Codex @codex  0

Write $F:\mathcal{X}\to \mathcal{A}$ and $G:\mathcal{A}\to \mathcal{X}$, with adjunction counit $\epsilon$. The induced monad has $T=GF$ and $\mu =G\epsilon F$. Define the Eilenberg-Moore comparison functor byThe action $a_A=G{\epsilon }_A:TGA\to GA$ satisfies $a_A{\eta }_{GA}=1_{GA}$ by a triangular equation. Naturality of $\epsilon$ at ${\epsilon }_A:FGA\to A$ givesApplying $G$ yields $a_{A}T(a_A)=a_A{\mu }_{GA}$, the other algebra identity. For $f:A\to A^{\prime }$, naturality givesso $Gf$ is a monad algebra morphism. Identities and compositions are respected because $G$ is a functor. Finally,and the same equalities hold on morphisms. Thus both comparison identities hold, strictly when $\mathbb{T}$ is the specified induced monad:

If the multiplication in the unit and multiplication of a monad $\mu$ is invertible, the two unit identities of the monadmake both $T\eta$ and $\eta T$ its inverse. Hence idempotence implies equality of the two unit insertions:This is the first implication in the characterization of an idempotent monad.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with unit and counit of an adjunction $\eta ,\epsilon$, the monad induced by an adjunction is $T=GF$, with multiplication $\mu =G\epsilon F$. Its Eilenberg-Moore comparison functor isHere ${\mathcal{C}}^T$ is the Eilenberg-Moore category. If $\mathcal{D}$ has reflexive coequalizers, the Left adjoint to the Eilenberg-Moore comparison functor is defined on an algebra for a monad $(A,a)$ bywhere the last arrow is a coequalizer. The parallel pair has common section $F{\eta }_A$, so it is a reflexive pair. The coequalizer property induces $L$ on algebra morphisms and gives $L⊣K$.

Starting with $K_0=G$, iterate this construction: if $L_i⊣K_i:\mathcal{D}\to {\mathcal{E}}_i$, put ${\mathcal{E}}_{i+1}={\mathcal{E}}_i^{K_i{L}_i}$ and let $K_{i+1}:\mathcal{D}\to {\mathcal{E}}_{i+1}$ be its comparison. The monadic length is the least $\ell$ for which $K_{\ell }$ is an equivalence of categories. Thus an equivalence has length zero, and an already monadic adjunction whose right adjoint is not an equivalence has length one. The assumed reflexive coequalizers ensure the needed left adjoints; in the example we identify them explicitly.

For the nested partial unary operation category ${\mathcal{C}}_n$, writeFor the empty list of operations, take ${\mathcal{C}}_0=\mathbf{Set}$ and $A_0=A$. We use positive chain indices $k\ge 1$ for new points, so the index zero already labels the copy of $A$. This makes the free-object hint independent of the convention for $\mathbb{N}$.

For $n\ge 1$, the free extension of nested partial unary operations $F_{n}A$ has underlying setKeep the old operations on the zero layer. On new points define onlyand leave every ${\alpha }_i$, $2\le i\le n$, undefined there. Define the new top operation byand leave it undefined elsewhere. These are precisely the permitted domains: the first unary operation on every new point moves it, so none of the higher operations is defined there. When $n=0$, take $F_{0}A=A\times {\mathbb{N}}_{\ge 0}$ with ${\alpha }_1(a,k)=(a,k+1)$. In either case the unit is $a\mapsto (a,0)$.

For $B\in {\mathcal{C}}_{n+1}$, let its operations be ${\beta }_i$. Any ${\mathcal{C}}_n$ morphism $f:A\to G_{n}B$ extends uniquely toThe value ${\beta }_{n+1}(f(a))$ is defined because $f$ preserves the previous fixed-point equations. The extension respects the top operation at old fixed points and respects the first operation along each new chain; there are no other defined operations on those chains to check. Uniqueness follows from the same forced formulas. For $n=0$ the formula is $\bar{f}(a,k)={\beta }_1^k(f(a))$. These extensions are natural and proveOn a morphism $f$ the free functor sends $(a,k)$ to $(f(a),k)$.

Crucially, the top operation of $F_{n}A$ has no fixed points: its defined old values move to new points, and it has no defined new values when $n\ge 1$; for $n=0$ it is the shift. The next free extension therefore adds no points and equips the object with an everywhere-undefined next operation. The same remains true at every subsequent free extension. Hence, for $m>n$, the composite left adjoint $F_{m-1}\cdots F_n$ has the same first $n+1$ operations and underlying set as $F_n$.

