Duality and Rigidity in Monoidal Categories

Reuse the same two morphisms ${\mathrm{ev}}_X:X^{\vee }\otimes X\to \mathbf{1}$ and ${\mathrm{coev}}_X:\mathbf{1}\to X\otimes X^{\vee }$, but now regard them as candidate right-dual data for the object $X^{\vee }$ with proposed right dual $X$.

The first right-dual zig-zag for $X^{\vee }$ is exactly the second left-dual zig-zag for $X$, and the second right-dual zig-zag for $X^{\vee }$ is exactly the first left-dual zig-zag for $X$. Both are identities by the hypotheses that $X^{\vee }$ is a left dual of $X$.

Hence $X$ is a right dual of $X^{\vee }$. The dual assertion is the same argument with left and right interchanged.

Take the evaluation to be ${\lambda }_{\mathbf{1}}:\mathbf{1}\otimes \mathbf{1}\to \mathbf{1}$ and the coevaluation to be ${\lambda }_{\mathbf{1}}^{-1}:\mathbf{1}\to \mathbf{1}\otimes \mathbf{1}$. By [L1], this is the same pair as ${\rho }_{\mathbf{1}}$ and ${\rho }_{\mathbf{1}}^{-1}$.

Substituting these maps into either zig-zag composite gives an instance of the triangle identity with every object equal to $\mathbf{1}$, so each composite is the identity of $\mathbf{1}$.

Therefore $\mathbf{1}$ is a left dual of itself, and because the same maps also satisfy the mirrored unit equations, it is a right dual of itself as well.

Suppose $Y$ is a left dual of $X$ in ${\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$, with evaluation $Y{\otimes }^{\mathrm{r}\mathrm{e}\mathrm{v}}X=X\otimes Y\to \mathbf{1}$ and coevaluation $\mathbf{1}\to X{\otimes }^{\mathrm{r}\mathrm{e}\mathrm{v}}Y=Y\otimes X$.

Writing the two left-dual zig-zag composites in ${\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$ and then translating them with [L1] replaces ${\otimes }^{\mathrm{r}\mathrm{e}\mathrm{v}}$ by reversed tensor order, ${\alpha }^{\mathrm{r}\mathrm{e}\mathrm{v}}$ by ${\alpha }^{-1}$, and the reverse unitors by the ordinary opposite ones. The result is exactly the pair of right-dual zig-zag composites for $Y$ as a right dual of $X$ in $\mathcal{C}$.

Therefore the left-dual axioms in ${\mathcal{C}}^{\mathrm{r}\mathrm{e}\mathrm{v}}$ are equivalent to the right-dual axioms in $\mathcal{C}$. The converse and the left/right-swapped statement are the same calculation in reverse.

Define $\varphi :X_1^{\vee }\to X_2^{\vee }$ by the composite $X_1^{\vee }\stackrel{{\rho }^{-1}}{\to }{X}_1^{\vee }\otimes \mathbf{1}\stackrel{1\otimes {\mathrm{coev}}_2}{\to }{X}_1^{\vee }\otimes (X\otimes X_2^{\vee })\stackrel{{\alpha }^{-1}}{\to }(X_1^{\vee }\otimes X)\otimes X_2^{\vee }\stackrel{{\mathrm{ev}}_1\otimes 1}{\to }\mathbf{1}\otimes X_2^{\vee }\stackrel{\lambda }{\to }{X}_2^{\vee },$ and define $\psi :X_2^{\vee }\to X_1^{\vee }$ by the same formula with the subscripts interchanged.

Postcomposing the definition of $\varphi$ with ${\mathrm{ev}}_2$ and precomposing it with ${\mathrm{coev}}_1$, then using the zig-zag identities from [L1], yields the two compatibility equations in the statement. The same calculation with $\psi$ gives the analogous equations for $\psi$.

