Monadicity and Beck's Theorem

Let $k:y\to w$ satisfy $kf=kg$. Define $u:=ks:z\to w$. Then $uh=ksh=kft=kgt=k$, using $ft=sh$, the equality $kf=kg$, and $gt=1_y$.

Let $H$ be any functor with the given category as domain. Functoriality sends the four equations in [L1] to $HhHf=HhHg$, $HhHs=1_{Hz}$, $HgHt=1_{Hy}$, and $HfHt=HsHh$, so the image diagram is again split.

If $v:z\to w$ also satisfies $vh=k$, then $v=vhs=ks=u$ because $hs=1_z$. Thus $h$ has the coequalizer universal property, including when $f=g$ or $h$ is an identity.

Repeating steps 1.1 and 2.1 in the target of $H$ shows that $Hh$ is a coequalizer of $Hf$ and $Hg$. Since $H$ was arbitrary, every functor preserves this coequalizer, so it is absolute by [L2].

By [L2], $T(a):(T^{2}A,{\mu }_{TA})\to (TA,{\mu }_A)$ is an algebra homomorphism. The monad associativity equation says that ${\mu }_A:(T^{2}A,{\mu }_{TA})\to (TA,{\mu }_A)$ is also an algebra homomorphism.

The structure map $a:(TA,{\mu }_A)\to (A,a)$ is an algebra homomorphism by the algebra associativity law, and that same law gives $aT(a)=a{\mu }_A$, so $a$ coequalizes the canonical pair.

Let $r:(TA,{\mu }_A)\to (B,b)$ be an algebra homomorphism with $rT(a)=r{\mu }_A$. Define $\bar{r}:=r{\eta }_A:A\to B$. Naturality of $\eta$ and the monad unit law give $\bar{r}a=r{\eta }_{A}a=rT(a){\eta }_{TA}=r{\mu }_A{\eta }_{TA}=r$.

Since $r$ is an algebra homomorphism, $r{\mu }_A=bT(r)$. Hence $bT(\bar{r})=bT(r)T({\eta }_A)=r{\mu }_{A}T({\eta }_A)=r=\bar{r}a$, so $\bar{r}$ is an algebra homomorphism.

If $c:(A,a)\to (B,b)$ satisfies $ca=r$, then $c=ca{\eta }_A=r{\eta }_A=\bar{r}$ by the algebra unit law. Therefore $a$ has the universal property in [L3], including for initial or degenerate algebra objects, and is the claimed coequalizer.

The algebra law gives $aT(a)=a{\mu }_A$ and $a{\eta }_A=1_A$. The monad unit law gives ${\mu }_A{\eta }_{TA}=1_{TA}$, while naturality of $\eta$ gives $T(a){\eta }_{TA}={\eta }_{A}a$. These are exactly the four equations of [L2] for $t={\eta }_{TA}$ and $s={\eta }_A$.

For the free-monoid monad, take the monoid $M=\{1,e\}$ with $e^2=e$. On the two-letter word $[e,e]$, the composite ${\eta }_{M}a$ first multiplies and gives the one-letter word $[e]$, whereas ${\mu }_{M}T({\eta }_M)$ gives the two-letter word $[e,e]$.

Therefore the underlying canonical presentation is split in $\mathcal{C}$.

The equality ${\eta }_{M}a={\mu }_{M}T({\eta }_M)$ is precisely the algebra-homomorphism equation for ${\eta }_M:(M,a)\to (TM,{\mu }_M)$, and step 1.2 shows it fails.

For that same algebra no algebra section exists at all. Under the isomorphism over $\mathbf{S}\mathbf{e}\mathbf{t}$ of [L3], an algebra map $s:(M,a)\to (TM,{\mu }_M)$ with $as=1_M$ is a monoid homomorphism $\sigma$ from $M$ to the free monoid on the set $M$ whose composite with word evaluation is the identity. Concatenation adds word lengths, so $w^2=w$ forces $\mathrm{\mid }w\mathrm{\mid }=0$ and the empty word is the only idempotent of that free monoid; since $e^2=e$ in $M$, $\sigma (e)$ is the empty word and evaluates to $1\mathrm{\ne }e$. Hence the presentation of $(M,a)$ is not split in ${\mathcal{C}}^T$.

Thus the canonical splittings always exist in the base by step 2.1, the canonical ones need not lift to algebra homomorphisms by step 2.2, and by step 2.3 the presentation itself need not be split in ${\mathcal{C}}^T$. No failure is asserted for every monad or every algebra.

