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A natural transformation between functors $F,G:\mathcal{C}\to \mathcal{D}$ assigns a categorical morphism ${\alpha }_A:F(A)\to G(A)$ to every object of a category $A$, withThis naturality condition says that transforming the input before transporting it gives the same result as transporting it first and transforming the output. It constrains the whole family of arrows, so unrelated choices of a map for each object generally fail it.

For a concrete example, compare the functor that leaves every set and function unchanged with the power set functor using images of subsets. The maps ${\eta }_X:X\to \mathcal{P}(X)$ given by $x\mapsto \{x\}$ form a natural transformation: for any function $f:X\to Y$,The right-hand panel of the category illustration checks the square at $x=3$ with $f(x)=x+1$: both routes end at $\{4\}$. The statement works for every function, rather than depending on a preferred labeling of a particular set.

Map two disjoint walking arrows to the two mutually inverse arrows of the walking isomorphism. Their images generate the entire target, so the functor is an extremal epimorphism. Its kernel pair only identifies the matching endpoint objects, not the inverse-composition relations. Mapping both arrows to a nonidentity idempotent in a one-object category equalizes that kernel pair but cannot factor through the target groupoid. Thus the functor is not a regular epimorphism, although the category of small categories has finite limits.

A functor with natural maps $F_2:F(X\otimes Y)\to FX\otimes FY$ and $F_0:FI\to I$ satisfying the reversed monoidal functor associativity and unit diagrams. No invertibility is required.

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.

The two formulas for the adjunction transposes give both triangle identities for an adjunction directly. Apply the inverse bijection to the transpose of $1_{FA}$:Apply the forward bijection to the inverse transpose of $1_{GB}$:These are equalities of components of natural transformations, soNo cancellation assumption on either functor is needed: both identities follow from the inverse hom-set bijections.

Use the preservation obstructions to extending an adjoint chain: a functor having a further left adjoint must preserve categorical limits, while one having a further right adjoint must preserve colimits.

At the left end of the presheaf category chain, $\Lambda$ does not preserve the terminal object. The terminal categorical presheaf has singleton value everywhere, whereas $(\Lambda 1)(S)=\varnothing$ because $S\ne \varnothing$. Thus $\Lambda$ cannot itself be a right adjoint. At the right end, $\nabla$ does not preserve the initial object: $(\nabla \varnothing )(\varnothing )=\{\ast \}$ since the empty open set is proper. Thus $\nabla$ cannot itself be a left adjoint.

For the category of small categories chain, take two functors from the terminal category $1$ to the two-object indiscrete category $I\{0,1\}$, selecting its distinct objects. Their equalizer is the empty category. But their images under $C={\pi }_0$ are the same function between singleton sets, whose equalizer is a singleton. Hence $C$ does not preserve equalizers and has no further left adjoint. Finally, $I1⨿I1$ is the two-object discrete category, while $I(1⨿1)$ has arrows between its distinct objects. The canonical comparison is not an isomorphism of categories, so $I$ fails to preserve this coproduct and has no further right adjoint.

Let $C\mathcal{A}={\pi }_0\mathcal{A}$ be the set of connected components of a category: two objects of a category are identified when joined by a finite zigzag of morphisms. Let $DA$ be the discrete category on $A$, and let $IA$ be the indiscrete category on $A$, with exactly one morphism for every ordered pair of objects.

A functor $\mathcal{A}\to DA$ must give the same object value along every arrow, hence along every zigzag. Conversely a function ${\pi }_0\mathcal{A}\to A$ uniquely defines such a functor. A functor $DA\to \mathcal{A}$ is just a choice of an object of a category for each element of $A$. A functor $\mathcal{A}\to IA$ is likewise determined by an arbitrary object function, since its arrow images are forced and always compose correctly. ThereforeThis is the adjoint chain for the objects of a category:

Let $\hat{\mathcal{O}}=[\mathcal{O}(S{)}^{\mathrm{op}},\mathbf{Set}]$, with categorical presheaf restrictions $X(U)\to X(V)$ when $V\subseteq U$. The five functors areIn $\Delta A$ all restriction maps are identities. In $\Lambda A$ and $\nabla A$, all nonidentity restrictions are the unique maps permitted by the displayed sets. These definitions are valid also when $A$ is empty: a restriction out of $(\nabla A)(S)$ still has singleton target. Maps of sets act at the displayed $A$-components, and natural transformations act on $\Pi$ and $\Gamma$ by evaluation.

