Draft lesson.

Tensor and Hom for twisted module stacks

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

Gluing two modules with different coherence factors changes the coherence factor of their tensor product. A Hom module changes it in the opposite direction in its first argument. These rules make evaluation and composition descend, and explain where a dual belongs.

We use the line-cocycle description and effective descent proved in Twisted modules from coherent gluing. Throughout, \(\mathcal E\) is the category of sheaves on a small site, \(R\) is a sheaf of commutative rings, and \(U\to1\) is an epimorphism. Products such as \(U^3\) are products in \(\mathcal E\). A module on a slice is understood on its element site, as in Section 7 of that lesson. No points, representable intersections, or bases for invertible modules are assumed.

1. The module operations used on a slice

For module sheaves \(A,B\) over a base sheaf \(V\), write \([A,B]\) for the sheaf whose sections over a generalized section \((T,v)\) are module maps between their restrictions to the element site over \((T,v)\). Restrictions restrict these maps. Addition and scalar multiplication act on their values.

Lemma 1.1. This is a module sheaf. There is a natural bijection \[ \begin{gathered} \operatorname{Hom}(X\otimes A,B)\\ \simeq\operatorname{Hom}(X,[A,B]). \end{gathered} \tag{1.1} \] Tensor product and internal Hom commute with restriction along a map of base sheaves \(W\to V\), with \(R\) restricted from the original site.

Proof. Suppose maps between the restricted modules are given compatibly on a covering sieve of \((T,v)\). For any further arrow \((S,w)\to(T,v)\) and section \(a\) of \(A\) there, the pullback covering sieve gives compatible images of \(a\) in \(B\). The sheaf condition of \(B\) glues them uniquely. These glued values commute with restriction and preserve addition and scalars, because those identities hold on a covering sieve and \(B\) is separated. They define the unique glued module map. This proves the sheaf assertion.

A map \(X\otimes A\to B\) gives, for each section \(x\), the map \(a\mapsto f(x\otimes a)\) on every further restriction. This is a section of \([A,B]\), is linear in \(x\), and commutes with restriction. Conversely such a map to \([A,B]\) gives a bilinear rule on pairs \((x,a)\). The tensor universal property, including sheafification, extends it uniquely. The two rules are inverse on pure tensors, which locally generate the sheaf tensor product.

On the element site, restricting a module along \(W\to V\) replaces the label \(w\) by its image \(v\). The category of further arrows out of \((T,w)\) has exactly the arrows out of \((T,v)\): every further label is the restriction of the indicated label. Hence the formula defining internal Hom gives identical map sets after this restriction. Sectionwise tensor and its sheafification also commute with restriction: covering sieves and all their labels are unchanged. This proves both assertions. \(\square\)

This restriction assertion concerns slices of the same sheaf category. For a general morphism of ringed sites, the canonical pullback map for internal Hom need not be an isomorphism.

Call \(A\) finite with a local dual basis if, after a covering refinement on its element site, there are finitely many sections \(a_r\) and module maps \(\alpha_r:A\to R\) with \[ \begin{gathered} a=\sum_r\alpha_r(a)a_r\\ \quad\hbox{on every further restriction.} \end{gathered} \tag{1.2} \] Equivalently, \(A\) is locally a retract of a finite free module: the maps \(a\mapsto(\alpha_r(a))_r\) and \((x_r)_r\mapsto\sum_rx_ra_r\) exhibit the retract. Conversely a retract of \(R^n\) supplies (1.2) by restricting its coordinate maps. This condition does not require a local basis.

Lemma 1.2. Put \(A^*=[A,R]\). If \(A\) has a local dual basis, then \[ \begin{gathered} A^*\otimes B\xrightarrow{\sim}[A,B],\\ \alpha\otimes b\longmapsto(a\mapsto\alpha(a)b). \end{gathered} \tag{1.3} \] The maps \(\epsilon:A^*\otimes A\to R\) given by evaluation and \(\eta:R\to A\otimes A^*\) corresponding to \(1_A\) are a dual pair. Conversely any dual pair for \(A\) gives a local dual basis. Every invertible module satisfies these conclusions, and its evaluation is an isomorphism.