By [L2], $q$, $Tq$, and $T^{2}q$ are coequalizers of the corresponding images of $f,g$. In particular each is an epimorphism, since every coequalizer is epic by its uniqueness clause.

Because $f$ and $g$ are algebra homomorphisms, $qb:TB\to C$ coequalizes $Tf$ and $Tg$. The universal property of $Tq$ gives a unique $c:TC\to C$ satisfying $cTq=qb$.

After precomposition with the epimorphism $q$, the unit equation $c{\eta }_C=1_C$ is the unit law for $b$. After precomposition with the epimorphism $T^{2}q$, the associativity equation $cT(c)=c{\mu }_C$ is the associativity law for $b$. Hence $(C,c)$ is a $T$-algebra.

The defining equation $cTq=qb$ says exactly that $q:(B,b)\to (C,c)$ is an algebra homomorphism.

If $r:(B,b)\to (D,d)$ is an algebra homomorphism with $rf=rg$, [L2] gives a unique underlying $u:C\to D$ with $uq=r$. Precomposing $uc$ and $dT(u)$ with the epimorphism $Tq$ gives the same map, so $u$ is an algebra homomorphism and is the unique algebraic factorization.

Any algebra structure $c^{\mathrm{\prime }}$ on the supplied apex for which $q$ is an algebra homomorphism satisfies $c^{\mathrm{\prime }}Tq=qb=cTq$, so $c^{\mathrm{\prime }}=c$ because $Tq$ is epic. Together with step 5.1 this is the unique on-the-nose lift required by [L1].

For a $T$-algebra $(A,a)$, the pair $F(TA)\rightrightarrows F(A)$ with arrows $F(a)$ and ${\epsilon }_{FA}$ has under $U$ the canonical pair $T(a),{\mu }_A:T^{2}A\rightrightarrows TA$. By [L2] and [L3] it is $U$-split.

Let $q_A:F(A)\to H(A,a)$ be the created coequalizer supplied for $(A,a)$. Its image is isomorphic to the canonical base coequalizer from step 1.1, and the universal property fixes the resulting comparisons.

An algebra homomorphism $r:(A,a)\to (B,b)$ carries the canonical pair for $A$ to that for $B$. The two coequalizer universal properties therefore define a unique morphism $H(r):H(A,a)\to H(B,b)$, and uniqueness proves preservation of identities and composition.

For $d\in \mathcal{D}$, the counit fork $F(TUd)\rightrightarrows F(Ud)\stackrel{{\epsilon }_d}{\to }d$ maps under $U$ to the canonical split presentation of $K(d)$. Creation reflects the lifted coequalizer, so comparison with $q_{K(d)}$ yields an isomorphism $HK(d)\cong d$.

The underlying fork of $K(q_A)$ is a split coequalizer, transported from the canonical split fork along the isomorphism supplied by ordinary creation. Since $K(q_A)$ is a lift of that fork, [L4] makes it a coequalizer in ${\mathcal{C}}^T$. The map $a:TA\to A$ is the other canonical algebra coequalizer by [L5], so their universal properties produce a unique isomorphism $KH(A,a)\cong (A,a)$.

The defining equations for the comparisons in steps 3.3 and 3.2 commute with every algebra homomorphism and every morphism of $\mathcal{D}$. Uniqueness of maps out of the coequalizers therefore makes both families natural, proving the two displayed natural isomorphisms.

For the forward direction, suppose $U$ is monadic. Then $K$ is an equivalence by [L1] and $U=U^{T}K$. Transporting the strict-creation result [L2] across $K$ by [L4] shows that $U$ creates coequalizers of $U$-split pairs in the ordinary sense.

For the converse, suppose $U$ creates coequalizers of $U$-split pairs and the stated family of created canonical coequalizers is supplied. By [L3], those data define a quasi-inverse to $K$.

A functor with a quasi-inverse and the two natural isomorphisms is an equivalence, so $K$ is an equivalence and $U$ is monadic by [L1].

Step 1.1 proves the monadic-to-creation implication, while steps 1.2 and 2.1 prove the data-supplied converse.

For the forward direction, if $K$ is an isomorphism then $U=U^{T}K$ on the nose. Transport through the inverse functor of $K$ preserves the exact apex, legs, and uniqueness in [L2], so $U$ has the strict-creation property in [L3].

