. These are cowedge equations, and every cowedge to $A$ factors uniquely through $Q$.

For small $\mathcal{C}$, if the needed products and equalizers exist, the end is the equalizer of ${\prod }_{C}T(C,C)\rightrightarrows {\prod }_{f:C\to D}T(C,D)$. If the needed coproducts and coequalizers exist, the coend is the coequalizer of ${\coprod }_{f:C\to D}T(D,C)\rightrightarrows {\coprod }_{C}T(C,C)$. The two arrows in each construction are precisely the two sides of the compatibility equation. This explains both the mixed variance and the universal property, rather than treating the integral notation as an ordinary numerical integral.