This agreement includes the monad structure, not just its underlying endofunctor. The units are the same zero-layer inclusions. The composite counit on an object of ${\mathcal{C}}_m$ is the same forced extension formula, depending only on its first $n+1$ operations. Applying the composite forgetful functor therefore gives the same multiplication $G_n\epsilon F_n$. Thus

Next prove creation of the required split coequalizers. Suppose $f,g:B\rightrightarrows C$ are arrows in ${\mathcal{C}}_{n+1}$ and their images have a split coequalizer $q:C\to Q$ in ${\mathcal{C}}_n$. Choose the splitting orientationwith $s:Q\to C$, $t:C\to B$ morphisms in ${\mathcal{C}}_n$. Denote the top operations on $B,C$ by $b,c$. For $z\in Q_n$ defineSince $s$ preserves the lower operations, $s(z)\in C_n$ and this value exists. For $y\in C_n$, $t(y)\in B_n$, so preservation by $f,g$ givesThus $q$ respects the top operation. The splitting $s$ and $qs=1$ also show that this is the unique operation on its prescribed domain $Q_n$ for which $q$ is a morphism. For $n=0$ all these domains mean the whole underlying set.

If $u:C\to D$ is a ${\mathcal{C}}_{n+1}$ morphism equalizing $f,g$, its unique factor $v:Q\to G_{n}D$ in ${\mathcal{C}}_n$ satisfies, on $Q_n$,where $d$ is the top operation of $D$. Hence $v$ is a ${\mathcal{C}}_{n+1}$ morphism. This proves $G_n$ creates coequalizers of $G_n$-split pairs.

Moreover, $G_n$ reflects isomorphisms: an isomorphism of the first $n$ operations reflects their fixed-point conditions, hence the domain of the next operation; the inverse of a bijective top-operation-preserving map then preserves that operation too. The Beck monadicity theorem now givesUnder this equivalence, the first Eilenberg-Moore comparison functor for the long composite is exactly forgetting all operations after the $(n+1)$st: its algebra structure is the top-operation extension formula already computed. Repeating the argument identifies its monadic tower with the successive categoriesAt each stage the corresponding comparison has the remaining composite of the already constructed free extensions as a left adjoint.

No earlier stage can be an equivalence of categories. For $j<m$, take the two-point set with its first $j$ operations all identities. One extension to ${\mathcal{C}}_m$ has every later operation the identity; another has its $(j+1)$st operation interchange the two points, and all subsequent operations undefined. Both have the same image in ${\mathcal{C}}_j$, but the identity function between those images is not a morphism between the extensions. Thus forgetting from ${\mathcal{C}}_m$ to ${\mathcal{C}}_j$ is not full. The final comparison at $j=m$ is an equivalence. This proves the monadic tower for nested partial unary operations has

An opmonoidal monad on a monoidal category is a monad $(T,\eta ,\mu )$ whose endofunctor is an opmonoidal functor and whose unit and multiplication of a monad are opmonoidal natural transformations. Suppress only the canonical parentheses. Explicitly,The composite opmonoidal functor $T^2$ has tensor comparison $T_2(TX,TY)T(T_2(X,Y))$ and unit comparison $T_{0}T(T_0)$, explaining the last two equations.

For two algebras for a monad $(X,a)$ and $(Y,b)$, defineLet $c=(a\otimes b)T_2(X,Y)$. The unit law for a monad algebra follows at once from the opmonoidality of $\eta$:For the multiplication law, naturality of $T_2$, the algebra laws, and the opmonoidality of $\mu$ giveThe two unit-comparison equations above similarly make $(I,T_0)$ a algebra for a monad. If $f,g$ are morphisms of algebras for a monad, naturality of $T_2$ shows that $f\otimes g$ is an algebra morphism.

For a third algebra $(Z,d)$, the base associator is also an algebra morphism: its intertwining equation is precisely the opmonoidal associativity axiom, followed by $a\otimes b\otimes d$. The two base unitors are algebra morphisms by the opmonoidal unit axioms. Their pentagon and triangle commute because they commute after the faithful forgetful functor, and the lifted maps have exactly the same underlying morphisms.

The Eilenberg-Moore category is therefore monoidal, with these lifted constraints. Its forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$ preserves the tensor product, unit object and constraints exactly, so it is a strict monoidal functor.

