The Duality Behind the Equivalence of Modules and Comodules

From Right Duality to Actions and Coactions

Duality in a monoidal category reverses both the direction of morphisms and the order of tensor factors. Consequently, if a monoid admits a right dual, its right dual naturally becomes a comonoid.

Duality also transforms actions of the monoid into coactions of the corresponding comonoid, while leaving the underlying objects unchanged.

In the monoidal category of endofunctors, these statements become:

• A right adjoint of a monad naturally carries a comonad structure.

• The category of algebras for the monad is isomorphic to the category of coalgebras for the corresponding comonad.

We first explain the general monoidal construction, then verify the correspondence between algebras and coalgebras directly.

The terminology requires some care. A monoid is an algebra object in a monoidal category. An algebra for a monad is an object equipped with an action of that monad. For example, algebras for the monad $A\otimes -$$A\otimes-$ are left $A$$A$-modules.

Right Duals and Conventions

Let $(\mathcal{M},\otimes ,\mathbf{1})$$(\mathcal M,\otimes,\mathbf 1)$ be a monoidal category. We suppress associativity and unit constraints, interpreting all expressions using monoidal coherence.

A right dual of an object $L$$L$ is an object $R$$R$ equipped with morphisms

$$\mathrm{ev}:L\otimes R⟶\mathbf{1},\mathrm{coev}:\mathbf{1}⟶R\otimes L$$

such that

$$(\mathrm{ev}\otimes {\mathrm{id}}_L)\circ ({\mathrm{id}}_L\otimes \mathrm{coev})={\mathrm{id}}_L$$

and

$$({\mathrm{id}}_R\otimes \mathrm{ev})\circ (\mathrm{coev}\otimes {\mathrm{id}}_R)={\mathrm{id}}_R.$$

These are the snake identities. We use this convention for right duals throughout.

Suppose $X$$X$ and $Y$$Y$ have chosen right duals. For a morphism $f:X\to Y$$f:X\to Y$, its right dual

$$f^r:Y^r⟶X^r$$

is the composite

$$\begin{array}{rl}{Y}^r& \stackrel{{\mathrm{coev}}_X\otimes \mathrm{id}}{\to }{X}^r\otimes X\otimes Y^r\\ & \stackrel{\mathrm{id}\otimes f\otimes \mathrm{id}}{\to }{X}^r\otimes Y\otimes Y^r\\ & \stackrel{\mathrm{id}\otimes {\mathrm{ev}}_Y}{\to }{X}^r.\end{array}$$

The snake identities imply

$$(g\circ f{)}^r=f^r\circ g^r,({\mathrm{id}}_X{)}^r={\mathrm{id}}_{X^r}.$$

There are also canonical identifications

$$(X\otimes Y{)}^r\cong Y^r\otimes X^r,{\mathbf{1}}^r\cong \mathbf{1}.$$

For example, evaluation between $X\otimes Y$$X\otimes Y$ and $Y^r\otimes X^r$$Y^r\otimes X^r$ first evaluates the middle pair $Y\otimes Y^r$$Y\otimes Y^r$, then evaluates $X\otimes X^r$$X\otimes X^r$. Thus, reversing the tensor order is intrinsic to duality; no braiding is required.

There are therefore two reversals to keep track of: duality reverses morphisms and reverses the order of tensor factors.

The Right Dual of a Monoid

Let $L$$L$ be a monoid with multiplication and unit

$$\mu :L\otimes L⟶L,u:\mathbf{1}⟶L.$$

Suppose $L$$L$ has a right dual $R$$R$. Taking right duals of $\mu$$\mu$ and $u$$u$, define

$$\mathrm{\Delta }:R\stackrel{{\mu }^r}{\to }(L\otimes L{)}^r\stackrel{\sim }{\to }R\otimes R$$

and

$$e:R\stackrel{u^r}{\to }{\mathbf{1}}^r\stackrel{\sim }{\to }\mathbf{1}.$$

We verify that these maps make $R$$R$ a comonoid.

Associativity of $\mu$$\mu$ says

$$\mu \circ (\mu \otimes {\mathrm{id}}_L)=\mu \circ ({\mathrm{id}}_L\otimes \mu ).$$

Under the canonical duality identifications,

$$(\mu \otimes {\mathrm{id}}_L{)}^r={\mathrm{id}}_R\otimes \mathrm{\Delta },$$

and

$$({\mathrm{id}}_L\otimes \mu {)}^r=\mathrm{\Delta }\otimes {\mathrm{id}}_R.$$

Taking right duals of the associativity equation therefore gives

$$({\mathrm{id}}_R\otimes \mathrm{\Delta })\circ \mathrm{\Delta }=(\mathrm{\Delta }\otimes {\mathrm{id}}_R)\circ \mathrm{\Delta }.$$

This is coassociativity.

