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Eta conversion expresses extensionality for a lambda abstraction whose fresh argument is only passed to a function. The side condition prevents removal of a genuine dependency. In a categorical model with $D\cong D^D$, the inverse laws for currying and evaluation validate this conversion as well as beta reduction.

The evaluation map is the counit representing application in an exponential object. For every $Z$, composition with evaluation gives the natural bijection $\mathrm{Hom}(Z,Y^X)\cong \mathrm{Hom}(Z\times X,Y)$. Its inverse is currying. Evaluation must be a morphism in the ambient category, such as an equivariant map in a category of group actions.

For categorical presheaves $P,Q$ on a small category, restrictions in the displayed exponential object act by precomposition with $H_u\times 1_P$. Evaluation sends $(\alpha ,x)$ at $c$ to ${\alpha }_c(1_c,x)$. The currying of $t:R\times P\to Q$ sends $z\in R(c)$ to the natural transformation whose value on $(v:a\to c,x\in P(a))$ is $t_a(R(v)z,x)$. Evaluation and currying are inverse and natural, proving the Cartesian closed category structure.

Give $M\times X$ the diagonal right monoid action. On its equivariant maps into $Y$, define $(ek)(g,x)=e(kg,x)$. The evaluation map of an exponential object is $e(1,x)$ and currying of $h:Z\times X\to Y$ is $\hat{h}(z)(g,x)=h(zg,x)$. Both maps are equivariant, and they are inverse under evaluation. The left multiplication $kg$ in this right action is essential in a noncommutative monoid.

An object isomorphic to its own exponential supplies mutually inverse folding and unfolding maps. Evaluation defines application, and folding after currying defines abstraction. The exponential universal property gives the beta rule, while the full isomorphism gives eta conversion. A mere retraction of the function object generally suffices for beta but not automatically eta.

The Scott continuous maps from $D$ to $E$, ordered pointwise. A nonempty directed family has pointwise supremum, and the constant-bottom function is least. Joint evaluation is Scott continuous: for a directed family of pairs $(f_i,x_i)$, directedness makes diagonal values cofinal among all $f_i(x_j)$. Evaluation and currying make the category of pointed complete partial orders and non-strict Scott continuous maps a Cartesian closed category.

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 a locally small category $\mathcal{C}$, the covariant Yoneda lemma gives a bijectionnatural in $c$ and the functor $H:\mathcal{C}\to \mathbf{Set}$. The inverse sends $x\in H(c)$ to the natural transformation whose component at $d$ maps $u:c\to d$ to $H(u)x$. Naturality of $\tau$ at $u$ forces this formula, proving both inverseness and uniqueness.

Write $\mathcal{E}=\int F$ for the category of elements of $F$. Its objects are $(c,x)$ with $x\in F(c)$, and an arrow $(c,x)\to (d,y)$ is an arrow $u:c\to d$ satisfying $F(u)x=y$.

Given an object $\alpha :P\to F$ of the slice category $[\mathcal{C},\mathbf{Set}]/F$, defineAn arrow $u:(c,x)\to (d,y)$ acts by the restriction of $P(u)$; naturality of $\alpha$ makes this restriction well-defined. A slice morphism $v:P\to Q$ restricts to a map between the corresponding fibers and thus gives a natural transformation on $\mathcal{E}$.

Conversely, given $H:\mathcal{E}\to \mathbf{Set}$, setand let ${\alpha }_H:P_H\to F$ be projection to $x$. The functor laws and naturality follow from those for $F$ and $H$. On morphisms use the maps on each summand. Taking a fiber of this disjoint union recovers $H(c,x)$ canonically. Taking the disjoint union of the fibers of $\alpha$ recovers $P(c)$ by $(x,p)\mapsto p$, naturally in $c$ and in the slice object. These are inverse natural isomorphisms, provingThis is the slice of a set-valued functor category construction.

Now assume $\mathcal{C}$ is a small category, so $\mathcal{E}$ is small. Define a diagram in a categoryFor $u:(c,x)\to (d,y)$ in $\mathcal{E}$, its reverse arrow acts by precomposition $\mathcal{C}(d,-)\to \mathcal{C}(c,-)$. There is a cocone to $F$ with component $g\mapsto F(g)x$. To verify its colimit property directly, use pointwise colimits of sets. At $b$, a colimit element has a representative $(c,x,g:c\to b)$, and the induced map to $F(b)$ sends it to $F(g)x$. This map is surjective, since $z\in F(b)$ is represented by $(b,z,1_b)$. Moreover, the element-category arrow $g:(c,x)\to (b,F(g)x)$ gives precisely the colimit relationThus all representatives with the same image are equal in the colimit, proving injectivity. The resulting bijections are natural in $b$. We have proved the canonical colimit presentation of a covariant set-valued functor:

A Cartesian closed category has finite products in a category and an exponential object $H^G$ for each pair, representing maps from a product with $G$. In $[\mathcal{C},\mathbf{Set}]$, the terminal functor and binary products are computed pointwise. Define the exponential of covariant set-valued functors bySmallness makes this a set. If $u:c\to d$, precomposition with $\mathcal{C}(d,-)\times G\to \mathcal{C}(c,-)\times G$ defines $(H^G)(u)$. Functoriality is immediate from associativity of composition. Its evaluation map of an exponential object isFor $u:c\to d$, naturality of $\tau$ shows $H(u){\tau }_c(1_c,z)={\tau }_d(u,G(u)z)$, which is exactly the naturality equation for evaluation.

Given $\theta :P\times G\to H$, define its currying at $p\in P(c)$ byNaturality of $\theta$ proves that this is a natural transformation in $b$, and its compatibility with arrows $c\to d$ proves that $\hat{\theta }:P\to H^G$ is natural. Evaluation recovers $\theta$ by setting $f=1_c$. Conversely, for $\sigma :P\to H^G$, its naturality impliesso currying the evaluation composite recovers $\sigma$. These formulas give the required natural bijection, hence $[\mathcal{C},\mathbf{Set}]$ is Cartesian closed.

Finally, for every $F$ the small category $\mathcal{E}=\int F$ has a Cartesian closed functor category by the same construction. The slice equivalence above transports its terminal object, products and exponentials to $[\mathcal{C},\mathbf{Set}]/F$. The original functor category also has pointwise finite limits. Therefore it is locally Cartesian closed. Every construction here is explicit; no adjoint functor theorem is used.

Here a pointed complete partial order is a partially ordered set $D$ with a least element ${\perp }_D$ in which every nonempty directed set has a least upper bound. Directed means that every pair of its elements has an upper bound within that set. A Scott continuous map is monotone and preserves these directed suprema. It is not required to preserve the least element; restricting to strict maps would change the categorical claim. If completeness is instead formulated only for increasing countable chains, the same arguments below work with those chains.

Order the function space of complete partial orders $[D\to E]$ pointwise: $f\le g$ means $f(x)\le g(x)$ for every $x$. Its least element is the constant function with value ${\perp }_E$. For a directed family $\mathcal{F}$ of continuous maps, define $h(x)={\sup }_{f\in \mathcal{F}}f(x)$. This is well-defined and is its pointwise least upper bound. It is monotone. For any directed $A\subseteq D$,The interchange is valid because the combined family $\{f(a):f\in \mathcal{F},a\in A\}$ is directed, and either iterated supremum is its least upper bound. Thus $h$ is a Scott continuous map, proving that $[D\to E]$ is a pointed complete partial order.

Finite products carry the product order, componentwise suprema and least element $({\perp }_D,{\perp }_E)$. The one-element order is a terminal object. For the exponential, use the function space with evaluation $\mathrm{ev}(f,x)=f(x)$. To check joint continuity, take a directed family of pairs $(f_i,x_i)$. ThenIndeed, directedness supplies a $k$ above the pairs indexed by both $i$ and $j$, giving $f_i(x_j)\le f_k(x_k)$. Thus the diagonal values are cofinal in the double family, and evaluation preserves its supremum.

For a continuous $h:A\times D\to E$, its currying is $\Lambda h:A\to [D\to E]$, $(\Lambda h)(a)(d)=h(a,d)$. Each section is continuous because $d\mapsto (a,d)$ is; pointwise directed suprema in the $a$ coordinate show that $\Lambda h$ is continuous. Conversely a continuous $g:A\to [D\to E]$ gives the continuous map $\mathrm{ev}\circ (g\times {\mathrm{id}}_D)$. The two constructions are inverse, because they agree on every pair $(a,d)$. This gives the natural bijectionwith the required evaluation map. Together with the terminal object and products, it proves cartesian closure of pointed complete partial orders with non-strict continuous maps.

Choose mutually inverse maps $i:D^D\to D$ and $j:D\to D^D$. This is an extensional reflexive object in the Cartesian closed category. Define its application morphismFor endomorphisms $f,g:D\to D$, application in the underlying set $\mathrm{End}(D)$ is $f\cdot g=\mathrm{app}\circ ⟨f,g⟩$.

It is useful to define term interpretation uniformly over any context object $A$. An environment assigns each free variable $x$ a morphism $\rho (x):A\to D$. Variables are interpreted by their assigned maps, and application is interpreted by $\mathrm{app}$ and pairing. For lambda abstraction, lift the old environment along ${\pi }_A:A\times D\to A$, interpret the new variable by ${\pi }_D$, and let $h:A\times D\to D$ interpret the body. Then setTaking $A=D$ gives interpretations in the requested set of endomorphisms. Other context objects are needed in the recursive definition to handle bound variables correctly.