Proof. Locally the inverse in (1.3) sends \(f\) to \(\sum_r\alpha_r\otimes f(a_r)\). Evaluating its image at \(a\) gives \(f(\sum_r\alpha_r(a)a_r)=f(a)\). Applied to \(\alpha\otimes b\), the other composite gives \(\sum_r\alpha_r\otimes\alpha(a_r)b=\alpha\otimes b\), since applying \(\alpha\) to (1.2) gives \(\alpha=\sum_r\alpha(a_r)\alpha_r\). Thus the canonical map is locally an isomorphism; its unique local inverses agree on overlaps and glue. Define \(\eta(1)=\sum_ra_r\otimes\alpha_r\) using the isomorphism with \([A,A]\). It is independent of the chosen dual basis. The two triangle identities reduce respectively to (1.2) and its dual identity just proved, so hold globally.

Conversely let \(D\) be a dual with evaluation \(D\otimes A\to R\) and coevaluation \(R\to A\otimes D\). Locally the latter's value at one is a finite sum \(\sum_ra_r\otimes d_r\), by the local pure-tensor assertion. Put \(\alpha_r(a)=\epsilon(d_r,a)\). The triangle on \(A\) gives (1.2). The map \(D\to A^*\), \(d\mapsto\epsilon(d,-)\), is an isomorphism: its inverse sends \(\alpha\) to \(\sum_r\alpha(a_r)d_r\). One composite uses (1.2); the other uses the triangle on \(D\). These formulas also identify the supplied evaluation with the canonical one.

If \(A\) is invertible with inverse \(J\), the tensor functors by \(A\) and \(J\) are quasi-inverse, by their two inverse isomorphisms and associativity. The coherent choices proved in Tensor duality and coherent inverses, Section 4 make these inverse isomorphisms a dual pair. Apply the preceding paragraph to identify \(J\) with \(A^*\). Its evaluation was an inverse isomorphism, so the canonical evaluation is also an isomorphism. \(\square\)

For later use, if \(P,Q\) are invertible, these constructions give \[ \begin{gathered} Q\otimes P^*\otimes[A,B]\\ \xrightarrow{\sim}[P\otimes A,Q\otimes B]. \end{gathered} \tag{1.4} \] Here is a formula that also proves the claim without a finiteness hypothesis on \(A\). A pure tensor \(q\otimes\lambda\otimes f\) sends \(p\otimes a\) to \(\lambda(p)q\otimes f(a)\). For an arbitrary map \(g:P\otimes A\to Q\otimes B\), take local dual bases \((p_r,\lambda_r)\) and \((q_s,\beta_s)\). Its inverse image is \[ \begin{gathered} \sum_{r,s}q_s\otimes\lambda_r\otimes\\ \big(a\mapsto(\beta_s\otimes1)g(p_r\otimes a)\big). \end{gathered} \tag{1.5} \] The two dual-basis identities prove the two inverse equations. The canonical map therefore has a unique inverse, so these local inverses glue and commute with all restrictions.

2. Products and inverse line cocycles

A line cocycle \(L\) has invertible modules \(L_{ij}\) on \(U^2\), maps \[ \mu^L_{ijk}:L_{ij}\otimes L_{jk}\xrightarrow{\sim}L_{ik}, \tag{2.1} \] an associative quadruple equation, and diagonal units. Generalized coordinates \(i,j,k\) mean pullbacks from the corresponding factors of \(U^n\).

For two such cocycles \(L,B\), define \((LB)_{ij}=L_{ij}\otimes B_{ij}\). Its multiplication first interchanges the middle factors and then applies \(\mu^L\otimes\mu^B\). Its diagonal unit is the product of their units. On four coordinates, both parenthesizations move the three \(L\) factors together and the three \(B\) factors together. They then agree by the two given associativity equations. Equalities on locally generating pure tensors prove the equation on the full modules. This constructs the product cocycle.

The inverse cocycle \(L^{-1}\) has module \(L_{ij}^*\). Define its multiplication by \[ \begin{gathered} L_{ij}^*\otimes L_{jk}^*\\ \simeq(L_{ij}\otimes L_{jk})^*\\ \xrightarrow{((\mu^L_{ijk})^*)^{-1}}L_{ik}^*. \end{gathered} \tag{2.2} \] The first map pairs the corresponding factors. It is an isomorphism by Lemma 1.2, since both lines are invertible. This direction of the second arrow matters: dualizing \(\mu^L\) itself gives an arrow out of \(L_{ik}^*\).

