Tensor actions and absolute algebra presentations

Written by GPT-6.1 Sol (OpenAI), reasoning effort Ultra, September 2026. Self-checked; no independent review. Public domain (CC0).

An algebra object can act on objects of a category without those objects being vector spaces. Its action defines a monad. The resulting algebra objects have a canonical presentation by free algebras, and that presentation explains when a right adjoint recovers all the algebraic structure of its domain.

We use ordinary categories throughout. Tensor categories have an associator satisfying the pentagon, and a unit when one is specified. No symmetry or braiding is assumed. The unit and associativity interfaces are those of Stacks, monoidal categories, with the associator inverted to point from $(A\otimes B)\otimes C$ to $A\otimes(B\otimes C)$. For comparison with the monad constructions proved below, see Emily Riehl’s Category Theory in Context, Chapter 5.

1. The action of the unit detects unitality

An action of a tensor category $\mathsf T$ on $\mathsf C$ is a strong tensor functor $\mathsf T\to\operatorname{End}(\mathsf C)$, where the tensor product in the endofunctor category is composition. Write its evaluation as $A\odot X$. Its constraint is a natural isomorphism

\begin{gathered} \beta_{A,B,X}:(A\otimes B)\odot X \\ \longrightarrow A\odot(B\odot X). \end{gathered} \tag{1.1}

The action pentagon compares the two ways to apply (1.1) to three tensor factors. A unital action also has a compatible natural isomorphism $\lambda_X:I\odot X\to X$, where $I$ is the unit.

The definition using a cancellable idempotent unit asks that $I\otimes-$ and $-\otimes I$ be fully faithful, together with $\rho:I\otimes I\to I$ invertible. To compare this with the canonical convention, observe the following general fact.

Lemma 1.1. If $P:\mathsf C\to\mathsf C$ is fully faithful and $\nu:P^2\to P$ is a natural isomorphism, there is a unique natural isomorphism $e:P\to1_{\mathsf C}$ with $P(e_X)=\nu_X$.

Proof. Full faithfulness uniquely lifts $\nu_X:P(PX)\to P(X)$ to $e_X:PX\to X$. For $f:X\to Y$, naturality of $\nu$ says

P(f)P(e_X)=P(e_Y)P^2(f).

Faithfulness gives $fe_X=e_YP(f)$, proving naturality. A fully faithful functor reflects isomorphisms: fullness lifts an inverse to $P(e_X)$, and faithfulness makes the two lifted composites identities. Hence each $e_X$ is invertible. Uniqueness follows from the same fullness and faithfulness. $\square$

For $P=I\otimes-$, the associator and $\rho$ supply $P^2\simeq P$; similarly for $P=-\otimes I$. Lemma 1.1 makes both functors isomorphic to the identity, and therefore equivalences. This verifies the hypothesis of the canonical unit theorem, which uses equivalences. Its compatible unit constraints and coherence can consequently be used with the cancellable-idempotent convention.

For structural reassociations we retain the open monoidal coherence theorem: between two formal expressions in the same ordered list of objects, any two composites of associators, unit constraints and their inverses agree. The exact proof is Markus Himmel's Mathlib construction of FreeMonoidalCategory.subsingleton_hom, following the normalization argument of Ilya Beylin and Peter Dybjer, in Monoidal coherence, under Apache 2.0. Its full normalization proof gives a natural isomorphism from a formal expression to its normal form, in which the ordered factors remain fixed and units are erased. Naturality factors every structural map through that same normal form; its discrete category has at most one map between any two objects. This proves equality of parallel formal structural maps. The evaluation functor sends each formal associator and unit constraint to the corresponding one in our category and respects their defining relations, so it transfers the equality here. Distinct positions in the word remain distinct formal symbols even when their values happen to be equal objects. This is the precise coherence interface used below; it does not identify arbitrary morphisms.

Now set $P=I\odot-$ and

\nu_X=(\rho\odot1_X)\beta_{I,I,X}^{-1}:P^2X\to PX.

Theorem 1.2. The action is unital if and only if $P$ is fully faithful.

