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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:

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.

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.

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:

Suppose $U(C)\cong \mathcal{C}(R,C)$ naturally and $\mathcal{C}$ has small coproducts in a category. DefineFor a function $v:X\to Y$, define $F(v)$ by sending the $x$-summand by the identity of $R$ to the $v(x)$-summand. The coproduct property gives the identity and composition laws, so $F$ is a functor. There are natural bijectionsThe first chooses a morphism on every summand; the second views that family as a function. Hence this is the left adjoint to a covariant representable functor:For empty $X$, the formula uses the empty coproduct, an initial object, and the same bijection remains valid.

If $F⊣U$, let $1$ denote a singleton set. The adjunction gives the natural bijectionsThus $U$ is represented by $F1$, proving $A\Rightarrow R$.

If $U\cong \mathcal{C}(R,-)$ is a covariant representable functor, a morphism $R\to {\lim }_j{C}_j$ is exactly a compatible family of morphisms $R\to C_j$. Consequently the comparison mapis a bijection for every existing small categorical limit. This proves $R\Rightarrow L$ by the universal property of a limit. In particular,

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying the unit and multiplication of a monad lawsAn algebra for a monad is an object $A$ equipped with $a:TA\to A$ such thatA morphism of algebras for a monad $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $fa=bT(f)$. Identities and composition obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^T$ and its forgetful functor $U^T(A,a)=A$.

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with induced monad $T=UF$, the Eilenberg-Moore comparison functor isThe functor $U$ is monadic when it has such a left adjoint and this comparison is an equivalence of categories; a stricter convention requires an isomorphism of categories over $\mathcal{C}$. We will identify which limit-lifting conclusion each convention supports in part (iii).

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

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying, at every object $A$,These are the associativity and unit laws. An algebra for a monad is $(A,a:TA\to A)$ withA monad algebra morphism $h:(A,a)\to (B,b)$ is a morphism $h:A\to B$ satisfying $ha=bTh$. Identity morphisms satisfy this equation, and if $h,k$ satisfy it, then $kha=kbTh=cTkTh=cT(kh)$. Hence these objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$, with the forgetful functor $U(A,a)=A$ and $Uh=h$.

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 $\mathcal{D}$.

For the Category of sets, take the monoidal tensor product to be the Cartesian product, $A\otimes B=A\times B$, with unit a singleton $1$. On morphisms use $(f\times g)(a,b)=(f(a),g(b))$. Define the associator by $((a,b),c)\mapsto (a,(b,c))$ and the unitors by $(\ast ,a)\mapsto a$ and $(a,\ast )\mapsto a$. These are natural isomorphisms, with inverses obtained by rebracketing or adjoining $\ast$. Both paths in the pentagon send $(((a,b),c),d)$ to $(a,(b,(c,d)))$, and both paths in the triangle send $((a,\ast ),b)$ to $(a,b)$. This proves the axioms elementwise.

There is a second structure: use disjoint union $A⨿B$ and unit $\varnothing$. Define the associator by preserving each element and its summand while changing the nested tags, and define the unitors by removing the empty summand. These are natural and invertible. In the pentagon, either path takes an element in any of the four summands to that same element in the same summand of $A⨿(B⨿(C⨿D))$; the triangle removes an empty intermediate summand on either path. Thus both axioms hold.

These product and coproduct monoidal structures on sets have different units, $1$ and $\varnothing$, which are not isomorphic. They are even inequivalent as monoidal categories: an equivalence of the underlying Category of sets preserves terminal objects, so it cannot take the singleton unit of the product structure to the empty unit of the coproduct structure. Hence the monoidal structure on sets is not unique.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a monoidal unit object $I$, and natural isomorphismsThe associator obeys the pentagon identity for a monoidal category,and the unitors obey the triangle identity for a monoidal category,Composition in a category here acts from right to left. The first equation compares two paths from $((A\otimes B)\otimes C)\otimes D$ to $A\otimes (B\otimes (C\otimes D))$; the second compares two ways of removing the intervening unit.

It is precisely a bicategory with one object $\ast$. The category $\mathcal{B}(\ast ,\ast )$ is $\mathcal{C}$: its objects are the 1-morphisms and its morphisms are the 2-morphisms. Horizontal composition in a category is $\otimes$, the identity 1-morphism is $I$, and the associator and unitors give the weak associativity and unit constraints. Vertical composition in a category is ordinary composition in a category in $\mathcal{C}$; the interchange law follows because $\otimes$ is a bifunctor. The bicategory pentagon and triangle axioms are exactly the two equations above. Conversely the sole hom-category of any one-object bicategory gives these data, with the convention that horizontal composition in a category uses the indicated ordering of tensor factors.

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 $Z:\mathcal{C}\to \mathbf{Set}$.

Here is a precise base change of a pullback square statement. Suppose $A=B{\times }_{D}C$ is a pullback in a category and $t:D^{\prime }\to D$ is a morphism. Assume the pullbacks in a categoryexist, where $A\to D$ is the common composite. The induced square with vertices $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ is a pullback in a category; equivalently,Indeed, compatible morphisms $T\to B^{\prime }$ and $T\to C^{\prime }$ give morphisms $b:T\to B$, $c:T\to C$, and a common $v:T\to D^{\prime }$, with the composites of $b,c$ to $D$ both equal to $tv$. The original pullback in a category gives a unique morphism $h:T\to A$ with projections $b,c$. Its composite to $D$ is $tv$, so the defining pullback in a category for $A^{\prime }$ gives a unique morphism $T\to A^{\prime }$ above $h,v$. Its maps to $B^{\prime }$ and $C^{\prime }$ are the prescribed ones by uniqueness in those pullbacks in a category. Conversely, these projections recover $b,c,v$, so every possible factorization is the same. This proves the universal property without requiring arbitrary categorical limits in the ambient category.