For the reverse direction, [L4] makes each lifted canonical pair $U$-split. Strict creation [L3] gives a unique object $H(A,a)$ of $\mathcal{D}$ on the prescribed underlying apex $A$ and a coequalizer whose underlying map is $a:TA\to A$. Uniqueness and the coequalizer universal property define $H$ on algebra homomorphisms.

The functor $K(H(A,a))$ and the given algebra $(A,a)$ are lifts of the same split base fork; [L2] and the canonical coequalizer [L5] make the lift unique, so $KH=1_{{\mathcal{C}}^T}$ on objects and morphisms. For $d\in \mathcal{D}$, its counit fork is the existing lift of the canonical split fork of $K(d)$, so strict uniqueness gives $HK(d)=d$ and the same equality on morphisms. Hence $H$ is a two-sided inverse of $K$, and $U$ is strictly monadic by [L1].

Step 1.1 proves the forward implication and steps 1.2 and 2.1 prove the reverse implication, establishing the biconditional.

For a $T$-algebra $(A,a)$, the pair $F(TA)\rightrightarrows F(A)$ used in canonical reconstruction has common section $F({\eta }_A)$: one composite is the algebra unit law and the other is the adjunction triangle identity. Hence it is reflexive by [L1].

Use the supplied coequalizer $q_A:F(A)\to H(A,a)$ of this reflexive pair. By hypothesis, applying $U$ preserves it.

The preserved coequalizer $U{q}_A$ and the split canonical base coequalizer in [L3] coequalize the same pair, so transport of the splitting makes the underlying fork of $K(q_A)$ split. By [L5], $K(q_A)$ is a coequalizer in ${\mathcal{C}}^T$; by [L4], so is the canonical fork ending at $(A,a)$. Their universal properties therefore give an isomorphism $KH(A,a)\cong (A,a)$.

For $d\in \mathcal{D}$, compare the coequalizer of the reflexive counit pair with the counit fork ending at $d$. Their images under $U$ are isomorphic canonical coequalizers, so the comparison morphism becomes an isomorphism under $U$ and is itself an isomorphism by [L2].

The supplied object assignment and coequalizer universality define $H$ on algebra homomorphisms, and uniqueness makes the comparisons in steps 3.1 and 4.1 natural. Thus $H$ is a quasi-inverse to $K$, so $K$ is an equivalence and $U$ is monadic.

By [L3], $K(G)=(UG,U{\epsilon }_G)$ and $K(h)=U(h)$. Under [L4] the counit ${\epsilon }_G$ corresponds to the identity of $UG$, so it is the unique group homomorphism $F(UG)\to G$ carrying each basis element $g$ to $g$; a homomorphism out of a free group is determined by its values on the basis, so ${\epsilon }_G$ evaluates a reduced word in the elements of $G$ to its product in $G$. Hence $K(G)$ is the underlying set of $G$ with word evaluation.

A function $f:UG\to UH$ commutes with word evaluation exactly when it is a group homomorphism: evaluating the words $[x,y]$, the empty word and $[x^{-1}]$ turns commutation into preservation of product, identity and inverse, and conversely a homomorphism preserves the value of every word. With [L5] and $K(h)=U(h)$ this makes $K$ bijective on the morphisms between any two groups.

$K$ is injective on objects, since step 1.1 recovers the product of $G$ from $U{\epsilon }_G$ as its value on two-letter words.

$K$ is surjective on objects. Let $(A,a)$ be an algebra for the free-group monad. Put $x\cdot y:=a([x,y])$, $1:=a([])$ and $x^{-1}:=a([x^{-1}])$. Writing a word as the concatenation of its first letter with its tail and applying the multiplication law $aT(a)=a{\mu }_A$ of [L5] to the corresponding word of words gives $a(w)=a([x_1])\cdot a(\text{tail})$, while the unit law $a{\eta }_A=1_A$ gives $a([x])=x$; induction on length therefore identifies $a$ with evaluation of words in the operations just defined. Substituting the group-word identities into the same multiplication law turns them into associativity, the unit laws and the inverse laws, so those operations make $A$ a group $G_A$ with $a=U{\epsilon }_{G_A}$, that is $(A,a)=K(G_A)$.