The composite $\psi \varphi$ is the unique morphism $X_1^{\vee }\to X_1^{\vee }$ compatible with ${\mathrm{ev}}_1$ and ${\mathrm{coev}}_1$, and the identity morphism has that same compatibility by [L1]. Expanding one copy of $\varphi$ and one copy of $\psi$ and then straightening with the zig-zag identities shows that $\psi \varphi =1_{X_1^{\vee }}$; similarly $\varphi \psi =1_{X_2^{\vee }}$. Thus $\varphi$ is an isomorphism with inverse $\psi$.

If $\chi :X_1^{\vee }\to X_2^{\vee }$ is any other morphism satisfying the two compatibility equations, insert $\chi$ into the formula of step 1.1 and use those compatibilities to collapse the same zig-zag composites; the result is $\chi =\varphi$. Hence the compatible isomorphism is unique. The right-dual statement is the mirror argument.

For each object $U$, define a unit ${\eta }_U:U\to V\otimes (V^{\vee }\otimes U)$ by $U\stackrel{{\lambda }_U^{-1}}{\to }\mathbf{1}\otimes U\stackrel{{\mathrm{coev}}_V\otimes 1_U}{\to }(V\otimes V^{\vee })\otimes U\stackrel{{\alpha }_{V,V^{\vee },U}}{\to }V\otimes (V^{\vee }\otimes U)\mathrm{.}$ For each object $W$, define a counit ${\epsilon }_W:V^{\vee }\otimes (V\otimes W)\to W$ by $V^{\vee }\otimes (V\otimes W)\stackrel{{\alpha }_{V^{\vee },V,W}^{-1}}{\to }(V^{\vee }\otimes V)\otimes W\stackrel{{\mathrm{ev}}_V\otimes 1_W}{\to }\mathbf{1}\otimes W\stackrel{{\lambda }_W}{\to }W\mathrm{.}$

The composite $(V\otimes {\epsilon }_W)\circ {\eta }_{V\otimes W}$ is exactly the first zig-zag for $V^{\vee }$ tensored with $W$, and the composite ${\epsilon }_{V^{\vee }\otimes U}\circ (V^{\vee }\otimes {\eta }_U)$ is exactly the second zig-zag for $V^{\vee }$ tensored with $U$. By [L2], both are identities.

Steps 1.1 and 2.1 provide an adjunction $V^{\vee }\otimes -⊣V\otimes -$ by [L1]. Transposition under this adjunction gives the displayed hom-set bijection, and the statement for right tensoring is the mirrored construction.

Under the tensor-by-composition convention, the data named in [L1] and [L2] are literally the same pair of natural transformations with the same sources and targets.

The two zig-zag identities for a left dual in the composition monoidal structure are exactly the two triangle identities for an adjunction, because both say that the composites $F\stackrel{\mathrm{coev}F}{\Rightarrow }FLF\stackrel{F\mathrm{ev}}{\Rightarrow }F\text{and}L\stackrel{L\mathrm{coev}}{\Rightarrow }LFL\stackrel{\mathrm{ev}L}{\Rightarrow }L$ are identities.

Therefore $L$ is a left dual of $F$ exactly when $L⊣F$. The same comparison with the mirrored data shows that $R$ is a right dual of $F$ exactly when $F⊣R$.

By [L1], the two adjunctions $L_1⊣F$ and $L_2⊣F$ make $L_1$ and $L_2$ into two left duals of the same object $F$ in the endofunctor composition monoidal category.

Applying [L2] to those two duals yields a unique compatible isomorphism $L_1\cong L_2$. Compatibility with the duality data is exactly compatibility with the units and counits of the two adjunctions by [L1].

Hence left adjoints of a fixed functor are unique up to unique compatible natural isomorphism. The right-adjoint statement is the same argument with right duals.

By [L3], choose a basis $(v_1,\dots ,v_n)$ of $V$. Let $(v_1^{\ast },\dots ,v_n^{\ast })$ be the dual basis from [L2], and define $\mathrm{ev}:V^{\ast }\otimes V\to k,f\otimes v\mapsto f(v),$ $\mathrm{coev}:k\to V\otimes V^{\ast },1\mapsto {\sum }_{i=1}^n{v}_i\otimes v_i^{\ast }\mathrm{.}$

For each basis vector $v_j$, the first zig-zag sends $v_j$ to ${\sum }_{i=1}^n{v}_i^{\ast }(v_j)v_i={\sum }_{i=1}^n{\delta }_{ij}{v}_i=v_j\mathrm{.}$ By linearity it is the identity on $V$.

