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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 diagram in a category of shape $J$ is a functor $D:J\to \mathcal{C}$. A categorical cone with vertex $X$ is a family ${\gamma }_j:X\to Dj$ satisfying $D(a){\gamma }_j={\gamma }_k$ for every $a:j\to k$. A morphism between categorical cones from $(X,\gamma )$ to $(Y,\delta )$ is a morphism $u:X\to Y$ with ${\delta }_{j}u={\gamma }_j$ for all $j$. A categorical limit is a terminal object in this category of cones: each cone has a unique such morphism to the limiting cone.

For a finite $J$, form the products in a categoryDefine $s,t:P\rightrightarrows Q$ with $a$-coordinates $D(a){\pi }_j$ and ${\pi }_k$. The equalizer $e:L\to P$ imposes exactly the cone equations. Therefore ${\pi }_{j}e$ gives a categorical limit of $D$, since maps into $P$ encode families of legs and factoring through $e$ encodes their compatibility. The empty product in a category is the terminal object, covering the empty diagram. This is the construction of small limits from products and equalizers, restricted to finite shapes.

Now take an initial functor $F:I\to J$ and a categorical cone ${\delta }_i:X\to DFi$ over $DF$. For each $j$ and each object $(i,u:Fi\to j)$ of the comma category $(F\downarrow j)$, consider $D(u){\delta }_i$. A morphism $a:(i,u)\to (i^{\prime },u^{\prime })$ there satisfies $u^{\prime }F(a)=u$, soSince $(F\downarrow j)$ is a nonempty connected category, this common value is independent of the object. Define it to be ${\gamma }_j$. This does not require choosing representatives: the value is uniquely determined.

For $b:j\to k$, replacing $(i,u)$ by $(i,bu)$ proves $D(b){\gamma }_j={\gamma }_k$. Taking $(i,1_{Fi})$ proves ${\gamma }_{Fi}={\delta }_i$. Conversely, extending a restricted cone recovers its original legs, since ${\gamma }_j=D(u){\gamma }_{Fi}$. A vertex morphism commuting with every ${\delta }_i$ also commutes with each ${\gamma }_j=D(u){\delta }_i$, and the converse follows by restriction. Thus extension and restriction are strictly inverse functors, not merely an equivalence of categories. This proves cone restriction along an initial functor.

If $\mathcal{C}$ has all categorical limits of shape $I$, transport a terminal object in the cone category of $DF$ across this isomorphism to obtain a categorical limit of $D$. Uniqueness of the induced comparison, and its compatibility with natural transformations of diagrams, givesFor the converse use the representable test for initial functors. For $j\in J$, the representable presheaf $J(-,j):J^{\mathrm{op}}\to \mathbf{Set}$ is equivalently a diagram in a category $D_j:J\to {\mathbf{Set}}^{\mathrm{op}}$. Its categorical limit in the opposite category is the colimit of $J(-,j)$ in sets. Elements of that colimit are connected components of the category of elements, equivalently of the opposite category of the slice $J\downarrow j$. This slice has terminal object $(j,1_j)$, so that colimit is a singleton.

For the restricted presheaf $J(F(-),j)$, the same description identifies its colimit with the connected-component set of $(F\downarrow j)$. Indeed a relation identifying $u^{\prime }:F{i}^{\prime }\to j$ with $u^{\prime }F(a):Fi\to j$ is exactly a generating edge of the zigzag relation in this comma category. The assumed isomorphism of limit functors forces this set to be a singleton. Therefore $(F\downarrow j)$ is nonempty and connected for every $j$, proving

A monad on $\mathcal{C}$ is an endofunctor $T$ with natural transformations $\eta :1\to T$ and $\mu :T^2\to T$ satisfyingAn algebra for a monad is $(A,\alpha :TA\to A)$ with $\alpha {\eta }_A=1_A$ and $\alpha T\alpha =\alpha {\mu }_A$. A morphism of algebras for a monad $f:(A,\alpha )\to (B,\beta )$ satisfies $f\alpha =\beta Tf$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$. Its free algebra functor is $FX=(TX,{\mu }_X)$, and its adjunction to the forgetful functor $U$ has the explicit bijection

Write $S=A+B$, $V=TA+TB$, and let ${\nu }_1,{\nu }_2$ be the coproduct in a category injections. SetHere ${\mu }_S:FTS\to FS$ is an algebra morphism, with underlying multiplication $T^{2}S\to TS$. For an algebra $(C,\gamma )$, an algebra morphism $h:FS\to (C,\gamma )$ corresponds to $u:S\to C$, and the transposes of $hr,hs$ are respectivelyFor the second formula, $({\mu }_{S}T\kappa ){\eta }_V=\kappa$ by naturality of $\eta$ and the monad unit law. Thus $hr=hs$ exactly when, writing $u=[a,b]$,These say precisely that $a$ and $b$ are morphisms of algebras for a monad. Consequently any coequalizer $q:FS\to Q$ of $r,s$ represents pairs of algebra morphisms out of $(A,\alpha )$ and $(B,\beta )$, proving the coproduct presentation for monad algebras:Its two algebra injections have underlying morphisms $q{\eta }_S{\nu }_1$ and $q{\eta }_S{\nu }_2$; the equivalence just proved verifies both the algebra equations and their universal property.