By steps 2.1, 2.2 and 2.3 the comparison is bijective on objects and on morphisms, hence an isomorphism of categories over $\mathbf{S}\mathbf{e}\mathbf{t}$; the isomorphism over $\mathbf{S}\mathbf{e}\mathbf{t}$ asserted by [L1] may therefore be taken to be $K$. So $U$ is strictly monadic by [L2], and strict monadicity implies monadicity.

The isomorphism in [L1] sends a module to its underlying set with its finite-linear-combination algebra structure and acts identically on underlying functions. By the comparison formula in [L3], it is the comparison for the free-module adjunction, including for the zero ring and zero module.

The comparison is therefore an isomorphism over $\mathbf{S}\mathbf{e}\mathbf{t}$, so the underlying-set functor is strictly monadic by [L2].

Regard ${\mathbb{Z}}^{(X^{\ast })}$ as the free abelian group on the words. Define the product by bilinearly extending concatenation of basis words, equivalently by the displayed coefficient formula.

A finite word $w$ has only finitely many cuts $w=uv$, and finite supports leave only finitely many nonzero summands. Hence every coefficient sum is finite, and the product has finite support contained in the finite set of concatenations of support words.

For $\alpha ,\beta ,\gamma$, both coefficients of $(\alpha \beta )\gamma$ and $\alpha (\beta \gamma )$ are the sum of $\alpha (u)\beta (v)\gamma (r)$ over triples with $uvr=w$. The two bracketings are bijective reindexings of the same finite set, so [L3] proves associativity.

Splitting finite sums termwise proves $(\alpha +\beta )\gamma =\alpha \gamma +\beta \gamma$ and $\alpha (\beta +\gamma )=\alpha \beta +\alpha \gamma$. Together with the pointwise abelian-group structure over $\mathbb{Z}$, these are both distributive laws.

Let ${\delta }_{[]}$ be $1$ on the empty word and $0$ elsewhere. The only contributing cut with a nonzero ${\delta }_{[]}$ factor is $w=[]w$ or $w=w[]$, so ${\delta }_{[]}\alpha =\alpha =\alpha {\delta }_{[]}$. This verifies the multiplicative identity and all ring laws. If $X=\mathrm{\varnothing }$, then $X^{\ast }=\{[]\}$ and the coefficient at $[]$ identifies the ring with $\mathbb{Z}$.

Send a word $[x_1,\dots ,x_n]$ to $u(x_1)\cdots u(x_n)$ in $R$, and send the empty word to $1_R$. The generator $x$ is represented by the basis word $[x]$.

Apply [L2] over $\mathbb{Z}$ to extend this word-evaluation function uniquely to a $\mathbb{Z}$-linear map $\bar{u}:{\mathbb{Z}}^{(X^{\ast })}\to R$.

Expanding convolution as a finite sum shows $\bar{u}(\alpha \beta )=\bar{u}(\alpha )\bar{u}(\beta )$, and the empty-word basis vector maps to $1_R$. Thus $\bar{u}$ is a unit-preserving ring homomorphism, including when $R$ is the zero ring.

Any ring homomorphism extending $u$ must send every word basis vector to the corresponding product and is additive, so it equals $\bar{u}$ by uniqueness in [L2].

Hence the generator inclusion is a universal arrow from every set $X$ to the underlying-set functor on rings. By [L3] these universal arrows assemble into the free-ring functor and the asserted adjunction; for $X=\mathrm{\varnothing }$ the free ring is $\mathbb{Z}$.

By [L4], each finite Cartesian power $q^n$ is the coequalizer of $f^n,g^n$.

Each basic ring operation on $B$, followed by $q$, coequalizes the appropriate finite powers of $f$ and $g$. It therefore descends uniquely to $Q$; explicitly $0_Q=q(0_B)$, $1_Q=q(1_B)$, and negation, addition, and multiplication are the unique operations making $q$ preserve them.

Each ring axiom in [L2] becomes true after precomposition with the relevant surjection $q^n$, because it then becomes the corresponding axiom in $B$. Hence the descended operations make $Q$ a unital ring, including the zero-ring case.

By construction $q$ preserves addition, multiplication, and one, so it is a ring homomorphism by [L3].

If a ring homomorphism $r:B\to C$ coequalizes $f,g$, the set coequalizer gives a unique $u:Q\to UC$ with $uq=r$. Precomposing the preservation equations for $u$ with the appropriate $q^n$ reduces them to those for $r$, so $u$ is a ring homomorphism and is the unique algebraic factor.