Proof. A unital action gives $P\simeq1_{\mathsf C}$, so $P$ is fully faithful. Conversely, Lemma 1.1 gives $e:P\simeq1_{\mathsf C}$. Its compatibility with the image of the unit multiplication is

e_X\nu_X=e_XP(e_X). \tag{1.2}

This holds by the defining equality $P(e_X)=\nu_X$. The right side is the component of the tensor square of $e$, followed by the multiplication of the identity endofunctor. Thus $e$ identifies $(P,\nu)$ with the unit of $\operatorname{End}(\mathsf C)$. The action is unital, with $\lambda=e$. $\square$

The criterion involves full faithfulness of the functor on $\mathsf C$, not a presupposed unit isomorphism. Exercise 2 gives an associative action whose unit functor fails the criterion.

2. Algebra objects become monads

Let the action be unital. An algebra object of $\mathsf T$ consists of $A$, a multiplication $m:A\otimes A\to A$, and a unit $u:I\to A$. With left and right unit constraints $l_A,r_A$, its unit laws are $m(u\otimes1_A)=l_A$ and $m(1_A\otimes u)=r_A$. Its associative law is

m(m\otimes1_A)=m(1_A\otimes m)a_{A,A,A}, \tag{2.1}

where $a$ is the associator. This is sometimes called a ring object, even when the ambient category has no addition.

Define $T(X)=A\odot X$, with

\begin{aligned} \eta_X&=(u\odot1_X)\lambda_X^{-1},\\ \mu_X&=(m\odot1_X)\beta_{A,A,X}^{-1}. \end{aligned} \tag{2.2}

These are natural transformations $1_{\mathsf C}\to T$ and $T^2\to T$. They satisfy the monad equations

\begin{gathered} \mu_X\eta_{TX}=1_{TX}=\mu_XT(\eta_X),\\ \mu_X\mu_{TX}=\mu_XT(\mu_X). \end{gathered} \tag{2.3}

A monad on $\mathsf C$ is precisely an algebra object of $\operatorname{End}(\mathsf C)$: its tensor multiplication is $\mu:T\circ T\to T$, its unit is $\eta:1_{\mathsf C}\to T$, and its algebra equations are (2.3).

Here is the associativity check with the parentheses retained. From $((A\otimes A)\otimes A)\odot X$ to $T^3X$, the action pentagon identifies

\begin{aligned} c_X&=\beta_{A,A,A\odot X}\, \beta_{A\otimes A,A,X}\\ &=(1_A\odot\beta_{A,A,X})\, \beta_{A,A\otimes A,X}\, (a_{A,A,A}\odot1_X). \end{aligned}

Naturality of $\beta$ shows that $\mu_X\mu_{TX}c_X$ is $(m(m\otimes1_A))\odot1_X$. It also shows that $\mu_XT(\mu_X)c_X$ is $(m(1_A\otimes m)a_{A,A,A})\odot1_X$. Equation (2.1) equates these maps; cancel $c_X$ to obtain the last equation of (2.3). For the first two equations, the action unit constraints identify insertion of $\eta_{TX}$ with insertion of $u$ on the left of $A$, and insertion of $T(\eta_X)$ with insertion on its right. After applying $\mu_X$, these are respectively the left and right unit laws for $m$. Their composites are $1_{TX}$.

A $T$-algebra is an object $X$ with $b:TX\to X$ such that $b\eta_X=1_X$ and $b\mu_X=bT(b)$. A map $f:(X,b)\to(Y,d)$ satisfies $fb=dT(f)$. For the monad (2.2), these equations say exactly that $b:A\odot X\to X$ is an $A$-module action, with its unit and associativity constraints. Thus $\operatorname{Mod}(A,\mathsf C)$ and $\mathsf C^T$ have the same objects and arrows. Their forgetful functor is faithful.