For each dual basis vector $v_j^{\ast }$, the second zig-zag sends $v_j^{\ast }$ to ${\sum }_{i=1}^n{v}_j^{\ast }(v_i)v_i^{\ast }={\sum }_{i=1}^n{\delta }_{ji}{v}_i^{\ast }=v_j^{\ast }\mathrm{.}$ By linearity it is the identity on $V^{\ast }$.

Thus $V^{\ast }$ is a left dual of $V$. The same formulas, read in the mirrored order, make $V^{\ast }$ a right dual of $V$ as well, so $V$ is rigid. Since $V$ was arbitrary and [L1] supplies the monoidal structure, every finite-dimensional object of ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$ is rigid.

Expanding the definition from [L1] with $f=1_X$ and then straightening the resulting coevaluation-evaluation pair by the zig-zag identities shows $(1_X{)}^{\vee }=1_{X^{\vee }}$.

The usual tensoring of the chosen dual pairs gives evaluation and coevaluation maps exhibiting $Y^{\vee }\otimes X^{\vee }$ as a left dual of $X\otimes Y$. Since $(X\otimes Y{)}^{\vee }$ is the chosen left dual of the same object, [L3] gives a unique compatible isomorphism $(X\otimes Y{)}^{\vee }\cong Y^{\vee }\otimes X^{\vee }\mathrm{.}$

For morphisms $X\stackrel{f}{\to }Y\stackrel{g}{\to }Z$, the defining composite for $(g\circ f{)}^{\vee }$ contains the block $g\circ f$ between one coevaluation and one evaluation. Splitting that block into $g$ followed by $f$ yields exactly the composite for $f^{\vee }\circ g^{\vee }$, so $(g\circ f{)}^{\vee }=f^{\vee }\circ g^{\vee }$. Hence $(-{)}^{\vee }$ is contravariant.

By [L4], the tensor unit $\mathbf{1}$ is itself a left dual of $\mathbf{1}$. Applying [L3] to the chosen left dual ${\mathbf{1}}^{\vee }$ and this canonical one gives a unique compatible isomorphism ${\mathbf{1}}^{\vee }\cong \mathbf{1}$. Together with step 1.2, this supplies the unit comparison, so $(-{)}^{\vee }$ is antimonoidal.

By [L1], $(-{)}^{\vee }:{\mathcal{C}}^{\mathrm{o}\mathrm{p}}\to \mathcal{C}$ is contravariant, so applying it twice yields a covariant endofunctor $(-{)}^{\vee \vee }:\mathcal{C}\to \mathcal{C}$.

Again by [L1], there are compatible isomorphisms $(X\otimes Y{)}^{\vee }\cong Y^{\vee }\otimes X^{\vee }$ and ${\mathbf{1}}^{\vee }\cong \mathbf{1}$. Dualizing once more reverses the order a second time, so the composite comparison gives $X^{\vee \vee }\otimes Y^{\vee \vee }\cong (Y^{\vee }\otimes X^{\vee }{)}^{\vee }\cong (X\otimes Y{)}^{\vee \vee }$ together with a unit isomorphism $\mathbf{1}\cong {\mathbf{1}}^{\vee \vee }$.

Since the monoidal comparison maps are obtained by composing those of the antimonoidal functor with itself, their coherence is inherited from the coherence in [L1]. Therefore $(-{)}^{\vee \vee }$ is a monoidal endofunctor.

Because $\mathcal{C}$ is rigid, choose for each object $X$ a left dual $X^{\vee }$. Since $F$ and $G$ are strong monoidal, they send the duality maps of $X$ to duality maps of $F(X)$ and $G(X)$: after transporting the images of ${\mathrm{ev}}_X$ and ${\mathrm{coev}}_X$ across the strong monoidal structure isomorphisms, $F(X^{\vee })$ is a left dual of $F(X)$ and $G(X^{\vee })$ is a left dual of $G(X)$.