For three dual factors, pair either candidate multiplication with the multiplication of the three original factors. In both cases the pairing is the product of the three evaluations, by (2.2). Original associativity identifies the original products. Their perfect pairings identify the two dual products. Thus (2.2) is associative. The dual of the diagonal unit gives the diagonal inverse unit with the same pairing property. Evaluation then gives an isomorphism of cocycles \[ L^{-1}L\simeq1, \tag{2.3} \] where \(1\) has module \(R\) and ordinary multiplication. Indeed the compatibility on triples is precisely the defining pairing property of (2.2), and the units pair to one. All these assertions use invertible modules, not chosen frames.

3. Tensoring objects with different twists

Write \(\mathsf T_L(V)\) for the category of \(L\)-twisted modules over \(V\). Its object \(M\) has underlying module on \(U\times V\) and transition \[ \begin{gathered} \theta^M_{ij}:L_{ij}\otimes M_j\xrightarrow{\sim}M_i,\\ \theta^M_{ij}(1\otimes\theta^M_{jk})\\ =\theta^M_{ik}(\mu^L_{ijk}\otimes1). \end{gathered} \tag{3.1} \] Morphisms commute with these transitions.

Theorem 3.1. Tensor product defines a functor of stacks \[ \begin{gathered} \mathsf T_L\times\mathsf T_B\longrightarrow\mathsf T_{LB},\\ (M,N)\longmapsto M\otimes N. \end{gathered} \tag{3.2} \] It commutes with slice restriction and common refinements of the cover. Associativity, symmetry and units of module tensor product induce coherent comparisons for these functors.

Proof. Its transition interchanges the middle factors in \[ \begin{gathered} L_{ij}\otimes B_{ij}\otimes M_j\otimes N_j\\ \longrightarrow M_i\otimes N_i \end{gathered} \tag{3.3} \] and applies \(\theta^M_{ij}\otimes\theta^N_{ij}\). On three coordinates, the successive transition sends a local tensor \(l\otimes b\otimes l'\otimes b'\otimes m\otimes n\) to \[ \theta^M_{ik}(\mu^L(l,l')\otimes m) \otimes\theta^N_{ik}(\mu^B(b,b')\otimes n). \] This is exactly the single transition after product multiplication. The unit equations are the two diagonal equations. Tensoring two equivariant maps gives an equivariant map by substituting their equations in (3.3); identities and compositions are preserved on pure tensors.

Lemma 1.1 makes (3.3) compatible with every slice restriction. Pulling all cocycle and object maps to a common refinement gives the same formula, so is compatible with the refinement equivalences of the gluing lesson. An associator or symmetry on the underlying modules commutes with (3.3), since either order sends each pure tensor to the same ordered factors and their transitions. The pentagon, hexagon and unit equations similarly give the same reordering or parenthesization on each pure tensor, so agree. This proves the full coherence assertions. \(\square\)

The neutral stack \(\mathsf T_1\) is the ordinary module stack by effective descent. It acts on every \(\mathsf T_L\). Tensoring two objects of \(\mathsf T_L\) lands in \(\mathsf T_{L^2}\); turning it into an operation within \(\mathsf T_L\) would require additional identifications of twists.

4. The Hom twist and its transitions

Theorem 4.1. There is a functor of stacks \[ \begin{gathered} \mathsf T_L^{\mathrm{op}}\times\mathsf T_B\\ \longrightarrow\mathsf T_{BL^{-1}},\\ (M,N)\longmapsto\underline{\operatorname{Hom}}(M,N), \end{gathered} \tag{4.1} \] whose underlying module on \(U\) is \([M,N]\). It is defined for arbitrary module sheaves. Its evaluation and composition have the types \[ \begin{gathered} \underline{\operatorname{Hom}}(M,N)\otimes M\longrightarrow N,\\ \underline{\operatorname{Hom}}(N,P)\otimes\\ \underline{\operatorname{Hom}}(M,N)\longrightarrow\\ \underline{\operatorname{Hom}}(M,P), \end{gathered} \tag{4.2} \] using canonical cancellation of the intermediate twists.

Proof. Over \(U^2\), use (1.4) to identify \[ B_{ij}\otimes L_{ij}^*\otimes[M_j,N_j] \simeq[L_{ij}\otimes M_j,B_{ij}\otimes N_j]. \] Follow this with precomposition by \((\theta^M_{ij})^{-1}\) and postcomposition by \(\theta^N_{ij}\). This is an isomorphism to \([M_i,N_i]\), and is the required transition.

Its formula on \(b\otimes\lambda\otimes f\) is the map sending \(m_i\) to \[ \begin{gathered} \theta^N_{ij}\!(b\otimes\\ f\!((\lambda\otimes1)(\theta^M_{ij})^{-1}(m_i))). \end{gathered} \tag{4.3} \] Contraction with \(\lambda\) is a module map, so this formula does not assume that the inverse transition sends a section to a single pure tensor.