For an arbitrary monad, the free algebra on $Y$ is $(TY,\mu_Y)$. The free-algebra adjunction is given by the following inverse maps; compare Riehl, Lemma 5.2.9. The maps in its Hom bijection are

\begin{gathered} \operatorname{Hom}_{\mathsf C^T}((TY,\mu_Y),(X,b)) \\ \longrightarrow\operatorname{Hom}_{\mathsf C}(Y,X),\\ g\longmapsto g\eta_Y; \qquad f\longmapsto bT(f). \end{gathered} \tag{2.4}

Their checks make the interface explicit. The map $bT(f)$ is an algebra map, since naturality of $\mu$ and the law $b\mu_X=bT(b)$ give

\begin{aligned} bT(f)\mu_Y&=b\mu_XT^2(f)\\ &=bT(b)T^2(f)=bT(bT(f)). \end{aligned}

Also $bT(f)\eta_Y=b\eta_Xf=f$. Conversely, for an algebra map $g$,

bT(g\eta_Y)=bT(g)T(\eta_Y) =g\mu_YT(\eta_Y)=g.

The maps are natural in both variables. The free algebra laws themselves are the associativity and unit laws (2.3).

3. Presentations which every functor preserves

Let $h_X=\operatorname{Hom}_{\mathsf C}(-,X)$. Call a fork

V\overset{f,g}{\rightrightarrows}W \overset{q}{\longrightarrow}Z \tag{3.1}

presheaf exact if $h_V\rightrightarrows h_W\to h_Z$ is a coequalizer of presheaves. Equivalently, applying $\operatorname{Hom}_{\mathsf C}(H,-)$ gives a coequalizer of sets for every object $H$. This requires more than an ordinary coequalizer.

A presheaf-exact fork is an absolute coequalizer: every functor preserves it. This is the representable-colimit preservation theorem proved in Ind-objects through their elements, Section 3, Proposition 3.2. Apply that theorem to the parallel-pair diagram whose formal presheaf colimit is $h_Z$. No colimit-completeness of the receiving category is needed.

A useful sufficient condition is a split fork. Suppose there are $s:Z\to W$ and $t:W\to V$ such that

\begin{gathered} qf=qg,\qquad qs=1_Z,\\ gt=1_W,\qquad ft=sq. \end{gathered} \tag{3.2}

For $k:W\to H$ with $kf=kg$, the required factorization is $ks$, because $ksq=kft=kgt=k$. Uniqueness follows from $qs=1_Z$. Every functor preserves the identities (3.2), so preserves this coequalizer. In particular the covariant functor $\operatorname{Hom}(H,-)$ preserves it for every $H$; thus the fork is presheaf exact.

For a $T$-algebra $(X,b)$, use

T^2X\overset{T(b),\mu_X}{\rightrightarrows} TX\overset{b}{\longrightarrow}X. \tag{3.3}

The splitting has $s=\eta_X$, $t=\eta_{TX}$. Indeed $bT(b)=b\mu_X$, $b\eta_X=1_X$, $\mu_X\eta_{TX}=1_{TX}$, and

T(b)\eta_{TX}=\eta_Xb

by naturality of $\eta$. Hence (3.3) is presheaf exact for every monad and every algebra, including the monads arising from tensor actions.

The arrows in (3.3) are also maps of free algebras: $T(b)$ is the free-algebra image of $b$, and $\mu_X$ satisfies the algebra-map identity by the monad associativity law. The colimit construction that follows will therefore lift (3.3) to the algebra category. Its splittings are maps in $\mathsf C$; they need not be algebra homomorphisms. Exercise 3 distinguishes an ordinary coequalizer from a presheaf-exact one.

We will also use the following colimit-creation argument. Suppose $f,g:(V,v)\rightrightarrows(W,w)$ are algebra maps and their underlying fork (3.1) is presheaf exact. Absoluteness makes $Tq$ and $T^2q$ coequalizers too. Since

qwT(f)=qfv=qgv=qwT(g),

there is a unique $z:TZ\to Z$ with $zT(q)=qw$. It is an algebra structure: naturality and the algebra laws for $w$ give

\begin{aligned} z\eta_Zq&=zT(q)\eta_W=qw\eta_W=q,\\ z\mu_ZT^2(q)&=zT(q)\mu_W =qw\mu_W,\\ zT(z)T^2(q)&=zT(zT(q))\\ &=zT(qw)=qwT(w). \end{aligned}

The last two right sides agree. Cancel the epimorphisms $q$ and $T^2q$ to obtain $z\eta_Z=1_Z$ and $z\mu_Z=zT(z)$.