Define ${\xi }_X:G(X)\to F(X)$ using the transported duality maps by the composite $G(X)\cong \mathbf{1}\otimes G(X)\stackrel{{\mathrm{coev}}_{F(X)}\otimes 1}{\to }F(X)\otimes F(X^{\vee })\otimes G(X)\stackrel{1\otimes {\eta }_{X^{\vee }}\otimes 1}{\to }F(X)\otimes G(X^{\vee })\otimes G(X)\stackrel{1\otimes {\mathrm{ev}}_{G(X)}}{\to }F(X)\otimes \mathbf{1}\cong F(X)\mathrm{.}$ This construction uses $F(X^{\vee })$ and $G(X^{\vee })$ as left duals of $F(X)$ and $G(X)$ from step 1.1; it does not identify a left dual of $F(X^{\vee })$ with $F(X)$.

Expand ${\xi }_X{\eta }_X$ using step 2.1. Naturality of $\eta$, together with its tensor and unit compatibility from [L2], moves ${\eta }_X$ across the coevaluation and evaluation; the remaining composite is the zig-zag identity for the dual pair $F(X^{\vee }),F(X)$. Hence ${\xi }_X{\eta }_X=1_{F(X)}$. The mirrored calculation uses the zig-zag identity for $G(X^{\vee }),G(X)$ and gives ${\eta }_X{\xi }_X=1_{G(X)}$. Thus every component ${\eta }_X$ is an isomorphism, so $\eta$ is a natural isomorphism.

The displayed composite is well typed because rigidity provides ${\mathrm{coev}}_{X^{\vee }}:\mathbf{1}\to X^{\vee }\otimes X^{\vee \vee }$ and ${\mathrm{ev}}_X:X^{\vee }\otimes X\to \mathbf{1}$, while the braiding supplies the middle swap. Naturality of the braiding in [L1] makes the assignment $X\mapsto u_X$ natural in $X$.

After inserting the monoidal comparison $d_{X,Y}$ from [L2], both sides of the displayed tensor relation are morphisms $X\otimes Y\to (X\otimes Y{)}^{\vee \vee }$. The relation in [F1] is exactly this compatibility in a convention that suppresses the comparison. In particular, it records the precise obstruction to $u$ being monoidal: the double braiding appears between $u_{X\otimes Y}$ and the transported map $u_X\otimes u_Y$.

The explicit inverse in [F1] is defined under the same braided-rigid hypotheses as the displayed map, so every $u_X$ is an isomorphism. Together with step 1.1, this makes $u$ a natural isomorphism in every braided rigid monoidal category; step 1.2 records why it need not be monoidal.

By [L1], both composites in the statement are well typed in any pivotal category, and the extra spherical hypothesis is exactly the one assumed in [F1].

Applying [F1] to the present spherical tensor category gives ${\mathrm{Tr}}_L(a_{X}x)={\mathrm{Tr}}_R(x{a}_X^{-1})\mathrm{.}$

Hence in a spherical category the left and right traces agree after inserting the pivotal comparison.

The formula in [L2] for ${\mathrm{Tr}}_L(a)$ is obtained by inserting $a$ between one coevaluation and one evaluation. Dualizing that composite and using the contravariant antimonoidality from [L1] reverses the order and turns it into the defining formula for ${\mathrm{Tr}}_R(a^{\vee })$, which is the first clause recorded in [F1].

The same proposition [F1] states that direct sums split the trace additively in additive categories, tensor products split it multiplicatively, and composing with an endomorphism may be cycled across the trace at the cost of a double dual: ${\mathrm{Tr}}_L(ac)={\mathrm{Tr}}_L(c^{\vee \vee }a)\mathrm{.}$

Therefore all four displayed identities hold, and the right-trace versions follow by applying the same argument to the mirrored formulas in [L2].

By [L1], the family $u_X$ is invertible, so every natural isomorphism ${\psi }_X:X\to X^{\vee \vee }$ can be written uniquely as $u_X{\theta }_X$ for a natural automorphism $\theta$ of the identity.