Similarly, the unit identities

$$\mu \circ (u\otimes {\mathrm{id}}_L)={\mathrm{id}}_L=\mu \circ ({\mathrm{id}}_L\otimes u)$$

become

$$({\mathrm{id}}_R\otimes e)\circ \mathrm{\Delta }={\mathrm{id}}_R=(e\otimes {\mathrm{id}}_R)\circ \mathrm{\Delta }.$$

Hence $(R,\mathrm{\Delta },e)$$(R,\Delta,e)$ is a comonoid.

Only the existence of a right dual of $L$$L$ is required. The ambient monoidal category need not be rigid, and $R$$R$ itself need not admit a further right dual.

Duality in the Category of Endofunctors

Consider the monoidal category

$$(\mathrm{End}(\mathcal{C}),\circ ,{\mathrm{Id}}_{\mathcal{C}}),$$

whose tensor product is composition of functors.

A right dual $R$$R$ of an endofunctor $L$$L$ consists of natural transformations

$$\epsilon :LR⟶{\mathrm{Id}}_{\mathcal{C}},\eta :{\mathrm{Id}}_{\mathcal{C}}⟶RL$$

satisfying

$$(\epsilon L)\circ (L\eta )={\mathrm{id}}_L,(R\epsilon )\circ (\eta R)={\mathrm{id}}_R.$$

These are exactly the data and triangle identities of an adjunction $L⊣R$$L\dashv R$.

Thus we have the following dictionary:

Monoidal notion	Interpretation in $\mathrm{End}(\mathcal{C})$$\operatorname{End}(\mathcal C)$
Tensor product	Composition of functors
Monoid	Monad
Comonoid	Comonad
Right dual	Right adjoint
Right dual of a morphism	Right mate of a natural transformation

Consequently, if $(L,\mu ,u)$$(L,\mu,u)$ is a monad and $L⊣R$$L\dashv R$, the preceding construction equips $R$$R$ with a comonad structure

$$(R,\delta ,e),\delta :R⟶R^2,e:R⟶{\mathrm{Id}}_{\mathcal{C}}.$$

We now verify that this construction also gives an isomorphism

$$\mathrm{Alg}(L)\cong \mathrm{Coalg}(R)$$

preserving underlying objects and morphisms.

Algebras and Coalgebras: A Direct Verification

Fix the adjunction $L⊣R$$L\dashv R$, with unit $\eta$$\eta$ and counit $\epsilon$$\varepsilon$. We continue to write $u$$u$ for the unit of the monad.

These are different natural transformations:

$$u:{\mathrm{Id}}_{\mathcal{C}}\to L,\eta :{\mathrm{Id}}_{\mathcal{C}}\to RL.$$

For $f:LX\to Y$$f:LX\to Y$, its adjoint transpose is

$${\mathrm{\Phi }}_{X,Y}(f)=R(f)\circ {\eta }_X:X⟶RY.$$

The inverse transpose sends $g:X\to RY$$g:X\to RY$ to

$${\epsilon }_Y\circ L(g):LX⟶Y.$$

Applying the adjunction twice gives a bijection

$$\begin{array}{rl}{\mathrm{\Psi }}_{X,Y}:\mathcal{C}(L^{2}X,Y)& \stackrel{\sim }{\to }\mathcal{C}(X,R^{2}Y),\\ h& ⟼R^2(h)\circ R({\eta }_{LX})\circ {\eta }_X.\end{array}$$

The induced comonad structure can be written explicitly as

$$e_Y={\epsilon }_Y\circ u_{RY},$$

and

$${\delta }_Y={\mathrm{\Psi }}_{RY,Y}({\epsilon }_Y\circ {\mu }_{RY}).$$

The latter is the double transpose of

$$L^{2}RY\stackrel{{\mu }_{RY}}{\to }LRY\stackrel{{\epsilon }_Y}{\to }Y.$$

Structure maps. Given a morphism $a:LX\to X$$a:LX\to X$, define

$$b=R(a)\circ {\eta }_X:X⟶RX.$$

Conversely, given $b:X\to RX$$b:X\to RX$, define

$$a={\epsilon }_X\circ L(b):LX⟶X.$$

These assignments are mutually inverse. Indeed,

$$\begin{array}{rl}{\epsilon }_X\circ L(R(a)\circ {\eta }_X)& ={\epsilon }_X\circ LR(a)\circ L({\eta }_X)\\ & =a\circ {\epsilon }_{LX}\circ L({\eta }_X)\\ & =a,\end{array}$$

where the second equality uses naturality of $\epsilon$$\varepsilon$, and the last uses a triangle identity.