A natural transformation $\Lambda A\to X$ is determined by an arbitrary map $A\to X(\varnothing )$, since all other source components are empty. A natural transformation $X\to \Delta A$ is determined by its empty-open component $X(\varnothing )\to A$: the component at $U$ must first restrict $X(U)\to X(\varnothing )$. Similarly, a natural transformation $\Delta A\to X$ is determined by $A\to X(S)$, followed by the restriction $X(S)\to X(U)$. Finally, a natural transformation $X\to \nabla A$ is determined by $X(S)\to A$, since all proper-open targets are singletons. These descriptions give the hom-set bijectionsThus the five adjoints for constant presheaves on open sets areThe values at the empty open set are genuine presheaf values here; no sheaf condition or sheafification is being imposed.

Here a universal element means a pair $(A,u)$, with $u\in X(A)$, such that for every $B$ and every $x\in X(B)$ there is a unique morphism $f:B\to A$ satisfying $X(f)(u)=x$. Equivalently,By the Yoneda lemma these component maps constitute a natural transformation $h_A\to X$. If $u$ is universal, every component is bijective, so this is a natural isomorphism and $X$ is a representable presheaf. Conversely a natural isomorphism $h_A\to X$ takes $1_A$ to an element $u$ with exactly this property. In the category of elements of the contravariant functor $X$, $(A,u)$ is a terminal object: every $(B,x)$ has a unique arrow to it.

Let $\mathcal{C}$ be a locally small category, let $h_A=\mathcal{C}(-,A)$ be a representable presheaf, and let $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$. The Yoneda lemma gives a bijection, natural in $A$ and $X$,For $x\in X(A)$, define a natural transformation ${\alpha }^x:h_A\to X$ byFor $v:C\to B$, the functor composition law gives $X(v){\alpha }_B^x(f)=X(fv)(x)={\alpha }_C^x(fv)$, proving naturality. Conversely, naturality of any $\alpha :h_A\to X$ along $f:B\to A$ givesThus $\alpha$ is determined by its displayed element, and ${\alpha }_A^x(1_A)=x$ proves that the constructions are inverse.

A natural transformation $\theta :X\to Y$ sends the element to ${\theta }_A(x)$, which corresponds to $\theta {\alpha }^x$. A morphism $u:A^{\prime }\to A$ induces $h_u:h_{A^{\prime }}\to h_A$; precomposition by $h_u$ sends the element to $X(u)(x)$. This proves naturality in both variables and completes the Yoneda lemma proof.

Use the explicit construction of small limits from products and equalizers from part (i), here with finite indexing sets. If $F$ preserves finite products in a category and equalizers, then $F(P)$ and $F(Q)$, with their image projections, are the products of the objects $FD(j)$ and of the target objects indexed by arrows of $J$.

The image maps $F(a),F(b)$ have componentsAlso $F(e):F(L)\to F(P)$ is their equalizer. Therefore the same universal property proof makes $F(L)$, with projections $F(p_j)$, a categorical limit of $FD$. Any other chosen limit of $D$ is uniquely isomorphic to this construction, and the functor carries that isomorphism to an isomorphism. Consequently $F$ preserves every finite limit. Preservation of finite products includes the empty product, so the terminal object is also preserved.