The beta reduction rule follows from $ji={\mathrm{id}}_{D^D}$ and the evaluation identity for currying. Applying the interpreted abstraction to a map $g:A\to D$ givesInduction on terms identifies this with interpretation of the capture-avoiding substitution of the argument for $x$.

For eta conversion, suppose $x$ is not free in $M$ and write $f:A\to D$ for the interpretation of $M$. The body $Mx$ on $A\times D$ is the uncurried form of $jf$. HenceThus beta-eta-equivalent terms have equal interpretations in $\mathrm{End}(D)$. The full isomorphism, rather than just a retract of the function object, provides the inverse law used in the eta step.

Write products of endomorphisms as composition, with the rightmost map applied first. Because $1$ is terminal, the only hom-sets that can contain multiple arrows are $\mathrm{Hom}(1,D)$ and $\mathrm{End}(D)$. If the latter contained only the identity, any point $p:1\to D$ would have $p!={\mathrm{id}}_D$, where $!:D\to 1$. Then any other point $q$ would satisfy $q=p!q=p$. Thus the category would be a preorder. The hypothesis therefore implies $|\mathrm{End}(D)|>1$.

If the product object $D\times D$ were $1$, its universal property applied to arrows from $D$ would give $\mathrm{End}(D{)}^2\cong \mathrm{Hom}(D,1)$, a singleton, which is impossible. Since there are only two objects, $D\times D=D$. Similarly, if $D^D=1$, the exponential adjunction at $1$ would giveagain impossible. Hence $D\times D=D^D=D$ as chosen objects.

Let $M=\mathrm{End}(D)$, let $\pi ,{\pi }^{\prime }$ be the product projections, and let $\epsilon :D^D\times D\to D$ be evaluation, all now endomorphisms. Pairing is $⟨-,-⟩$, and for $x:D\times D\to D$ write $x^{\ast }$ for its currying. The product identities and the two inverse exponential identities giveThis is the structure of a cartesian closed monoid.

For the converse, start only from the monoid and these identities. The first three make pairing a bijection from $M^2$ to $M$, inverse to $z\mapsto (\pi z,{\pi }^{\prime }z)$. In particularas follows by applying both projections and using uniqueness. Define $U(y)=\epsilon ⟨y\pi ,{\pi }^{\prime }⟩$. The last two identities say that $U$ and $(-{)}^{\ast }$ are inverse bijections. These facts also derive, rather than assume, the naturality lawIndeed $U(x^{\ast }z)=\epsilon ⟨x^{\ast }z\pi ,{\pi }^{\prime }⟩=U(x^{\ast })⟨z\pi ,{\pi }^{\prime }⟩=x⟨z\pi ,{\pi }^{\prime }⟩$; applying the inverse bijection proves the law.

Put $e=({\pi }^{\prime }{)}^{\ast }$. The law applied to $x={\pi }^{\prime }$ gives, for every $z\in M$,Taking $z=e$ proves $e^2=e$. This stronger absorption identity is what makes the split object terminal.

Assume first that $M$ is nontrivial; the degenerate case is handled below. Construct the required category using the two idempotents $d=1_M$ and $e$, as a two-object part of the Karoubi envelope. Call their objects $D$ and $1$. A morphism from the object with idempotent $a$ to the object with idempotent $b$ is an element $f\in M$ satisfying $bfa=f$. Composition is monoid multiplication, and the identity on each object is its idempotent. These laws are well-defined and associative. Its hom-sets are concretelyFor example, $ef=e$ for every $f$, so the condition for $D\to 1$ forces $f=e$. Thus $1$ is terminal, and the original monoid appears unchanged at $D$.

Set $D\times D=D$ with projections $\pi ,{\pi }^{\prime }$ and the given pairing. The product property works for arrows from either object: for global arrows $x,y$ satisfying $xe=x$, $ye=y$, pairing also satisfies $⟨x,y⟩e=⟨x,y⟩$. The original projection identities give existence and uniqueness. Products with $1$ are the other object, with its identity and the unique terminal arrow as projections; $1\times 1=1$.

For the exponential set $D^D=D$ with evaluation $\epsilon$. For a parameter object $D$, the given $(-{)}^{\ast }$ and $U$ already establish the universal property. We must additionally verify it for parameter object $1$. An arrow $f:1\times D=D\to D$ is an arbitrary element of $M$. Define its transpose byNaturality gives $\kappa (f)z=\kappa (f)$ for every $z$, because $f{\pi }^{\prime }⟨z\pi ,{\pi }^{\prime }⟩=f{\pi }^{\prime }$. In particular $\kappa (f)e=\kappa (f)$, so it really is an arrow $1\to D$.