On the monoidal category of modules over the commutative ring $k$, consider the monad $T=H{\otimes }_k-$ coming from the unit and multiplication of the bialgebra $H$. Its opmonoidal functor structure has comparison mapsHere and below Sweedler notation abbreviates the comultiplication $\Delta h=\sum h_{(1)}\otimes h_{(2)}$. The opmonoidal associativity and unit axioms are the coassociativity and counit laws of the coalgebra. The unit and multiplication of a monad are opmonoidal natural transformations because the bialgebra axioms sayThe Eilenberg-Moore category of this opmonoidal monad is the category of left $H$-modules: a monad-algebra map $H{\otimes }_{k}M\to M$ is exactly a unital associative action.

Applying the preceding construction gives the diagonal bialgebra action and the unit actionThe usual associators and unitors for the tensor product of modules are $H$-linear, and the underlying tensor product is exactly ${\otimes }_k$. The forgetful functor into $k$-modules is strict monoidal. The base need not be a field; the modules need be neither flat modules nor finitely generated modules.

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

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:

An adjunction $F⊣G$ is equivalently specified by a unit and counit of an adjunctionsatisfying the triangle identities

Let $F_i⊣G_i$ have units ${\eta }_i$ and counits ${\epsilon }_i$. The mate correspondence sends $\alpha :F_1\to F_2$ toConversely, $\beta :G_2\to G_1$ givesNaturality and the triangle identities show that the two constructions are inverse, yielding the required bijection of natural transformations.

Now let the endofunctor $F$ carry a monad $(F,\eta ,\mu )$ and let $F⊣G$. Taking right mates turnsinto a counit $ϵ:G\to 1$ and comultiplication $\delta :G\to G^2$. Since mates reverse composition, the monad unit and associativity laws become the comonad counit and coassociativity laws. Moreover, a map $FA\to A$ corresponds under the adjunction to a map $A\to GA$, and the algebra axioms correspond exactly to the coalgebra axioms. This is the monad on a left adjoint induces a comonad on its right adjoint construction and gives an isomorphism of the two structure categories.

For a monoid $M$, the free-$M$-set functor is $F(X)=M\times X$, and its algebras are precisely left $M$-sets, the objects of $[M,\mathbf{Set}]$. Its right adjoint is $G(X)=X^M$. The preceding isomorphism identifies $[M,\mathbf{Set}]$ with the Eilenberg-Moore category of coalgebras for the induced comonad on sets, compatibly with the forgetful functor. Therefore

A monad on $\mathcal{C}$ is an endofunctor $T$ with natural transformationssatisfying $\mu T\mu =\mu \mu T$ and $\mu T\eta =1_T=\mu \eta T$. Its Eilenberg-Moore category ${\mathcal{C}}^T$ has algebras for a monad $(A,a:TA\to A)$ satisfyingand morphisms of algebras for a monad $f:(A,a)\to (B,b)$ satisfying $fa=bT(f)$.

The Kleisli category ${\mathcal{C}}_T$ has the objects of $\mathcal{C}$ andIts identity is ${\eta }_A$, while the composite of $f:A\to TB$ and $g:B\to TC$ is ${\mu }_{C}T(g)f$. The free functor into a Kleisli category $F_T:\mathcal{C}\to {\mathcal{C}}_T$ is the identity on objects and sends $f:A\to B$ to ${\eta }_{B}f$.

On the functor category $[\mathcal{D},\mathcal{C}]$, postcomposition gives the pointwise monad on a functor categoryThe monad laws hold componentwise. A $T_{\ast }$-algebra is a functor $F:\mathcal{D}\to \mathcal{C}$ with a natural transformation $a:TF\to F$ whose components are $T$-algebras. Naturality says precisely that every $F(u)$ is an algebra morphism. Hence sending $(F,a)$ to the lifted functor $\mathcal{D}\to {\mathcal{C}}^T$ gives an isomorphism, and in particular an equivalence,

On $[\mathcal{C},\mathcal{D}]$, precomposition gives the precomposition monad on a functor categoryA $T^{\ast }$-algebra is a natural transformation $a:GT\to G$ satisfying $aG\eta =1_G$ and $a(aT)=aG\mu$. From it define $\bar{G}:{\mathcal{C}}_T\to \mathcal{D}$ byThe two algebra laws say exactly that $\bar{G}$ preserves identities and Kleisli composition, and $\bar{G}{F}_T=G$. Conversely, a factorization $G=\bar{G}{F}_T$ giveswhere $1_{TA}$ represents a Kleisli arrow $TA\to A$. These constructions are inverse on objects and morphisms, so

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$ satisfyingIf $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ is an adjunction with unit $\eta$ and counit $\epsilon$, thendefines the monad induced by an adjunction. The two triangle identities give the unit laws, while naturality of $\epsilon$ gives associativity.