The descended operations are uniquely forced by the requirement that the supplied $q$ be a ring homomorphism. Thus the lift has exactly the same apex and legs and is unique, which is strict creation.

By [L6], the comparison for the free-monoid adjunction is $K(M)=(UM,U{\epsilon }_M)$ with $K(h)=U(h)$. Under [L5] the counit ${\epsilon }_M$ corresponds to the identity of $UM$, so it is the unique monoid homomorphism $(UM{)}^{\ast }\to M$ carrying each one-letter word $[m]$ to $m$; a homomorphism out of a free monoid is determined on the letters, so ${\epsilon }_M$ evaluates a word in the elements of $M$ to its product.

For rings, [L2] supplies the left adjoint and [L3] supplies strict creation of the required split coequalizers.

Applying strict Beck [L4] to step 1.2 makes the ring underlying-set functor strictly monadic. The construction includes the empty generating set and the zero ring.

A function $f:UM\to UN$ commutes with word evaluation exactly when it is a monoid homomorphism, by evaluating the two-letter words and the empty word in one direction and every word in the other; with [L7] and $K(h)=U(h)$ this makes $K$ bijective on morphisms. It is injective on objects because step 1.1 recovers the product of $M$ from $U{\epsilon }_M$ on two-letter words.

$K$ is surjective on objects: for an algebra $(A,a)$ of the free-monoid monad put $x\cdot y:=a([x,y])$ and $1:=a([])$. Splitting a word into its first letter and its tail and applying the multiplication law $aT(a)=a{\mu }_A$ of [L7] gives $a(w)=a([x_1])\cdot a(\text{tail})$, while the unit law gives $a([x])=x$; induction on length identifies $a$ with evaluation of words in these operations, and substituting the monoid-word identities into the same law gives associativity and the unit laws. Hence $A$ is a monoid $M_A$ with $a=U{\epsilon }_{M_A}$, that is $(A,a)=K(M_A)$.

By steps 2.2 and 2.3 the monoid comparison is bijective on objects and morphisms, hence an isomorphism of categories over $\mathbf{S}\mathbf{e}\mathbf{t}$ — the isomorphism [L1] asserts may be taken to be it — so the monoid underlying-set functor is strictly monadic. With step 2.1 this proves the assertion for both concrete categories, and strict monadicity implies monadicity.

Define $F(X)=G\times X$ with action $h\cdot (g,x)=(hg,x)$. A function $u:X\to UY$ into a $G$-set extends uniquely to the equivariant map $\bar{u}(g,x)=g\cdot u(x)$, whose inverse correspondence evaluates at $(e,x)$. This natural bijection gives the free-action adjunction $F⊣U$.

Its induced monad is $T(X)=G\times X$, with $T(f)=1_G\times f$, ${\eta }_X(x)=(e,x)$, and ${\mu }_X(g,(h,x))=(gh,x)$. The group identity and associativity laws verify the monad equations.

A map $a:G\times X\to X$ satisfies the two algebra laws in [L3] exactly when $a(e,x)=x$ and $a(g,a(h,x))=a(gh,x)$, which are the action laws in [L1]. This includes the empty set, the trivial group, and trivial actions.

The algebra-homomorphism equation is $f(a(g,x))=b(g,f(x))$, exactly the equivariance condition in [L2].

By steps 3.1 and 4.1, the comparison for the adjunction constructed in step 1.1 is bijective on objects and morphisms, with inverse given by the same action structure and underlying functions. It is therefore an isomorphism over $\mathbf{S}\mathbf{e}\mathbf{t}$, so the underlying-set functor is strictly monadic by [L4].

If $y\in q[p^{-1}[A]]$, then some $w\in P$ satisfies $q(w)=y$ and $p(w)\in A$. The pullback equation gives $g(y)=gq(w)=fp(w)$, so $g(y)\in f[A]$ and $y\in g^{-1}[f[A]]$.

Conversely, if $y\in g^{-1}[f[A]]$, choose $x\in A$ with $f(x)=g(y)$. The pullback universal property supplies the unique $w\in P$ with $p(w)=x$ and $q(w)=y$, so $y\in q[p^{-1}[A]]$. If the fibre is empty, both existential conditions fail.

Steps 1.1 and 1.2 prove equality. Identity squares give the identity direct and inverse images; if one map is a section of the other, the same equality specializes to the usual section–retraction image formulas.