Similarly,

$$\begin{array}{rl}R({\epsilon }_X\circ L(b))\circ {\eta }_X& =R({\epsilon }_X)\circ RL(b)\circ {\eta }_X\\ & =R({\epsilon }_X)\circ {\eta }_{RX}\circ b\\ & =b.\end{array}$$

Here we use naturality of $\eta$$\eta$ and the other triangle identity.

The unit axiom. For corresponding structure maps $a$$a$ and $b$$b$, we have

$$\begin{array}{rl}{e}_X\circ b& =e_X\circ R(a)\circ {\eta }_X\\ & =a\circ e_{LX}\circ {\eta }_X\\ & =a\circ {\epsilon }_{LX}\circ u_{RLX}\circ {\eta }_X\\ & =a\circ {\epsilon }_{LX}\circ L({\eta }_X)\circ u_X\\ & =a\circ u_X.\end{array}$$

The second line uses naturality of $e$$e$, the fourth uses naturality of $u$$u$, and the final line uses a triangle identity.

Therefore,

$$a\circ u_X={\mathrm{id}}_X⟺e_X\circ b={\mathrm{id}}_X.$$

The associativity axiom. The algebra associativity axiom compares two morphisms $L^{2}X\to X$$L^2X\to X$:

$$a\circ {\mu }_X=a\circ L(a).$$

We transpose each side separately.

Since $a={\epsilon }_X\circ L(b)$$a=\varepsilon_X\circ L(b)$, naturality of $\mu$$\mu$ gives

$$\begin{array}{rl}a\circ {\mu }_X& ={\epsilon }_X\circ L(b)\circ {\mu }_X\\ & ={\epsilon }_X\circ {\mu }_{RX}\circ L^2(b).\end{array}$$

By naturality of $\mathrm{\Psi }$$\Psi$ in its source variable and the definition of $\delta$$\delta$,

$$\begin{array}{rl}{\mathrm{\Psi }}_{X,X}(a\circ {\mu }_X)& ={\mathrm{\Psi }}_{RX,X}({\epsilon }_X\circ {\mu }_{RX})\circ b\\ & ={\delta }_X\circ b.\end{array}$$

For the other side, substitute directly into the formula for $\mathrm{\Psi }$$\Psi$:

$$\begin{array}{rl}{\mathrm{\Psi }}_{X,X}(a\circ L(a))& =R^2(a)\circ R^{2}L(a)\circ R({\eta }_{LX})\circ {\eta }_X\\ & =R^2(a)\circ R({\eta }_X)\circ R(a)\circ {\eta }_X\\ & =R(R(a)\circ {\eta }_X)\circ (R(a)\circ {\eta }_X)\\ & =R(b)\circ b.\end{array}$$

The second equality follows from naturality of $\eta$$\eta$ applied to $a:LX\to X$$a:LX\to X$:

$$RL(a)\circ {\eta }_{LX}={\eta }_X\circ a.$$

Since ${\mathrm{\Psi }}_{X,X}$$\Psi_{X,X}$ is a bijection,

$$a\circ {\mu }_X=a\circ L(a)⟺{\delta }_X\circ b=R(b)\circ b.$$

The equation on the right is precisely the coalgebra coassociativity axiom.

Together with the unit calculation, this proves that $a$$a$ defines an $L$$L$-algebra structure if and only if $b$$b$ defines an $R$$R$-coalgebra structure.

Morphisms. Suppose $(X,a)$$(X,a)$ and $(Y,a^{\prime })$$(Y,a')$ correspond to $(X,b)$$(X,b)$ and $(Y,b^{\prime })$$(Y,b')$. For a morphism $f:X\to Y$$f:X\to Y$, naturality of the adjunction gives

$${\mathrm{\Phi }}_{X,Y}(f\circ a)=R(f)\circ b,$$

and

$${\mathrm{\Phi }}_{X,Y}(a^{\prime }\circ L(f))=b^{\prime }\circ f.$$

Therefore,

$$f\circ a=a^{\prime }\circ L(f)⟺R(f)\circ b=b^{\prime }\circ f.$$

Thus the same underlying map is an algebra morphism exactly when it is a morphism between the corresponding coalgebras.

The mutually inverse assignments

$$(X,a)⟼(X,R(a)\circ {\eta }_X)$$

and

$$(X,b)⟼(X,{\epsilon }_X\circ L(b))$$

act identically on underlying morphisms. They therefore define an isomorphism

$$\mathrm{Alg}(L)\cong \mathrm{Coalg}(R).$$

Modules and Comodules in a Monoidal Category

Return to a general monoidal category $\mathcal{M}$$\mathcal M$. Let $L$$L$ be a monoid with right dual $R$$R$, equipped with the comonoid structure constructed above.