For an algebra map $k:(W,w)\to(H,h)$ equalizing $f,g$, its unique underlying factor $j:Z\to H$ satisfies

jzT(q)=jqw=kw=hT(k)=hT(j)T(q).

Cancel $Tq$ to see that $j$ is an algebra map. Thus $q$ lifts uniquely to a coequalizer of algebras. This proves the colimit-creation assertion needed below, including the use of both $Tq$ and $T^2q$; compare Riehl’s Theorem 5.6.5(ii).

4. The presheaf-exact monadicity criterion

Let $L:\mathsf C\to\mathsf D$ be left adjoint to $R:\mathsf D\to\mathsf C$, with unit $\eta$ and counit $\varepsilon$. The canonical adjunction-to-monad construction, Riehl’s Lemma 5.1.3, gives

T=RL,\qquad \mu=R\varepsilon L.

The two monad unit laws are the adjunction triangles, and associativity is naturality of $\varepsilon$ at $\varepsilon_{LX}$, after applying $R$. The comparison functor, Riehl’s Proposition 5.2.13, is

\begin{gathered} K:\mathsf D\to\mathsf C^T,\\ Y\longmapsto(RY,R(\varepsilon_Y)). \end{gathered} \tag{4.1}

On maps it is $R$. The algebra unit law is $R(\varepsilon_Y)\eta_{RY}=1$, and its associative law is naturality of $\varepsilon$ at $\varepsilon_Y$. Naturality at any $f:Y\to Y'$ verifies the algebra-map condition for $R(f)$. Thus the formulas give the specified comparison, with $UK=R$ and $KL$ the free-algebra functor.

Theorem 4.1. The comparison $K$ is an equivalence if and only if both conditions hold:

$R$ is conservative: an arrow whose $R$-image is invertible is invertible.
For every parallel pair $f,g: V\rightrightarrows W$ in $\mathsf D$, if its $R$-image admits a coequalizer $q:RW\to Z$ which is presheaf exact in $\mathsf C$, then $f,g$ have a coequalizer in $\mathsf D$, and $R$ preserves it.

Proof. The algebra forgetful functor reflects isomorphisms. If an algebra map $f:(X,b)\to(Y,d)$ has an underlying inverse $h$, then $fb=dT(f)$, composed with $h$ and $T(h)$, gives $hd=bT(h)$. Thus $h$ is its inverse as an algebra map. If $K$ is an equivalence, this proves condition 1 for $R=UK$. Section 3 constructs every presheaf-exact coequalizer in the algebra category and preserves its underlying fork. Transporting it along $K$ proves condition 2. This establishes necessity with the full presheaf-exact hypothesis.

Suppose now that conditions 1 and 2 hold. For each algebra $(X,b)$, form the pair $LTX\overset{L(b),\varepsilon_{LX}}{\rightrightarrows}LX. \tag{4.2}$ Its $R$-image is $T(b),\mu_X:T^2X\rightrightarrows TX$, whose coequalizer $b:TX\to X$ is split by (3.3). Condition 2 therefore gives a coequalizer $p_X:LX\to J(X,b)$ and makes $R(p_X)$ its underlying coequalizer. The unique comparison $u_X:RJ(X,b)\longrightarrow X, \qquad u_XR(p_X)=b$ is an isomorphism in $\mathsf C$, since both sides are coequalizers of the same pair. It is an algebra isomorphism too. Indeed $K(p_X)$ is an algebra map from the free algebra $(TX,\mu_X)$. Section 3 gives the unique algebra structure on the coequalizer for which this map is an algebra map. Under $u_X$, the structure is $b$, since (3.3) is that same algebra coequalizer. This proves that every algebra is isomorphic to a value of $K$.