Its uncurrying is $\epsilon ⟨\kappa (f),1_M⟩=f$. To see this, start from $U(\kappa (f))=f{\pi }^{\prime }$ and compose with $⟨e,1_M⟩$. Pairing naturality, ${\pi }^{\prime }⟨e,1_M⟩=1_M$, and the absorption property of $\kappa (f)$ give precisely the claimed equality. Conversely, if $g:1\to D$ satisfies $ge=g$, then $gz=gez=ge=g$ for every $z$. ThereforeThus transpose and uncurrying are inverse for the terminal parameter object too. This verifies the full exponential property, rather than only its endomorphism instance.

The remaining exponentials are $D^1=D$, $1^D=1$, and $1^1=1$, by the unit product law and terminality. Hence the constructed two-object category is cartesian closed and $\mathrm{End}(D)=M$.

For nontrivial $M$, the constructed $D$ is nonterminal because it has more than one endomorphism, and $e\ne 1_M$ by the absorption identity. This is the case corresponding to the forward non-preorder hypothesis.

If $M$ has one element, all the algebraic identities still hold, but splitting $e=1_M$ alone would make the original object terminal. To satisfy the literal converse's requirement of a nonterminal object, handle this case separately: take the two-object poset category $D<1$. Products are meets, $D^D=1$, $1^D=1$, $D^1=D$, and $1^1=1$, which verify the exponential adjunction on this two-element order. It is cartesian closed, $D$ is nonterminal, and its one-element endomorphism monoid is $M$. It is a preorder, so it does not recover the forward non-preorder setting or the self-exponential equality. Thus the converse's stated endomorphism realization works also in the degenerate case, while the stronger non-preorder reconstruction uses nontrivial $M$.

A Cartesian closed category has finite categorical products and, for every $P,Q$, an exponential object $Q^P$ with an evaluation morphism $\mathrm{ev}:Q^P\times P\to Q$ such that composition with evaluation gives natural bijectionsFinite categorical products in the presheaf category are pointwise; in particular its terminal object is the constant singleton categorical presheaf. For categorical presheaves $P,Q$ defineFor $u:d\to c$, restriction sends $\alpha$ to $\alpha \circ (H_u\times 1_P)$. Composition and identity laws follow from those of the Yoneda embedding, so this is a categorical presheaf.

Define evaluation at $c$ by ${\mathrm{ev}}_c(\alpha ,x)={\alpha }_c(1_c,x)$. For $u:d\to c$, naturality of $\alpha$ gives $Q(u){\alpha }_c(1_c,x)={\alpha }_d(u,P(u)x)$, which is the evaluation of the restricted pair. Hence evaluation is a natural transformation.

Given $t:R\times P\to Q$, its currying sends $z\in R(c)$ to the transformation whose component at $a$ isNaturality in $a$ follows from that of $t$, and naturality in $c$ follows by composing $v$ with the relevant arrow. Evaluating $\hat{t}$ at $(1_c,x)$ recovers $t_c(z,x)$. Conversely, if $s:R\to Q^P$, naturality of $s$ along $v:a\to c$ says that evaluating $s_a(R(v)z)$ at $(1_a,x)$ gives $(s_c(z){)}_a(v,x)$. Thus currying the evaluation of $s$ recovers $s$. These constructions are inverse and natural, proving the universal property. This is the exponential in a presheaf category, and proves the presheaf category is Cartesian closed.

Let $M\times X$ have the diagonal right monoid action $(g,x)m=(gm,xm)$, and put $E={\mathrm{Hom}}_M(M\times X,Y)$. Define the right action on this set of equivariant maps of monoid sets byIt really stays in $E$: $(ek)(gm,xm)=e(kgm,xm)=e(kg,x)m$. Also $e1=e$ and $((ek)l)(g,x)=e(klg,x)=(e(kl))(g,x)$, so it satisfies the right-action law. Left multiplication in the first argument is intentional; no commutativity of $M$ has been assumed.

Define the evaluation map of an exponential objectIt is equivariant, becauseFor any right $M$-set $Z$ and equivariant map $h:Z\times X\to Y$, define its currying byFor fixed $z$, this is equivariant in the diagonal variables: $h(zgm,xm)=h(zg,x)m$. The map $z\mapsto \hat{h}(z)$ is also equivariant, sinceIt satisfies $\mathrm{ev}(\hat{h}(z),x)=h(z,x)$.

Conversely, given an equivariant $u:Z\to E$, set $h(z,x)=u(z)(1,x)$. Evaluation makes this equivariant, and currying recovers $u$:These constructions are inverse and natural in $Z$. Therefore they establish the exponential object universal propertyThis is the exponential of right monoid actions, not the set of ordinary maps $X\to Y$ with an arbitrarily guessed action.