The displayed pair is a reflexive pair, with common sectionIndeed $rt=1_{FS}$ by the algebra unit laws, whileNow suppose $\mathcal{C}$ has all finite colimits and $T$ preserves reflexive coequalizers. The permitted creation theorem gives reflexive coequalizers in ${\mathcal{C}}^T$: the underlying pair is reflexive and its coequalizer exists and is preserved by $T$. The preceding construction therefore gives binary algebra coproducts in a category. The initial object is $F0$, since the free algebra functor is a left adjoint and $0$ is initial in $\mathcal{C}$.

For any algebra pair $a,b:X\rightrightarrows Y$, the augmented pairis a reflexive pair, with common section the injection of $Y$, and has exactly the same coequalizing morphisms as $a,b$. Hence all coequalizers exist in ${\mathcal{C}}^T$. The construction of finite colimits from coproducts and reflexive coequalizers now gives

Assume (c), and take an epimorphism $p:M\twoheadrightarrow h_A$ with $M$ a monofunctor. Since $h_A$ is a projective object in a category, there is a natural transformation $s:h_A\to M$ with $ps=1_{h_A}$. A retract in a category of a monofunctor is a monofunctor: if $h_A(f)(u)=h_A(f)(v)$, then naturality of $s$ gives $M(f)s(u)=M(f)s(v)$, injectivity of $M(f)$ gives $s(u)=s(v)$, and applying $p$ gives $u=v$. Thus (c) implies (b), completing

The stronger assertion that every set-valued functor is a monofunctor holds exactly when $\mathcal{C}$ is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.

For necessity, fix $f:A\to B$, and form the pointwise pushout in a categoryusing twice the natural transformation $h_B\to h_A$ given by precomposition with $f$. At an object $X$, $P(X)$ consists of two copies of $\mathcal{C}(A,X)$, with the two copies of $kf$ identified for every $k:B\to X$. No different underlying elements are identified: each generating relation simply joins the two copies of one element. In $P(B)$, the two copies of $f$ coincide, since $f=1_{B}f$. Hence $P(f)$ sends the two copies of $1_A$ in $P(A)$ to the same element. If $P$ is a monofunctor, those copies of $1_A$ coincide. The description of the pushout implies $rf=1_A$ for some $r:B\to A$. Thus every morphism is a split monomorphism.

Apply the same conclusion to $r$: there is $t:A\to B$ with $tr=1_B$. Then $t=trf=f$, and so $fr=1_B$ as well. Every arrow is invertible, proving

Use the square's notation $HF=KG$. For $E\in \mathcal{E}$, choose $(C,u:E\to GC)$ in the nonempty connected category $(E\downarrow G)$. Lift $Ku:KE\to HFC$ through the discrete fibration $H$, obtaining the uniquely determined arrowDefine $LE=D_u$. If $a:C\to C^{\prime }$ satisfies $Gau=u^{\prime }$, then $Fa\tilde{u}$ is a lift of $K{u}^{\prime }$ with codomain $F{C}^{\prime }$. Uniqueness makes it exactly ${\tilde{u}}^{\prime }$, including equality of its domain. Connectedness of the comma category therefore makes $LE$ independent of the chosen object, with strict equality rather than merely an isomorphism.

For $v:E\to E^{\prime }$, choose $(C,u:E^{\prime }\to GC)$ and lift $Kv:KE\to K{E}^{\prime }=HL{E}^{\prime }$ with codomain $L{E}^{\prime }$. Write this lift as $w:D\to L{E}^{\prime }$. Its composite with $\tilde{u}$ is the lift of $K(uv)$ to $FC$, so its domain is $LE$. Set $Lv=w$. The lift of an identity is an identity, and the composite of two lifts is the lift of the composite with the specified codomain. Thus $L$ is a functor, and $HL=K$.