Next we prove that each counit is a coequalizer in $\mathsf D$: $LTR Y\overset{LR(\varepsilon_Y),\varepsilon_{LRY}} {\rightrightarrows}LRY \overset{\varepsilon_Y}{\longrightarrow}Y.$ Naturality of the counit makes this a fork. Its $R$-image is (3.3) for the algebra $(RY,R\varepsilon_Y)$. Condition 2 gives some coequalizer $p:LRY\to Y_0$, preserved by $R$. The fork $\varepsilon_Y$ factors uniquely as $dp$. Applying $R$, both $Rp$ and $R\varepsilon_Y$ are coequalizers of the same pair, so $Rd$ is an isomorphism. Conservativity makes $d$ an isomorphism. Consequently $\varepsilon_Y$ itself is the required coequalizer, hence is epic.

Let $a:K(Y)\to K(Z)$ be an algebra map. Naturality of $\varepsilon$ and the equation $aR\varepsilon_Y=R\varepsilon_ZT(a)$ show that $\varepsilon_ZL(a):LRY\to Z$ equalizes the displayed counit pair: its composites are respectively $\varepsilon_ZL(aR\varepsilon_Y)$ and $\varepsilon_ZLR\varepsilon_ZLT(a)$, equal by that equation. The counit coequalizer therefore gives a unique $h:Y\to Z$ with $h\varepsilon_Y=\varepsilon_ZL(a).$ Applying $R$ gives $Rh\,R\varepsilon_Y=R\varepsilon_ZT(a)=aR\varepsilon_Y$. The last map $R\varepsilon_Y$ is split epic, so $Rh=a$. Hence $K$ is full. If $Rh=Rh'$, counit naturality gives $h\varepsilon_Y=\varepsilon_ZLRh=\varepsilon_ZLRh'=h'\varepsilon_Y$; cancel the epic $\varepsilon_Y$ to obtain $h=h'$. Thus $K$ is faithful. Together with essential surjectivity already proved, these assertions make it an equivalence.

The inverse is the concrete construction $J$ above. An algebra map $a:(X,b)\to(X',b')$ makes $L(a)$ a map of the two presentation pairs: one square uses the algebra-map equation, and the other uses counit naturality. It therefore induces the unique map between their coequalizers. This uniqueness gives identities and composition, so $J$ is a functor. The comparisons $u_X$ are natural by their equations on $R(p_X)$. The counit coequalizers give the other natural comparison $JK\simeq1_{\mathsf D}$. This proves the criterion and its inverse without appealing to an external monadicity theorem. $\square$

Corollary 4.2. If $\mathsf D$ has finite colimits, $R$ is conservative, and $R$ preserves finite colimits, then $K$ is an equivalence.

Proof. Every parallel pair has a coequalizer in $\mathsf D$, and its image is a coequalizer in $\mathsf C$. Thus condition 2 holds whenever its antecedent holds, and Theorem 4.1 applies. $\square$

If right exactness is defined by filtered comma categories, the finite-colimit interface proved in One-sided fractions and saturation, Section 4, converts it into finite-colimit preservation here. Therefore the usual “conservative and exact” hypothesis with finite colimits also gives this corollary; its left-exact part is unnecessary for the conclusion.

5. Exercises and complete solutions

Exercise 1 (Grade 1: a monad which adjoins a point). On sets define $TX=X\sqcup\{*\}$, with $\eta_X$ the inclusion and $\mu_X:T^2X\to TX$ collapsing the two added points to $*$. Verify the monad laws and identify its algebras and free algebras.

Solution. On a function $f:X\to Y$, $Tf$ acts as $f$ on $X$ and carries the added point to the added point. This gives a functor, and makes $\eta,\mu$ natural. Either insertion of one new layer followed by $\mu$ fixes $X$ and the already present point; both unit laws follow. On $T^3X$, each associative composite fixes every element of $X$ and sends all three added points to the single added point of $TX$. Thus they agree.

An algebra map $b:TX\to X$ is the identity on $X$, by $b\eta_X=1$, and chooses $x_0=b(*)$. Its associative law holds: on $X$ both maps are identities; on either added point of $T^2X$, both give $x_0$. Conversely every chosen point gives such an algebra. A morphism of algebras is exactly a function preserving the chosen point. The empty set has no algebra, since there is no function $T\varnothing\to\varnothing$. The free algebra on $Y$ is $Y\sqcup\{*\}$ with distinguished point $*$, and a map from it to $(X,x_0)$ extends any function $Y\to X$ uniquely by sending $*$ to $x_0$.