The duality data give an adjunction

$$L\otimes -⊣R\otimes -.$$

Algebras for the monad $L\otimes -$$L\otimes-$ are left $L$$L$-modules, while coalgebras for the corresponding comonad $R\otimes -$$R\otimes-$ are left $R$$R$-comodules. Hence

$${}_L\mathrm{Mod}(\mathcal{M})\cong {}^R\mathrm{Comod}(\mathcal{M}).$$

Explicitly, a left action $a:L\otimes X\to X$$a:L\otimes X\to X$ corresponds to the left coaction

$$\begin{array}{rl}\rho :X& \stackrel{\mathrm{coev}\otimes {\mathrm{id}}_X}{\to }R\otimes L\otimes X\\ & \stackrel{{\mathrm{id}}_R\otimes a}{\to }R\otimes X.\end{array}$$

The inverse construction is

$$\begin{array}{rl}a:L\otimes X& \stackrel{{\mathrm{id}}_L\otimes \rho }{\to }L\otimes R\otimes X\\ & \stackrel{\mathrm{ev}\otimes {\mathrm{id}}_X}{\to }X.\end{array}$$

More generally, let $\mathcal{D}$$\mathcal D$ be a left $\mathcal{M}$$\mathcal M$-module category, with action denoted by $\odot$$\odot$. The same duality data give

$$L\odot -⊣R\odot -,$$

and consequently identify left $L$$L$-module objects in $\mathcal{D}$$\mathcal D$ with left $R$$R$-comodule objects in $\mathcal{D}$$\mathcal D$.

The object $X$$X$ on which the action is defined does not need to have a dual.

An Example and a Check of Conventions

Let $A$$A$ be an algebra over a field $k$$k$. On ${\mathrm{Vect}}_k$$\operatorname{Vect}_k$, there is an adjunction

$$A\otimes -⊣{\mathrm{Hom}}_k(A,-).$$

The action of a left $A$$A$-module $M$$M$ corresponds to

$$\beta :M⟶{\mathrm{Hom}}_k(A,M),\beta (m)(a)=a\cdot m.$$

The corresponding comonad structure is

$$e_V(f)=f(1_A),{\delta }_V(f)(a)(b)=f(ba).$$

The counit axiom for $\beta$$\beta$ is therefore

$$1_A\cdot m=m.$$

Evaluating its coassociativity axiom at $a,b\in A$$a,b\in A$ gives

$$(ba)\cdot m=b\cdot (a\cdot m).$$

These formulas hold for arbitrary $A$$A$, with no finite-dimensionality assumption. Here the relevant right dual is the right adjoint of the endofunctor $A\otimes -$$A\otimes-$.

If $A$$A$ is finite-dimensional, then $A$$A$ itself also has a right dual $A^{\ast }$$A^*$ in ${\mathrm{Vect}}_k$$\operatorname{Vect}_k$, and

$${\mathrm{Hom}}_k(A,V)\cong A^{\ast }\otimes V.$$

At this point, the order of tensor factors matters.

Using the order-reversing right-dual identification

$$(A\otimes A{)}^r\cong A^r\otimes A^r,$$

the resulting comultiplication satisfies

$$\sum {\phi }_{(1)}(a){\phi }_{(2)}(b)=\phi (ba).$$

By contrast, the usual linear-dual coalgebra structure ${\mathrm{\Delta }}_{\mathrm{std}}$$\Delta_{\mathrm{std}}$ is characterized by

$$\sum {\phi }_{[1]}(a){\phi }_{[2]}(b)=\phi (ab).$$

The two differ by the symmetry:

$$\mathrm{\Delta }=\tau \circ {\mathrm{\Delta }}_{\mathrm{std}}.$$

Thus the construction above directly identifies left $A$$A$-modules with left comodules over the opposite coalgebra

$$(A^{\ast },{\mathrm{\Delta }}_{\mathrm{std}}{)}^{\mathrm{cop}}.$$

Applying the symmetry to the two factors of the coaction yields the familiar formulation

$${}_A\mathrm{Mod}\cong {\mathrm{Comod}}^{(A^{\ast },{\mathrm{\Delta }}_{\mathrm{std}})}.$$

The correspondence between left modules and right comodules is therefore consistent with the left-action/left-coaction formulation above. The difference is accounted for by the reversal of tensor order in right duality.

References

• P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, §2.10, especially Definition 2.10.2, Exercises 2.10.4 and 2.10.7, and Proposition 2.10.8. Author's version.

• G. Böhm, T. Brzeziński, and R. Wisbauer, Monads and comonads in module categories, §§2.2 and 2.6. arXiv:0804.1460.