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.

Bending a crossing. For any object $Y$, naturality at evaluation and coevaluation, followed by the two hexagons and the zig-zags, gives $c_{X^{\vee },Y}=({\mathrm{ev}}_X\otimes 1_Y\otimes 1_{X^{\vee }})(1_{X^{\vee }}\otimes c_{X,Y}^{-1}\otimes 1_{X^{\vee }})(1_{X^{\vee }}\otimes 1_Y\otimes {\mathrm{coev}}_X)\mathrm{.}$ Indeed expand $c_{Y,X\otimes X^{\vee }}=(1_X\otimes c_{Y,X^{\vee }})(c_{Y,X}\otimes 1_{X^{\vee }})$ in naturality at ${\mathrm{coev}}_X$, then tensor on the left with $X^{\vee }$ and apply ${\mathrm{ev}}_X$ to cancel the adjacent cup and cap. Replacing $c_{A,B}$ throughout by $c_{B,A}^{-1}$ gives the companion identity for $c_{Y,X^{\vee }}^{-1}$. These are algebraic identities, independent of a graphical calculus.

A second left dual of $X^{\vee }$. Put $e_X^{\mathrm{\prime }}={\mathrm{ev}}_X{c}_{X,X^{\vee }}:X\otimes X^{\vee }\to \mathbf{1}$ and $h_X^{\mathrm{\prime }}=c_{X^{\vee },X}^{-1}{\mathrm{coev}}_X:\mathbf{1}\to X^{\vee }\otimes X$. These give $X$ as a left dual of $X^{\vee }$. To check the first zig-zag, expand $(e_X^{\mathrm{\prime }}\otimes 1_X)(1_X\otimes h_X^{\mathrm{\prime }})$; move the crossing incident to the cup across that cup by step 1.1, cancel the resulting inverse crossing pair, and the remaining composite is $(1_X\otimes {\mathrm{ev}}_X)({\mathrm{coev}}_X\otimes 1_X)=1_X$. For the other zig-zag expand $(1_{X^{\vee }}\otimes e_X^{\mathrm{\prime }})(h_X^{\mathrm{\prime }}\otimes 1_{X^{\vee }})$; the companion bending identity cancels the crossing pair and leaves $({\mathrm{ev}}_X\otimes 1_{X^{\vee }})(1_{X^{\vee }}\otimes {\mathrm{coev}}_X)=1_{X^{\vee }}$.

Comparison and inverse. If $(A,e_A,h_A)$ and $(B,e_B,h_B)$ are left duals of an object $Z$, the maps $(e_A\otimes 1_B)(1_A\otimes h_B):A\to B$ and $(e_B\otimes 1_A)(1_B\otimes h_A):B\to A$ are inverse: substitution of either composite permits first the $Z,B$ zig-zag and then the $A,Z$ zig-zag, leaving the identity. They are the unique maps intertwining the evaluations, since tensoring that intertwining equation with $h_B$ recovers the displayed formula. Apply this to $Z=X^{\vee }$, $A=X$ from step 2.1, and $B=X^{\vee \vee }$ with its chosen duality. The comparison is precisely the displayed $u_X$, and its inverse is $v_X=({\mathrm{ev}}_{X^{\vee }}\otimes 1_X)(1_{X^{\vee \vee }}\otimes c_{X^{\vee },X}^{-1})(1_{X^{\vee \vee }}\otimes {\mathrm{coev}}_X)\mathrm{.}$ Thus $v_X{u}_X=1_X$ and $u_X{v}_X=1_{X^{\vee \vee }}$. For $f:X\to Y$, the mate equations for $f,f^{\vee }$, and naturality of $c$ show that both $u_{Y}f$ and $f^{\vee \vee }{u}_X$ intertwine the same evaluation pairing; uniqueness of the comparison gives $u_{Y}f=f^{\vee \vee }{u}_X$.

Tensor relation. Work first with the canonical tensor dual $(X\otimes Y{)}^{\vee }=Y^{\vee }\otimes X^{\vee }$. Its evaluation is ${\mathrm{ev}}_Y(1_{Y^{\vee }}\otimes {\mathrm{ev}}_X\otimes 1_Y)$, and the coevaluation of its dual is $(1_{Y^{\vee }}\otimes {\mathrm{coev}}_{X^{\vee }}\otimes 1_{Y^{\vee \vee }}){\mathrm{coev}}_{Y^{\vee }}$. Insert these into the formula for $u_{X\otimes Y}$. Expand its middle braiding using, successively, $c_{A\otimes B,C}=(c_{A,C}\otimes 1_B)(1_A\otimes c_{B,C})$ and $c_{A,B\otimes C}=(1_B\otimes c_{A,C})(c_{A,B}\otimes 1_C)$. This produces the four crossings of $X,Y$ with $Y^{\vee },X^{\vee }$. Bend the $X,Y^{\vee }$ crossing across its $Y$ evaluation and the $Y,X^{\vee }$ crossing across its $X$ evaluation by the two identities of step 1.1. The remaining $X,X^{\vee }$ and $Y,Y^{\vee }$ crossings and their cups and caps are respectively $u_X$ and $u_Y$, while the two bent crossings on the input are $c_{X,Y}^{-1}{c}_{Y,X}^{-1}$. Consequently the expanded composite is $u_{X\otimes Y}=(u_X\otimes u_Y)c_{X,Y}^{-1}{c}_{Y,X}^{-1}$. Reinsert the chosen-dual comparison $d_{X,Y}$ from [L2] and multiply on the right by $c_{Y,X}{c}_{X,Y}$ to obtain exactly the stated identity. Steps 3.1 and 4.1 prove the natural-isomorphism and obstruction claims.

Sliding a morphism. The defining mate identities for $f:U\to V$ are $(f\otimes 1_{U^{\vee }}){\mathrm{coev}}_U=(1_V\otimes f^{\vee }){\mathrm{coev}}_V$ and ${\mathrm{ev}}_U(f^{\vee }\otimes 1_U)={\mathrm{ev}}_V(1_{V^{\vee }}\otimes f)$. Apply the first identity to $f=a:X\to X^{\vee \vee }$. Postcomposing with ${\mathrm{ev}}_{X^{\vee }}$ gives ${\mathrm{Tr}}_L(a)={\mathrm{ev}}_{X^{\vee }}(1_{X^{\vee \vee }}\otimes a^{\vee }){\mathrm{coev}}_{X^{\vee \vee }}={\mathrm{Tr}}_R(a^{\vee })$, where the last trace is taken on $X^{\vee }$. This proves clause 1 without identifying a double dual with the original object.

Biproducts. In an additive rigid category, tensoring by a fixed object has both adjoints, furnished by its two duals, and hence preserves finite biproducts; duality carries a biproduct to the corresponding biproduct of duals. Write $i_U,p_U$ for the injections and projections of $Z=X\oplus Y$. Relative to $Z^{\vee }=X^{\vee }\oplus Y^{\vee }$, its coevaluation is ${\sum }_{U=X,Y}(i_U\otimes p_U^{\vee }){\mathrm{coev}}_U$, and its evaluation pairs matching summands by ${\mathrm{ev}}_U$ and different summands by zero. These formulas satisfy the zig-zags since $p_U{i}_V$ is the identity for $U=V$ and zero otherwise, so uniqueness of duals permits their use for the chosen dual. Inserting $a\oplus b$ leaves exactly the two diagonal trace composites. Their sum is ${\mathrm{Tr}}_L(a)+{\mathrm{Tr}}_L(b)$, proving clause 2.

Nested tensor trace. Use $(X\otimes Y{)}^{\vee }=Y^{\vee }\otimes X^{\vee }$. The coevaluation of $X\otimes Y$ is $(1_X\otimes {\mathrm{coev}}_Y\otimes 1_{X^{\vee }}){\mathrm{coev}}_X$. The evaluation of its dual $Y^{\vee }\otimes X^{\vee }$ first pairs the middle $Y^{\vee \vee },Y^{\vee }$ and then the outer $X^{\vee \vee },X^{\vee }$. Substitute $a\otimes b$ into this composite. Evaluating the middle pair gives $s={\mathrm{Tr}}_L(b)$ situated between $X^{\vee \vee }$ and $X^{\vee }$. Thus the result is exactly ${\mathrm{ev}}_{X^{\vee }}(r_{X^{\vee \vee }}(s)a\otimes 1_{X^{\vee }}){\mathrm{coev}}_X={\mathrm{Tr}}_L(r_{X^{\vee \vee }}(s)a)$. If $r_{X^{\vee \vee }}(s)={\mathrm{\ell }}_{X^{\vee \vee }}(s)$, the inserted scalar instead acts on the left of the whole evaluation input. Naturality of that left action at ${\mathrm{ev}}_{X^{\vee }}$ moves it to the output unit, giving $s{\mathrm{Tr}}_L(a)$. Endomorphisms of the unit commute under composition by interchange, so this equals the asserted product. No equality of left and right actions is assumed in the general nested formula.

Cyclicity, including different objects. More generally, for $f:U\to V$ and $h:V\to U^{\vee \vee }$, the first mate identity in step 1.1 and then the second identity applied to $f^{\vee }$ give ${\mathrm{ev}}_{U^{\vee }}(hf\otimes 1){\mathrm{coev}}_U={\mathrm{ev}}_{U^{\vee }}(h\otimes f^{\vee }){\mathrm{coev}}_V={\mathrm{ev}}_{V^{\vee }}(f^{\vee \vee }h\otimes 1){\mathrm{coev}}_V$. Thus ${\mathrm{Tr}}_{L,U}(hf)={\mathrm{Tr}}_{L,V}(f^{\vee \vee }h)$. Taking $U=V=X$, $f=c$ and $h=a$ proves clause 4. The mirrored mate identities give the corresponding right-trace cyclicity.