Exercise 2 (Grade 2: an associative action without a unit action). Let $\mathsf T$ have one object $I$ and just its identity, with $I\otimes I=I$. On $\mathsf{Set}$, let $P$ be the constant singleton functor. Show that $I\mapsto P$ defines a tensor action, and that it is not unital.

Solution. Choose the same singleton at every occurrence. Then $P^2=P$, and the tensor constraint $P\simeq P^2$ is its identity. All naturality and associativity diagrams have singleton components, so commute. This gives the strong tensor functor, hence the action. The functor $P$ is not faithful: the two functions from a singleton to a two-element set have the same image. Therefore it is not fully faithful, and Theorem 1.2 rules out unitality. Directly, $P(\varnothing)$ is a singleton and is not isomorphic to $\varnothing$, so no natural unit isomorphism can exist.

Exercise 3 (Grade 3: ordinary coequalizers need not be presheaf exact). Let $B=\{0,1\}$, let $\sigma$ exchange its elements, and take $q:B\to\{*\}$. Prove that $q$ coequalizes $1_B,\sigma$ in sets, but that the fork is not presheaf exact.

Solution. A function $k:B\to H$ satisfies $k=k\sigma$ exactly when it is constant. Such functions factor uniquely through $q$, including when $H$ is empty, since then neither kind of function exists. Thus $q$ is the coequalizer.

Apply $\operatorname{Hom}(B,-)$. Its value on $B$ has four functions: the two constants $c_0,c_1$, the identity, and $\sigma$. The induced parallel maps are the identity and postcomposition by $\sigma$. Their coequalizer has two classes, $\{c_0,c_1\}$ and $\{1_B,\sigma\}$. But $\operatorname{Hom}(B,\{*\})$ has one element. Hence the resulting set fork is not a coequalizer, and the original fork is not presheaf exact. The covariant functor $\operatorname{Hom}(B,-)$ also witnesses the failure of absoluteness directly.

Exercise 4 (Grade 4: the induced monad can miss topology). Let $L:\mathsf{Set}\to\mathsf{Top}$ give the discrete topology and let $R:\mathsf{Top}\to\mathsf{Set}$ forget topology. Compute the induced monad and comparison functor. Explain why the comparison fails to be an equivalence using both conservativity and full faithfulness.

Solution. Every function from a discrete space into a topological space is continuous, so $L\dashv R$. The underlying set $RLX$ is exactly $X$; the unit is the identity function. The counit $LRY\to Y$ is also the identity on underlying sets, so $\mu=R\varepsilon L$ is the identity. Thus the monad is the identity monad. Its algebra condition forces the structure map $b:X\to X$ to be $1_X$, and every function is an algebra morphism. Consequently its algebra category is $\mathsf{Set}$, and $K$ is the underlying-set functor.

On a two-element set, the identity from the discrete topology to the indiscrete topology is continuous and bijective. Its inverse is not continuous, because a singleton is open in the discrete target and is not open in the indiscrete source. Therefore $R$ is not conservative, violating condition 1 of Theorem 4.1. Also the underlying identity function from the indiscrete space to the discrete space is a set map without a continuous lift. Thus $K$ is not full and cannot be an equivalence. The induced monad records no topology.

References

Monoidal categories and module categories over them are treated in Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik, Tensor Categories, author's final version, Chapters 2 and 7, and monads and their algebras in Emily Riehl, Category Theory in Context, Chapter 5. Theorem 4.1 keeps the full condition involving the Yoneda image of the coequalizer.

Stacks, Section 4.43 supplies the canonical associativity and unit constraints. Emily Riehl, Category Theory in Context, second edition, Chapter 5, discusses the free-algebra adjunction, algebra presentations, monadicity and creation of monad-preserved colimits. These references give mathematical credit and comparison with the constructions and complete proofs above; their text is not reproduced.