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Restricting the functor of points to affine schemes gives a fully faithful functor from schemes to covariant set-valued functors on commutative rings. Given a natural transformation, apply it to the inclusions of an affine open cover of $X$. Naturality on affine open subschemes of overlaps makes the resulting maps to $Y$ agree, so they glue. For an arbitrary map from an affine scheme to $X$, cover its inverse images of these charts by affine open subschemes and apply naturality again. This proves that the glued map induces the prescribed natural transformation, and uniqueness follows from the original cover. In particular, the restricted functor of points determines the scheme up to an isomorphism of schemes.

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.

A morphism of schemes $f:X\to Y$ defines a natural transformation $g_f:g_X\to g_Y$ by $u\mapsto f\circ u$. We construct the inverse correspondence.

Let $\alpha :g_X\to g_Y$ be a natural transformation, and choose an affine open cover $(j_i:U_i\hookrightarrow X)$. Regard $j_i$ as an element of $g_X(U_i)$ and setHere and below the equivalence of categories in the question lets us write the restricted functor of points on affine schemes, contravariantly, rather than on commutative rings, covariantly. If $W\subseteq U_i\cap U_j$ is an affine open subscheme, naturality for its two inclusions givesSuch $W$ cover the overlap even when that overlap is not an affine scheme. Consequently the $f_i$ agree and glue to a morphism of schemes $f:X\to Y$.

It remains to check all elements of the restricted functor of points, not only the chosen inclusions. Given an affine scheme $T$ and $u:T\to X$, cover each $u^{-1}(U_i)$ by affine open subschemes $V$. On such a $V$, factor $u{|}_V=j_i\circ v$. Applying naturality first to $v:V\to U_i$ and then to $V\hookrightarrow T$ givesThese local equalities prove ${\alpha }_T(u)=f\circ u$. The glued $f$ is unique, since its restrictions to all $U_i$ are prescribed. Henceso the assignment is fully faithful.

If $g_X\cong g_Y$, the natural isomorphism and its inverse correspond to morphisms of schemes $f:X\to Y$ and $h:Y\to X$. Their compositions induce the identity natural transformations, and faithfulness implies $hf={\mathrm{id}}_X$, $fh={\mathrm{id}}_Y$. Conversely, an isomorphism of schemes induces a natural isomorphism. Thus

Define maps between the hom-sets byUsing naturality of the proposed counit of an adjunction and then the first triangle identity for an adjunction,Using naturality of the proposed unit of an adjunction and then the other triangle identity for an adjunction,Hence $\Phi$ and $\Psi$ are inverse bijections. They are natural in both variables: for $a:A^{\prime }\to A$ and $b:B\to B^{\prime }$, the unit of an adjunction identity givesThis constructs an adjunction $F⊣G$. Taking $f=1_{FA}$ and $g=1_{GB}$ recovers the prescribed unit of an adjunction and counit of an adjunction. Conversely, part (i) shows that any adjunction with those components must have exactly these transpose formulas. The adjunction exists and is unique with the prescribed unit and counit.

Write the adjunction as the natural bijectionThe unit of an adjunction and counit of an adjunction are the natural transformations whose components areNaturality of the hom-set bijections gives, for $f:FA\to B$ and $g:A\to GB$,For example, postcomposition by $f$ in the first hom-set corresponds to postcomposition by $Gf$ in the second, giving the first formula; naturality in $A$ gives the second. The same naturality says that, for $a:A\to A^{\prime }$ and $b:B\to B^{\prime }$,Thus these components do define the asserted natural transformations $1_{\mathcal{C}}\to GF$ and $FG\to 1_{\mathcal{D}}$.

An isomorphism $f:A\to B$ gives a natural isomorphism $h_f:h_A\to h_B$ by postcomposition, with inverse $h_{f^{-1}}$.

Conversely suppose $\alpha :h_A\to h_B$ is a natural isomorphism, with inverse $\beta$. Since the Yoneda embedding is a full and faithful functor, there are unique $f:A\to B$ and $g:B\to A$ with $h_f=\alpha$ and $h_g=\beta$. Their composites satisfyFaithfulness gives $gf=1_A$ and $fg=1_B$. ThusThe naturality requirement ensures that all incoming morphisms are respected; unrelated componentwise bijections alone would not justify the conclusion.

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.

Let $a:X\to Y$ be a natural transformation of categorical presheaves. If every $a_C:X(C)\to Y(C)$ is a surjective function, and $r,s:Y\rightrightarrows Z$ satisfy $ra=sa$, then $r_C{a}_C=s_C{a}_C$ at every $C$. Surjectivity gives $r_C=s_C$ for all $C$, hence $r=s$. Thus $a$ is an epimorphism.

Conversely suppose $y\in Y(C)$ is outside the image of some $a_C$. Construct the pushout witness for failure of pointwise surjectivity explicitly: at each $D$ takeRestrictions send $[(z,k)]$ to $[(Y(f)z,k)]$; naturality of $a$ ensures that every generating identification is respected. Thus $P$ is a categorical presheaf, and the two inclusions $j_0,j_1:Y\to P$ are natural transformations with $j_{0}a=j_{1}a$.

All identifications concern the same element $z$ in its two copies, so an element outside the image is never identified with its other copy. Consequently $j_{0,C}(y)\ne j_{1,C}(y)$, proving $a$ is not an epimorphism. ThereforeThis is the pointwise epimorphism in a functor category criterion. It requires neither a componentwise choice of preimages nor a natural section.