At $E=GC$, choose $(C,1_{GC})$; its lift is $1_{FC}$, giving $LGC=FC$. For $a:C\to C^{\prime }$, the unique lift of $KGa=HFa$ with codomain $F{C}^{\prime }$ is $Fa$, giving $LG=F$ on arrows as well. Finally, any other filler $L^{\prime }$ must send $u$ to the same lift of $Ku$ with codomain $FC$, so $L^{\prime }E=LE$; it must also send $v$ to the unique lift of $Kv$ with codomain $L{E}^{\prime }$. This proves orthogonality of final functors and discrete fibrations:

Use $\mathbb{N}=\{0,1,\dots \}$, as required by the formula at zero, and composition of functions as multiplication in $M$. The displayed $T$ preserves identities, and for $n>0$,both sides send zero to zero. Thus $T$ is a functor on the one-object category.

For the shift monad on order-preserving maps of natural numbers, setThese are order-preserving functions. The equations $Tf\eta =\eta f$ and $\mu T^{2}f=Tf\mu$ hold pointwise, giving the required natural transformations. For the second equation, both sides are zero at $n=0,1$, and are $f(n-2)+1$ for $n\ge 2$. The two unit laws are $\mu \eta =\mu T\eta =1$. Associativity is checked byConsequently these maps define a monad. It is not an idempotent monad, since $\mu (0)=\mu (1)$, so $\mu$ cannot be invertible.

An algebra for a monad is a map $a\in M$ with $a\eta =1$ and $aTa=a\mu$. The first equation forces $a(n+1)=n$; monotonicity then forces $a(0)=0$. Thus $a=\mu$, which satisfies the second equation by the monad associativity law. The Eilenberg-Moore category therefore has exactly one object. Its endomorphisms $h$ satisfy $h\mu =\mu Th$, and this equation holds precisely when $h(0)=0$: evaluate at zero for necessity, and at $n>0$ both sides equal $h(n-1)$.

The Kleisli comparison functor takes its only object to this only algebra and sends $f\in M$ toIt is a bijection from the Kleisli arrows to the algebra endomorphisms, with inverse $h\mapsto (n\mapsto h(n+1))$. It preserves identities and composition by the comparison construction; directly, $K(\eta )=1$ and $K(g\star f)=K(g)K(f)$. Bijectivity on objects and arrows makes it an isomorphism of categories:

A monad is an endofunctor $T:\mathcal{C}\to \mathcal{C}$ with natural transformations $\eta :1_{\mathcal{C}}\to T$ and $\mu :T^2\to T$ satisfyingIts Kleisli category has the same objects as $\mathcal{C}$, with ${\mathcal{C}}_T(A,B)=\mathcal{C}(A,TB)$, identity ${\eta }_A$, and compositeThe unit laws follow from naturality of $\eta$ and the two monad unit laws. For $h:C\to TD$, the two triple composites reduce, using naturality of $\mu$, tothey agree by the monad associativity law. Thus this is a category.

Define the free functor into a Kleisli category $J$ and the right-hand functor $U$ byThe identity and composition formulas for $J$ areFor $U$, $U({\eta }_A)={\mu }_{A}T{\eta }_A=1_{TA}$, andwhere the last step is naturality of $\mu$. Hence both mappings are functors. The hom-set identity ${\mathcal{C}}_T(JA,B)=\mathcal{C}(A,UB)$ gives the adjunction $J⊣U$. Its unit is $\eta$, and its counit at $B$ is the Kleisli arrow represented by $1_{TB}:TB\to TB$. Applying $U$ to that counit gives ${\mu }_B$, so the induced monad is the specified one.

For any $L⊣R:\mathcal{D}\to \mathcal{C}$ inducing this same monad, with counit $\epsilon$, define the Kleisli comparison functorThe triangular identities give $K({\eta }_A)=1_{LA}$, and naturality of the counit together with $\mu =R\epsilon L$ gives $K(g\star f)=K(g)K(f)$. Explicitly, naturality moves ${\epsilon }_{LB}$ past $L(g)$, then moves the resulting counit past ${\epsilon }_{LC}$, reducing the composite to ${\epsilon }_{LC}L({\mu }_{C}T(g)f)$. Thus $K$ is a functor, with $KJ=L$ and $RK=U$, and it sends the Kleisli counit to ${\epsilon }_L$.

An adjunction morphism here is required to commute with the left and right adjoints and preserve their adjunction structure. Such a morphism must have the above object map; every Kleisli arrow $f$ factors as its counit after $J(f)$, so its arrow map is forced as well. This proves initiality of the Kleisli adjunction:

Fix $A$, and let $S(B)=B⨿A$ be the functor obtained from the chosen binary coproduct in a category construction. Its arrow map is $S(v)=v⨿1_A$. For $h_A=\mathcal{C}(A,-)$, the binary coproduct property gives $h_B\times h_A\cong h_{B⨿A}$. The Yoneda lemma and part (i) therefore give, naturally in $B,F$,Thus exponentiation by $h_A$ is precomposition $S^{\ast }:F\mapsto F\circ S$.