The tensor relation for $u$ from the Drinfeld-morphism theorem inserts exactly one double braiding between $u_{X\otimes Y}$ and $u_X\otimes u_Y$. Therefore the condition that $\psi$ be monoidal is equivalent to the condition that $\theta$ absorb that double braiding, namely ${\theta }_{X\otimes Y}=({\theta }_X\otimes {\theta }_Y)\circ c_{Y,X}\circ c_{X,Y},$ which is the twist law from [L3].

Hence $\psi$ is a pivotal structure exactly when $\theta$ is a twist.

Let $k$ be a field and let ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$ be the category of all $k$-vector spaces with the usual tensor product and unit object $k$. This is a monoidal category, and $V:=k[x]$ is infinite-dimensional by [L1].

Assume for contradiction that $V$ had a left dual $(V^{\vee },\mathrm{ev},\mathrm{coev})$. Because $\mathrm{coev}(1)\in V\otimes V^{\vee }$ is a single tensor in an algebraic tensor product, it is a finite sum ${\sum }_{i=1}^m{v}_i\otimes f_i$ with $f_i\in V^{\vee }$. Applying the first zig-zag identity to any $v\in V$ gives $v={\sum }_{i=1}^m\mathrm{ev}(f_i\otimes v)v_i,$ so every vector of $V$ lies in the span of the finite set $\{v_1,\dots ,v_m\}$.

Step 1.2 makes $V$ finite-dimensional, contradicting that $k[x]$ has no finite basis. Therefore $V$ has no left dual, hence is not rigid, so ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$ is a monoidal category that is not rigid.

Let $\mathbb{Z}$ be the ordered set of integers regarded as a small category, so there is a unique morphism $m\to n$ exactly when $m\le n$. Let $D(n)=2n$, $L(n)=\lceil n\mathrm{/}2\rceil$, $R(n)=\lfloor n\mathrm{/}2\rfloor$, and $H(n)=2n-1$. Directly from the defining inequalities, $L(m)\le n⟺m\le D(n),D(m)\le n⟺m\le R(n),H(m)\le n⟺m\le L(n)\mathrm{.}$ Thus $L⊣D⊣R$ and $H⊣L$.

Let $\mathcal{R}$ be the full subcategory of $\mathrm{End}(\mathbb{Z})$ generated under finite composition by the identity functor and all endofunctors in the bi-infinite adjoint chain obtained by repeatedly taking left and right adjoints of $D$. It is a monoidal subcategory by construction. Each generator has both adjacent adjoints in the chain, and a finite composite of functors with left and right adjoints has the corresponding reversed composites as its left and right adjoints. Those composites again belong to $\mathcal{R}$, and their units and counits lie in the full subcategory. Hence every object of $\mathcal{R}$ has both duals and $\mathcal{R}$ is rigid.

In the endofunctor category of a poset, a natural transformation $F\Rightarrow G$ exists exactly when $F(n)\le G(n)$ for all $n$, so two endofunctors are isomorphic exactly when they are equal pointwise. Here $L(1)=1>0=R(1)$, so there is no natural transformation $L\Rightarrow R$ and therefore no isomorphism $L\cong R$. Thus the object $D$ has nonisomorphic left and right duals.

If the chosen left duality on $\mathcal{R}$ uses left adjoints, then $D^{\vee }=L$ and $D^{\vee \vee }=H$. Since $H(0)=-1\mathrm{\ne }0=D(0)$, the endofunctors $H$ and $D$ are not isomorphic. Therefore in this rigid monoidal category neither left/right duals nor double duals are forced to collapse.

By the hypothesis ladder in What is needed before a trace can be written, categorical trace is first defined on a morphism $a:X\to X^{\vee \vee }$ or $b:X^{\vee \vee }\to X$, so rigidity is already required before any trace expression is even well typed.

Even in a rigid category, an endomorphism $x:X\to X$ does not itself have the required source and target. One needs a pivotal structure to insert a comparison $a_X:X\to X^{\vee \vee }$ or $a_X^{-1}:X^{\vee \vee }\to X$, as recorded in Pivotal structure.