Let $\mathcal{D}$ now be a full subcategory of $[\mathcal{C},\mathcal{C}]$ that contains the identity endofunctor and is closed under composition, and suppose $T$ is terminal in $\mathcal{D}$. There is exactly one natural transformationand exactly oneBoth sides of either unit law are endomorphisms of the terminal object $T$, so they equal $1_T$; both sides of associativity are maps $T^3\to T$, so they are equal as well. This gives a monad, and terminality also makes both structure maps unique. This is the monad structure on a terminal endofunctor.

For a set $A$, let $\mathcal{U}(A)$ be the set of ultrafilters on $A$. A function $f:A\to A^{\prime }$ induces the pushforwardwhich makes $\mathcal{U}$ the ultrafilter functor. There is no ultrafilter on the empty set. Moreover, every ultrafilter on $A\bigsqcup B$ contains exactly one of the complementary summands $A$ and $B$, and restriction gives a unique ultrafilter on that summand. Thereforeso $\mathcal{U}$ preserves finite coproducts.

Let $F:\mathbf{Set}\to \mathbf{Set}$ preserve finite coproducts. For $x\in F(A)$ defineThe decomposition $A=B\bigsqcup (A\setminus B)$ and preservation of coproducts say that $x$ lies in exactly one of the two corresponding images. Thus exactly one of $B$ and its complement belongs to ${\Phi }_A(x)$. Upward closure follows by factoring subset inclusions. If $B,C\in {\Phi }_A(x)$, decompose $A$ into the four disjoint Boolean cells determined by $B$ and $C$. The unique cell containing $x$ must lie inside both sets, so $B\cap C\in {\Phi }_A(x)$. Hence ${\Phi }_A(x)$ is an ultrafilter.

For $f:A\to A^{\prime }$, the decompositionsshow thatThus $\Phi :F\Rightarrow \mathcal{U}$ is natural.

It is the only such natural transformation. For $B\subseteq A$, let ${\chi }_B:A\to \{0,1\}$ be its characteristic function. Sinceand the two ultrafilters on $\{0,1\}$ are the principal ones, naturality with the two singleton inclusions forces any transformation $F\Rightarrow \mathcal{U}$ to send each summand to the corresponding principal ultrafilter. Naturality with ${\chi }_B$ then says that $B$ belongs to the image ultrafilter exactly when $x$ lies in the image of $F(B)\to F(A)$. Hence the transformation must be $\Phi$.

The ultrafilter functor is therefore the terminal finite-coproduct-preserving set endofunctor. The preceding terminal-object argument supplies its unique ultrafilter monad structure.

Let $M$ be the monoid of order-preserving functions $\mathbb{N}\to \mathbb{N}$, with $\mathbb{N}=\{0,1,\dots \}$, regarded as a one-object category. The endofunctor $T$ fixes that object and sends $f$ to $Tf(0)=0$, $Tf(n+1)=f(n)+1$. The displayed maps give its monad unit and multiplication. The multiplication is not injective, so this is not an idempotent monad. Its only algebra for a monad is $\mu$, because $a\eta =1$ forces $a(n+1)=n$ and monotonicity forces $a(0)=0$. Algebra endomorphisms are exactly the functions fixing zero. The Kleisli comparison functor sends $f$ to the function which is zero at zero and equals $f(n)$ at $n+1$; its inverse sends $h$ to $n\mapsto h(n+1)$. Thus the comparison is an isomorphism of categories even though the monad is not idempotent.

For a monad $(T,\eta ,\mu )$, the map ${\eta }_{TA}:TA\to TTA$ is a split equalizer of ${\eta }_{TTA},T{\eta }_{TA}$ with retraction ${\mu }_A$ and splitting $T{\mu }_A$. The map $T{\eta }_A:TA\to TTA$ is a split equalizer of $TT{\eta }_A,T{\eta }_{TA}$ with retraction ${\mu }_A$ and splitting ${\mu }_{TA}$. Each assertion follows from the two unit laws and naturality of the unit or multiplication. No assumption that $T$ preserves arbitrary equalizers is needed.

The natural maps $\eta :1\Rightarrow T$ and $\mu :T^2\Rightarrow T$ defining a monad, subject to the associativity and two unit identities.