This precomposition has a right adjoint given by Right Kan extension. Explicitly, for $K:\mathcal{C}\to \mathbf{Set}$ putThe comma category is small, so this categorical limit exists in the Category of sets. A map $v:B\to B^{\prime }$ gives $(RK)(v)$ by taking a compatible family indexed by $u:B\to SC$ and selecting its entries at $u^{\prime }v$ for the indices $u^{\prime }:B^{\prime }\to SC$.

A natural transformation $\alpha :F\circ S\to K$ determines $F\to RK$ by sending $x\in F(B)$ to the family ${\alpha }_C(F(u)x)$ indexed by $(C,u)$. Conversely, a map $F\to RK$ determines $F(SC)\to K(C)$ by projecting to the index $(C,1_{SC})$. Naturality and the limit's compatibility relations make these constructions inverse. Hence $S^{\ast }⊣R$, proving the tiny covariant representable functor on a category with binary coproducts result:

For a tiny object $A$, the exponential functor $E_A=(-{)}^A$ has a right adjoint, say $R_A$. The defining exponential adjunction gives the natural isomorphismsince $(X^B{)}^A\cong X^{A\times B}$. If $A,B$ are tiny objects, the composite on the right is a left adjoint with right adjoint $R_B{R}_A$. Thus $A\times B$ is tiny. The terminal object is tiny because exponentiation by it is naturally the identity, which is its own right adjoint. Induction, including this empty-product case, proves tiny objects are closed under finite products:

A Cartesian closed category has finite products in a category and, for every $F$, an exponential object $G^F$ characterized naturally byIn $[\mathcal{C},\mathbf{Set}]$, finite products and the terminal functor are computed pointwise. With $h_B=\mathcal{C}(B,-)$, defineThis is a set because $\mathcal{C}$ is a small category. For $v:B\to B^{\prime }$, precomposition $h_{B^{\prime }}\to h_B$ induces $(G^F)(v)$ by precomposition of natural transformations, making $G^F$ a covariant functor.

Given $\alpha :K\times F\to G$, define its curry $\hat{\alpha }:K\to G^F$ byFor $a:C\to C^{\prime }$, naturality of $\alpha$ gives $G(a){\alpha }_C(K(u)y,x)={\alpha }_{C^{\prime }}(K(au)y,F(a)x)$, proving that this is a natural transformation $h_B\times F\to G$. The same formula with $u$ precomposed proves naturality of $\hat{\alpha }$ in $B$.

Conversely, from $\beta :K\to G^F$ defineNaturality of $\beta$ and its values proves naturality of $\alpha$. Currying this result recovers $\beta$, since ${\beta }_C(K(u)y)=(G^F)(u)({\beta }_B(y))$; uncurrying a curry recovers $\alpha$ by setting $u=1_B$. These mutually inverse constructions are natural in $K,G$, giving the adjunction required for an exponential object. Therefore

The simplex category $\Delta$ has objects $[n]=\{0<1<\cdots <n\}$ for $n\ge 0$ and order-preserving maps as morphisms. A simplicial set is a functor $X:{\Delta }^{\mathrm{op}}\to \mathbf{Set}$.

The standard simplex is the representable simplicial setFor $0\le i\le n$, the simplicial horn ${\Lambda }_i^n\subseteq {\Delta }^n$ is the union of the images of all coface maps ${\Delta }^{n-1}\to {\Delta }^n$ except the face opposite vertex $i$.

For a smooth projective geometrically connected algebraic curve $C/k$, this functor assigns to a scheme $T$ the line bundles on $C\times T$ having degree $d$ on each geometric fibre, modulo bundles pulled back from $T$. One generally takes the associated sheaf for faithfully flat covers. A rational point of $C$ permits normalization of every line bundle along that point; normalized bundles have no automorphisms preserving their normalization and satisfy effective descent. The resulting functor is represented by the degree-$d$ component of the Picard scheme of a curve.

A natural transformation between functors to the Category of sets is a monomorphism in the functor category exactly when every component is an injective function. Componentwise cancellation proves sufficiency. For necessity, the Yoneda lemma converts two elements with equal component images into two transformations from a covariant representable functor; monicity cancels these transformations, forcing the elements to coincide. The argument remains valid for a locally small category in an ambient universe where the functor category is formed.