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:

For $F⊣U$ with unit of an adjunction $\eta$ and counit of an adjunction $\epsilon$, putThe triangle identities for an adjunction giveFor associativity, naturality of $\epsilon$ at ${\epsilon }_{FA}:FUFA\to FA$ saysApply $U$ to obtain ${\mu }_{A}T({\mu }_A)={\mu }_A{\mu }_{TA}$. Thus this is the monad induced by an adjunction.

Conversely, start with a monad $(T,\eta ,\mu )$. The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad laws ensure that these are algebras for a monad and morphisms of algebras for a monad. For an algebra $(B,b)$, there are inverse bijectionsThe inverse really is an algebra map becauseFor a free-algebra map $g$, its algebra equation and the monad unit law giveFor a base map $f$, naturality and the algebra unit law give $bT(f){\eta }_A=b{\eta }_{B}f=f$. The formulas are natural, establishing the free-forgetful Eilenberg-Moore adjunction $F^T⊣U^T$. Its unit is ${\eta }_A$, and its counit at $(B,b)$ is $b$. Hence its induced multiplication is $\mu$, and every monad arises from an adjunction.

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 ${\mathcal{C}}^T$.

Define the free algebra functor $F^T$ by $F^{T}X=(TX,{\mu }_X)$ and $F^{T}f=Tf$. The monad laws make ${\mu }_X$ an algebra action, and naturality of $\mu$ makes $Tf$ a monad algebra morphism. For an algebra for a monad $(A,a)$, definewith proposed inverse $g\mapsto aTg$. This inverse is a monad algebra morphism, sinceOne composite is $aTg{\eta }_X=a{\eta }_{A}g=g$. For a monad algebra morphism $h$, the other is $aT(h{\eta }_X)=aThT{\eta }_X=h{\mu }_{X}T{\eta }_X=h$. Both formulas commute with composition in a category in $X$ and with monad algebra morphisms in $(A,a)$, so the bijection is natural. Thus

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.

Define the Yoneda embedding $H_{\bullet }:\mathcal{C}\to [{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ by $H_A(U)=\mathcal{C}(U,A)$ and $H_A(k)(h)=hk$. For $f:A\to B$, define $H_{\bullet }(f{)}_U(h)=fh$. Associativity proves naturality in $U$, and preservation of identity morphisms and composition in a category proves that $H_{\bullet }$ is a functor.

Apply the Yoneda lemma to $X=H_B$. It givesThe natural transformation corresponding to $f$ has component $h\mapsto fh$, exactly $H_{\bullet }(f)$. Therefore the induced map on every hom-set is a bijection: the Yoneda embedding is full and faithful.

For a categorical presheaf $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$ and $A\in \mathcal{C}$, the Yoneda lemma gives a natural bijectionTo construct its inverse, take $x\in X(A)$ and set ${\theta }_U^x(h)=X(h)x$ for $h:U\to A$. For $k:V\to U$, the functor law gives $X(k){\theta }_U^x(h)=X(hk)x={\theta }_V^x(hk)$, so ${\theta }^x$ is a natural transformation. Conversely, naturality of any $\theta$ at $h:U\to A$ gives ${\theta }_U(h)=X(h){\theta }_A(1_A)$. Thus the two constructions are inverse, proving the bijection.

The bijection is natural in $X$: a natural transformation $t:X\to Y$ sends the distinguished element $x$ to $t_A(x)$ on either side. It is natural in $A$: for $f:A\to B$, precomposing a transformation $\mathcal{C}(-,B)\to X$ with postcomposition by $f$ sends its distinguished element $x\in X(B)$ to $X(f)x$. These equations prove the claimed naturality, rather than just an objectwise correspondence. Reversing morphisms gives the covariant form $\mathrm{Nat}(\mathcal{C}(A,-),Z)\cong Z(A)$ for $Z:\mathcal{C}\to \mathbf{Set}$.

For $D:\mathcal{I}\to \mathcal{C}$, define the functor $\mathrm{Cone}(-,D):{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$ bywhere $\Delta X$ is the constant diagram functor; a morphism into $X$ acts by precomposition on every categorical cone leg. A categorical limit is a representation of a functor of the formThe image of $1_L$ is the universal categorical cone $\lambda :\Delta L\to D$. By naturality of the representing isomorphism, the image of $h:X\to L$ is $\lambda \circ \Delta h$. Thus this definition is exactly existence and uniqueness of factorization of every categorical cone through $\lambda$. Conversely that universal property gives the representing natural bijection.

Given the two natural transformations, defineNaturality of $\eta$ and the second triangle identity for an adjunction giveNaturality of $\epsilon$ and the first triangle identity for an adjunction giveThe maps are natural in $C,D$, since the defining compositions commute with precomposition in $C$ and postcomposition in $D$ by naturality of $\eta ,\epsilon$. They therefore define an adjunction $F⊣G$. Their values on identity morphisms recover precisely $\eta ,\epsilon$. Conversely any adjunction with these unit and counit of an adjunction must have the same transpose formulas, so it is unique with the prescribed unit and counit.

Write the adjunction as a natural bijection ${\alpha }_{C,D}:\mathcal{D}(FC,D)\to \mathcal{C}(C,GD)$. The adjunction unit and adjunction counit are defined byNaturality of $\alpha$ implies $GF(u){\eta }_C={\eta }_{C^{\prime }}u$ for $u:C\to C^{\prime }$ and $v{\epsilon }_D={\epsilon }_{D^{\prime }}FG(v)$ for $v:D\to D^{\prime }$, so these are natural transformations $1_{\mathcal{C}}\to GF$ and $FG\to 1_{\mathcal{D}}$. More explicitly, naturality at $f:FC\to D$ and $g:C\to GD$ gives the two transpose formulasThese formulas will also prove the triangle identities for an adjunction.