Let $D:J\to [\mathcal{C},\mathcal{S}]$ be a small diagram in a category. Since $\mathcal{S}$ is a complete category, choose at each $C\in \mathcal{C}$ a categorical limitFor $f:C\to C^{\prime }$, the arrows $D_j(f)p_{j,C}$ form a cone over a diagram at $C^{\prime }$, because the diagram arrows $D_j\to D_k$ are natural transformations. There is therefore a unique arrow $L(f):L(C)\to L(C^{\prime })$ withThe universal property immediately gives $L(1_C)=1_{L(C)}$ and $L(gf)=L(g)L(f)$: both sides have the same composites with every limit projection. Thus $L$ is a functor, and the displayed equations make the $p_j$ natural transformations.

Given any categorical cone $q_j:X\to D_j$ in the functor category, its components induce unique $q_C:X(C)\to L(C)$. For $f:C\to C^{\prime }$, compare $L(f)q_C$ and $q_{C^{\prime }}X(f)$ after each $p_{j,C^{\prime }}$; naturality of $q_j$ makes their composites equal. Hence $q$ is a natural transformation, and uniqueness is componentwise. This proves that $L$ is a categorical limit in $[\mathcal{C},\mathcal{S}]$.

The functor category is complete, and every evaluation functor preserves limits, since evaluating the constructed limit at $C$ gives exactly the chosen limit in $\mathcal{S}$. This is the pointwise limits in a functor category construction; it also covers the empty diagram and its terminal object.

A Cartesian closed category has finite products in a category and, for all $X,Y$, an exponential object $Y^X$ with a natural bijectionEquivalently the functor $-\times X$ has a right adjoint. Its counit of an adjunction is the evaluation map of an exponential object $Y^X\times X\to Y$.

For the presheaf category, define the exponential in a presheaf category byFor $f:D\to C$, its restriction sends $\alpha$ to $\alpha \circ (h_f\times 1_X)$. Composition of these restriction maps follows from composition of the $h_f$, so this is a categorical presheaf. Define evaluation map of an exponential object byFor $f:D\to C$, naturality of $\alpha$ givesThus $e$ is a natural transformation.

Given a natural transformation $\beta :Z\times X\to Y$, define its currying byNaturality of $\beta$ proves that ${\hat{\beta }}_C(z)$ is a natural transformation $h_C\times X\to Y$. For $f:C^{\prime }\to C$, the same formula proves that $\hat{\beta }:Z\to Y^X$ is natural. Evaluating at $g=1_C$ gives $e(\hat{\beta }\times 1_X)=\beta$.

Conversely, if $\gamma :Z\to Y^X$ and $\beta =e(\gamma \times 1_X)$, thenby naturality of $\gamma$. Hence the two constructions are inverse and natural in $Z,Y$. Products are available pointwise, including the terminal singleton presheaf, so the presheaf category is Cartesian closed. The exponential uses natural transformations out of $h_C\times X$; simply exponentiating the individual values generally does not give this object.

For $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$, let $\int X$ be its category of elements. Its objects are $(A,a)$ with $a\in X(A)$; an arrow $(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $X(f)b=a$. This is a small category, because $\mathcal{C}$ is small and all values of $X$ are sets.

Send $(A,a)$ to the representable presheaf $h_A$, and send an arrow $f$ to postcomposition $h_f$. There is a cocone to $X$ whose map at $(A,a)$ isThe functor laws give its naturality and the required cocone compatibility.

Compute the colimit pointwise as a disjoint union modulo its diagram identifications. At $C$ a representative is a triple $(A,a,g:C\to A)$, and it maps to $X(g)a\in X(C)$. Every $x\in X(C)$ is the image of $(C,x,1_C)$. Moreover $g$ is an arrow $(C,X(g)a)\to (A,a)$ in $\int X$, so its diagram relation identifiesThus every representative is identified with the canonical representative of its image. Two representatives have equal images exactly when their canonical representatives agree. This proves a componentwise bijection; restriction maps respect it. The resulting natural isomorphism is the canonical colimit presentation of a presheaf: