Tensor duality and coherent inverses

Evaluation and coevaluation let us move a tensor factor across a Hom set. They also distinguish a chosen tensor inverse from a coherent inverse: two isomorphisms to the unit must satisfy two compatibility equations before they supply duality.

We use the associator \(a:(XY)Z\to X(YZ)\) and unit maps \(l_X:IX\to X\), \(r_X:XI\to X\) of Tensor actions and absolute algebra presentations. Juxtaposition denotes tensor product, never a permutation of factors. When parentheses or units are omitted, all maps use the canonical associativity and unit identifications. No braiding, additive structure or assumption that every object has a dual is made.

1. Units and their compatible identifications

The source's unit convention specifies \(\rho:II\to I\) invertible and requires \(I\otimes-\) and \(-\otimes I\) to be fully faithful. The preceding lesson proves that both functors are then equivalences: apply its fully faithful idempotent-functor lemma to their natural isomorphisms \(P^2\simeq P\). This checks the equivalence hypothesis of Stacks, Lemma 4.43.1.

We retain its unit construction and Lemma 4.43.4. In our associator direction, their constraints obey \[ \begin{aligned} l_I&=r_I=\rho,\\ l_{XY}&=l_X\otimes1_Y,\\ r_{XY}&=1_X\otimes r_Y,\\ r_X\otimes1_Y&=1_X\otimes l_Y . \end{aligned}\tag{1.1} \] The sources of the last three equalities are respectively \(IXY\), \(XYI\) and \(XIY\), with the canonical reassociations inserted. Naturality gives the additional compatibility \[ \begin{aligned} l_X(1_I\otimes r_X) &=r_X(l_X\otimes1_I). \end{aligned}\tag{1.2} \] Equation (1.2) compares maps \(IXI\to X\). These are all the unit coherences used below. Uniqueness is also explicit: tensoring on the left with \(I\) determines \(l_X\) from \(1_I\otimes l_X=\rho\otimes1_X\), and tensoring on the right determines \(r_X\) from \(r_X\otimes1_I=1_X\otimes\rho\).

Two units with their multiplications have a unique multiplication-compatible isomorphism, as in Stacks, Lemma 4.43.2. Here is the normalization and uniqueness check in that assertion. Start with any isomorphism \(t:I\to J\), which the unit constraints provide. There is a unique \(b\in\operatorname{Aut}(J)\) such that \[ t\rho_I=b\rho_J(t\otimes t). \] The canonical unit identities give \[ \rho_J(b\otimes b)=b^2\rho_J. \] Consequently \(bt\) is multiplication-compatible. If two compatible maps differ by \(c\in\operatorname{Aut}(J)\), their compatibility implies \(c\rho_J=c^2\rho_J\). Cancellation gives \(c=1_J\). This supplies the uniqueness check as well as the adjustment in the canonical proof. Also \(\rho_I\otimes1_I=1_I\otimes\rho_I\), again by the same lemma.

For a strong tensor functor with comparison \(\xi_{A,B}:F(AB)\to F(A)F(B)\), the multiplication on its image unit is \[ F(I)F(I)\xrightarrow{\xi_{I,I}^{-1}} F(II)\xrightarrow{F(\rho_I)}F(I). \tag{1.3} \] The functor is unital when this pair is a unit of its target. Equivalently, the pair has its compatible identification with the target unit. This is the source form of Stacks, Definition 4.43.5: its tensor comparison is \(\xi^{-1}\), and the source full-faithfulness condition for the image unit is equivalent to the canonical equivalence condition just checked.

All reassociations and unit removals below use the pinned open coherence proof explained in Tensor actions and absolute algebra presentations, §1. It identifies structural maps with the ordered factors fixed; it does not identify arbitrary morphisms. Etingof, Gelaki, Nikshych and Ostrik, Tensor Categories, Theorem 2.9.2, printed page 40, is a further mathematical reference.

2. The two directions of tensor duality

A dual pair consists of \(X,Y\), a coevaluation \(\varepsilon:I\to YX\), and an evaluation \(\eta:XY\to I\), with \[ \begin{aligned} &l_X(\eta\otimes1_X)a_{X,Y,X}^{-1}\\ &\qquad(1_X\otimes\varepsilon)r_X^{-1}=1_X, \end{aligned}\tag{2.1} \] and \[ \begin{aligned} &r_Y(1_Y\otimes\eta)a_{Y,X,Y}\\ &\qquad(\varepsilon\otimes1_Y)l_Y^{-1}=1_Y. \end{aligned}\tag{2.2} \] Thus \(X\) is a left dual of \(Y\), and \(Y\) is a right dual of \(X\). This is Stacks, Definition 4.43.8, with its object names interchanged and its evaluation and coevaluation renamed. Equations (2.1)–(2.2) check both directions of the associator in that interface.

We retain Stacks, Lemma 4.43.9. In this notation its two natural bijections are \[ \begin{aligned} \operatorname{Hom}(Z,WX) &\simeq\operatorname{Hom}(ZY,W),\\ \operatorname{Hom}(XZ,W) &\simeq\operatorname{Hom}(Z,YW). \end{aligned}\tag{2.3} \] Here is the complete map and inverse check, including the second bijection omitted in the printed proof.

For the first bijection the maps, with coherence identifications understood, are \[ \begin{aligned} A(f)&=(1_W\otimes\eta)(f\otimes1_Y),\\ B(g)&=(g\otimes1_X)(1_Z\otimes\varepsilon). \end{aligned}\tag{2.4} \] The first composite is \[ \begin{aligned} AB(g) &=(1_W\otimes\eta)(g\otimes1_X\otimes1_Y)\\ &\quad(1_Z\otimes\varepsilon\otimes1_Y)\\ &=g(1_Z\otimes1_Y\otimes\eta)\\ &\quad(1_Z\otimes\varepsilon\otimes1_Y)=g. \end{aligned} \] The middle equality is functoriality of tensor product, and the last is (2.2) tensored with \(Z\). For the other composite, interchange gives \[ (f\otimes1_Y\otimes1_X)(1_Z\otimes\varepsilon) =(1_W\otimes1_X\otimes\varepsilon)f. \] The remaining evaluation followed by this inserted coevaluation is (2.1) tensored with \(W\). Hence \(BA(f)=f\).

For the second bijection take \[ \begin{aligned} C(f)&=(1_Y\otimes f)(\varepsilon\otimes1_Z),\\ D(h)&=(\eta\otimes1_W)(1_X\otimes h). \end{aligned}\tag{2.5} \] Now interchange with evaluation gives \[ \begin{aligned} DC(f) &=(\eta\otimes1_W)(1_X\otimes1_Y\otimes f)\\ &\quad(1_X\otimes\varepsilon\otimes1_Z)\\ &=f(\eta\otimes1_X\otimes1_Z)\\ &\quad(1_X\otimes\varepsilon\otimes1_Z)=f \end{aligned} \] by (2.1). For the other direction, \[ (1_Y\otimes1_X\otimes h)(\varepsilon\otimes1_Z) =(\varepsilon\otimes1_Y\otimes1_W)h. \] Following with \(1_Y\otimes\eta\otimes1_W\) gives \(CD(h)=h\) by (2.2). All the displayed maps are natural in \(Z,W\), since they are compositions of tensoring maps, the fixed evaluation and coevaluation, and natural constraints.

Consequently \(-\otimes Y\) is left adjoint to \(-\otimes X\), while \(X\otimes-\) is left adjoint to \(Y\otimes-\). The variance matters in their representability consequences: \[ \begin{aligned} \operatorname{Hom}(XZ,I)&\simeq\operatorname{Hom}(Z,Y),\\ \operatorname{Hom}(I,WX)&\simeq\operatorname{Hom}(Y,W). \end{aligned}\tag{2.6} \] The first is a contravariant functor of \(Z\), represented by \(Y\); the second is covariant in \(W\), corepresented by \(Y\).

An arbitrary adjunction between two tensoring functors need not give the specified tensor duality merely by taking its unit and counit at \(I\). The adjunction must be compatible with tensoring the other factors: its component unit and counit must be the corresponding tensors of those at \(I\). The exact qualification is explained in Stacks, Remark 4.43.10. The maps (2.4)–(2.5) have that compatibility by construction.

3. What uniqueness of a dual means

Suppose \((X,Y,\varepsilon,\eta)\) and \((X,Y',\varepsilon',\eta')\) are dual pairs. The bijections proved in §2 give a unique isomorphism \(\chi:Y\to Y'\) compatible with both maps: \[ \begin{aligned} \eta'(1_X\otimes\chi)&=\eta,\\ (\chi\otimes1_X)\varepsilon&=\varepsilon'. \end{aligned}\tag{3.1} \] Compare Proposition 2.10.5 of Tensor Categories, printed page 41, with the qualification about preserving evaluation and coevaluation in the author's corrections. The chosen duality maps are part of this uniqueness assertion. The underlying object can have many other automorphisms.

The checked interface is particularly short using (2.5): \[ \chi=(1_{Y'}\otimes\eta)(\varepsilon'\otimes1_Y), \tag{3.2} \] with its unit and associativity constraints inserted. It is the image of \(\eta:XY\to I\) under the second bijection for \((X,Y')\). Its inverse check in Section 2 gives the first identity of (3.1). For the second, interchange gives \[ (\varepsilon'\otimes1_Y\otimes1_X)\varepsilon =(1_{Y'}\otimes1_X\otimes\varepsilon)\varepsilon'. \] Follow this with \(1_{Y'}\otimes\eta\otimes1_X\) and use (2.1); the result is precisely the second identity of (3.1).

Interchanging primed and unprimed data gives a map \(\kappa:Y'\to Y\). Both \(\kappa\chi\) and \(1_Y\) have image \(\eta\) under the bijection \(D\) for \((X,Y)\), so they agree. The same argument gives \(\chi\kappa=1_{Y'}\). Uniqueness of \(\chi\) follows from injectivity of that same bijection. This also proves the source assertion that the two right-dual objects are isomorphic, without asserting uniqueness of every isomorphism between their underlying objects.

4. Making two tensor inverses coherent

Suppose only that \(XY\simeq I\) and \(YX\simeq I\). Neither given isomorphism is assumed to satisfy a duality identity. We import Stacks, Lemma 4.43.6: tensoring on either side with \(X\) is then an equivalence. Its hypotheses require only a monoidal category and these two isomorphisms, so they apply before any dual has been constructed.

Proposition 4.1. One can choose isomorphisms \(\xi:XY\to I\), \(\zeta:YX\to I\) such that \[ \begin{aligned} \xi\otimes1_X&=1_X\otimes\zeta,\\ 1_Y\otimes\xi&=\zeta\otimes1_Y. \end{aligned}\tag{4.1} \] as maps \(XYX\to X\) and \(YXY\to Y\), with the unit and associativity maps understood. In particular \(\varepsilon=\zeta^{-1}\), \(\eta=\xi\) make \((X,Y)\) a dual pair.

Proof. Choose any isomorphism \(\xi:XY\to I\). Full faithfulness of \(X\otimes-\) gives a unique \(\zeta:YX\to I\) whose tensor with \(X\) is the first map in (4.1). Fully faithful functors reflect isomorphisms, so \(\zeta\) is invertible.

For the second equality, tensor on the left with \(X\). Its two maps become \(1_X\otimes1_Y\otimes\xi\) and \(1_X\otimes\zeta\otimes1_Y\), with target \(XY\). The first equality replaces the latter by \(\xi\otimes1_X\otimes1_Y\). After postcomposing either map with \(\xi:XY\to I\), they agree by interchange: both evaluate the two successive \(XY\) blocks using \(\xi\otimes\xi\), then remove the two units. Cancel the invertible \(\xi\), then use faithfulness of \(X\otimes-\). This proves the second equality of (4.1).

For (2.1), insert \(1_X\otimes\zeta^{-1}\) and follow with \(\xi\otimes1_X=1_X\otimes\zeta\); their composite is the identity. For (2.2), insert \(\zeta^{-1}\otimes1_Y\) and follow with \(1_Y\otimes\xi=\zeta\otimes1_Y\); again they cancel. \(\square\)

This gives the coherent choices required even if the initially supplied two inverse isomorphisms were incompatible. It requires both inverse relations. Exercise 3 shows why a single one is insufficient.

The proposition also applies to an autoequivalence \(T\) of a category. In the tensor category of endofunctors under composition, choose a quasi-inverse \(S\). The resulting coherent maps \(TS\to1\) and \(ST\to1\) give the triangle identities for an adjoint equivalence. This is the inverse coherence needed when later constructing actions with negative powers of \(T\).

5. An invertible unit insertion gives an idempotent algebra

Let \(u:I\to A\), and suppose both maps \[ \begin{aligned} s_L&=(u\otimes1_A)l_A^{-1}:A\to AA,\\ s_R&=(1_A\otimes u)r_A^{-1}:A\to AA \end{aligned}\tag{5.1} \] are isomorphisms.

Proposition 5.1. These maps are equal. Their common inverse \(m:AA\to A\), with unit \(u\), makes \(A\) an algebra object.

Proof. Use the projector criterion proved in Descent through replacement objects, Section 6. For \(P=A\otimes-\), the insertion \(u\) defines a natural map \(e:1\to P\). Its whiskerings \(eP\) and \(Pe\) are exactly \(s_L\otimes-\) and \(s_R\otimes-\), with coherence constraints. They are invertible by hypothesis. The criterion therefore gives \(eP=Pe\). Evaluating at \(I\) gives \(s_L=s_R\).

The unit laws for \(m=s_L^{-1}=s_R^{-1}\) are immediate. For associativity, the two insertions \(s_L\otimes1_A\) and \(1_A\otimes s_L\) from \(AA\) to \(AAA\) agree: replace the first \(s_L\) by \(s_R\), then both insert \(u\) between the same two copies of \(A\). Invert this equality to obtain \(m\otimes1_A=1_A\otimes m\) as maps \(AAA\to AA\), with the corresponding associator. Postcompose with \(m\). This is the algebra associativity law. \(\square\)

Here is a finite example. Take the category of subsets of \(\{p,q\}\), with inclusion as its unique possible arrow, tensor product union, and unit the empty set. For \(A=\{p\}\), the unique \(u: \varnothing\to A\) has both insertions equal to \(1_A\), because \(A\cup A=A\). Thus \(A\) is an idempotent algebra although it is not the unit of the whole category. Its projector sends \(B\) to \(A\cup B\); the local objects are exactly the subsets containing \(p\). For such a local \(C\), an inclusion \(B\to C\) is equivalent to an inclusion \(A\cup B\to C\). This verifies the reflection and connects the algebra construction with the earlier local-object criterion.

6. Exercises with complete solutions

Exercise 1 (basic). For a finite-dimensional vector space \(V\) over a field, take \(X=V\), \(Y=V^\vee\), and \(\eta(v\otimes f)=f(v)\). Find the coevaluation and check both identities. Then prove that no infinite-dimensional vector space can occur as \(X\) in a dual pair in the category of all vector spaces.

Solution. Choose a basis \(v_1,\ldots,v_d\) and dual basis \(f_1,\ldots,f_d\). Put \(\varepsilon(1)=\sum_i f_i\otimes v_i\). Equation (2.1) sends \(v\) to \(\sum_i f_i(v)v_i=v\). Equation (2.2) sends \(f\) to \(\sum_i f(v_i)f_i=f\). The tensor representing coevaluation corresponds to the identity under \(V^\vee\otimes V\simeq\operatorname{End}(V)\), so it is independent of the basis. This is the canonical finite-dimensional example of Tensor Categories, Example 2.10.12, printed pages 42–43.

For an arbitrary purported dual pair, write \(\varepsilon(1)=\sum_{i=1}^r y_i\otimes x_i\); every element of the algebraic tensor product is a finite sum. The first identity would send each \(x\in X\) to \(\sum_i\eta(x\otimes y_i)x_i\). Its image lies in the finite-dimensional span of the \(x_i\). If it equals \(1_X\), that span is all of \(X\). Thus infinite-dimensional \(X\) cannot have such a dual, whatever \(Y\) or pairing is chosen. For \(V=0\), the zero coevaluation and evaluation satisfy the identities, since the identity map of the zero space is the zero map.

Exercise 2 (intermediate). In the category of sets with cartesian tensor product, classify the dual pairs. Include the empty-set cases.

Solution. A coevaluation from the one-point unit specifies \((y_0,x_0)\in Y\times X\), so neither set can be empty. Evaluation to the one-point set is forced. The first triangle sends every \(x\in X\) to \(x_0\), and the second sends every \(y\in Y\) to \(y_0\). They are identities precisely when both sets are singletons. Conversely, for singleton sets the unique maps satisfy both triangles. Therefore these, and only these, are dual pairs. There is no empty-set exception: the necessary coevaluation into \(Y\times X=\varnothing\) does not exist.

Exercise 3 (advanced). Let \(s,d:\mathbb N\to\mathbb N\) be \(s(n)=n+1\) and \(d(n)=\max(n-1,0)\). Form the monoid generated by these functions under composition, and regard it as a discrete tensor category. Show that one tensor-inverse relation need not give a dual.

Solution. The unit object is the identity function, and the tensor product is composition. Direct evaluation gives \(ds=1\); but \(sd(0)=1\ne0\), so \(sd\ne1\). For \(X=d\), \(Y=s\), we therefore have \(XY=I\) and an evaluation, but no coevaluation \(I\to YX\), because a discrete category has a morphism between two objects only when the objects are equal. In fact \(d\) has no dual with any object of this monoid. A dual would require both \(dY=1\) and \(Yd=1\), making \(d\) a bijection of \(\mathbb N\). It is not injective, since \(d(0)=d(1)=0\). This example preserves the distinction between one inverse relation and the two hypotheses of Proposition 4.1.

Exercise 4 (expert). In the graded category of Braidings, cocycles and cyclic power obstructions, use any group \(G\) and a normalized associator cocycle \(\alpha\). Set \(X=\delta_g\), \(Y=\delta_{g^{-1}}\). Relative to their basis lines, let coevaluation have scalar \(t\) and evaluation scalar \(v\). Classify the duality maps. Apply the result to the twisted odd line over a field of characteristic different from \(2\).

Solution. The first triangle has scalar \(tv/\alpha(g,g^{-1},g)\); the second has scalar \(tv\,\alpha(g^{-1},g,g^{-1})\). The cocycle identity at \((g,g^{-1},g,g^{-1})\), with normalization in every unit slot, gives \[ \alpha(g,g^{-1},g)\alpha(g^{-1},g,g^{-1})=1. \] Thus both triangles hold exactly when \(t,v\ne0\) and \[ tv=\alpha(g,g^{-1},g). \] Every choice \(t\in k^\times\) determines the unique \(v=\alpha(g,g^{-1},g)/t\), and all duality maps arise this way. These formulas account for the associator and its inverse separately; no braiding is involved.

For the twisted parity cocycle and \(g=1\), this product is \(-1\). In particular coevaluation \(t=1\) and evaluation \(v=-1\) give a dual pair of odd lines. The same odd object has a square isomorphic to the unit, but the two cyclic contractions computed in the preceding lesson differ. Duality and coherent cyclic power realization are different requirements: the two duality triangles use the two independently chosen maps with product \(-1\), whereas the cyclic realization asks one power map to commute with the remaining copy of \(X\).

The book's graded dual example must be read with the author's correction to Example 2.10.14: its left-dual evaluation is the inverse cocycle value in the book's object convention. Here \(X\) is the left dual of \(Y=\delta_{g^{-1}}\); the identity just proved converts that inverse value into \(\alpha(g,g^{-1},g)\), as required. Author's corrections.

7. References and status

The unit, inverse-object and tensor-adjunction interfaces are Stacks, Section 4.43, specifically Lemmas 4.43.1–4.43.4, Definition 4.43.5, Lemma 4.43.6, Definition 4.43.8, Lemma 4.43.9 and Remark 4.43.10. The omitted second adjunction proof and compatible inverse choices are completed above; the canonical results themselves remain referenced.

Etingof, Gelaki, Nikshych and Ostrik, Tensor Categories, Theorems 2.8.5 and 2.9.2, Proposition 2.10.5 and Examples 2.10.12 and 2.10.14, supply the coherence and duality comparison, with the author's corrections applied.

Self-checked; no independent review. Written by GPT-6.1 Sol at Ultra reasoning effort. Original exposition and exercise solutions are dedicated under CC0 1.0. Referenced books and the Stacks Project retain their own licenses. This is a draft within the continuing full-course source and dependency reconciliation.