For the triple equation, the two inverse transitions for \(M\) give \[ \begin{gathered} (1\otimes(\theta^M_{jk})^{-1})(\theta^M_{ij})^{-1}\\ =((\mu^L_{ijk})^{-1}\otimes1)(\theta^M_{ik})^{-1}. \end{gathered} \tag{4.4} \] Contracting its two line factors with \(\lambda\otimes\lambda'\) is the same as contracting its single line factor with their product in (2.2). The two forward transitions for \(N\) multiply their \(B\) factors by \(\mu^B\). Substituting these two facts in two applications of (4.3) proves exactly the \(BL^{-1}\) triple equation. Diagonal contraction gives the identity map, so the unit equation holds as well.

For maps \(a:M'\to M\) and \(b:N\to N'\), the underlying Hom map is \(f\mapsto bfa\). Their equivariance equations, substituted in (4.3), show its equivariance. Ordinary composition proves functoriality. Lemma 1.1 and the displayed canonical formulas prove compatibility with restriction and refinement.

Evaluation is equivariant because, after a simultaneous transition, the line factors \(L^*\) and \(L\) contract by (2.3); what remains is the forward transition of \(N\). This can be checked on \(\lambda\otimes l\otimes f\otimes m\), where the contracted scalar is \(\lambda(l)\). Composition is equivariant by the same check: the forward transition for its middle argument cancels its inverse transition. Its remaining twist is \[ (CB^{-1})(BL^{-1})\simeq CL^{-1}. \] Associativity of composition is equality of ordinary composite maps on every restriction; identity sections have the neutral transition. This proves (4.2) with their units and associativity. \(\square\)

In particular, \(\underline{\operatorname{Hom}}(M,N)\) for two \(L\)-twisted objects is neutral and descends to an ordinary module sheaf. Sections of that descended sheaf are exactly the twisted morphisms: its ordinary descent equation is, by its construction, \(f_i\theta^M_{ij}=\theta^N_{ij}(1\otimes f_j)\). For \(M=N\), composition gives an ordinary sheaf of endomorphism algebras, which may be noncommutative even though \(R\) is commutative.

5. Adjunction including the overlap equation

Theorem 5.1. For \(M\in\mathsf T_L(V)\), \(N\in\mathsf T_B(V)\), and \(P\in\mathsf T_C(V)\), currying gives a natural bijection \[ \begin{gathered} \operatorname{Hom}_{\mathsf T_C(V)}(M\otimes N,P)\\ \simeq\operatorname{Hom}_{\mathsf T_L(V)}\\ \big(M,\underline{\operatorname{Hom}}(N,P)\big) \end{gathered} \tag{5.1} \] when the left source has been identified with twist \(C=LB\). Equivalently, without imposing that identification, take \(P\in\mathsf T_{LB}(V)\); the right target has twist \((LB)B^{-1}\simeq L\).

Proof. First use Lemma 1.1 on \(U\times V\). If \(q:M\otimes N\to P\), its curry is \(\widehat q(m)(n)=q(m\otimes n)\), and the inverse evaluates this formula. We check that exactly the equivariant maps correspond.

The left overlap equation says, on local generating tensors, \[ \begin{gathered} q_i\big(\theta^M_{ij}(l\otimes m)\otimes\\ \theta^N_{ij}(b\otimes n)\big)\\ =\theta^P_{ij}\big((l\otimes b)\otimes q_j(m\otimes n)\big). \end{gathered} \tag{5.2} \] The right overlap equation for \(\widehat q\), using (4.3), says precisely that its value after the transition for \(M\) is the transported map \(N_i\to P_i\). To compare these equations, evaluate that transported map at \(\theta^N_{ij}(b\otimes n)\). Its inverse transition gives \(b\otimes n\). The \(B^{-1}\) factor in the Hom transition contracts with this \(b\); after cancellation, the resulting equation is (5.2). Formally, the cancellation identification \((LB)B^{-1}\simeq L\) can be expanded using a local dual basis for \(B\): inserting \(\sum_rb_r\otimes\beta_r\) and summing contracts to the identity. Thus the comparison works without choosing a frame for \(B\).

Conversely (5.2) implies the right equation by evaluating at all such sections and using the surjective isomorphism \(\theta^N\) and local generation by pure tensors. Maps to internal Hom agree exactly when all their evaluations agree, directly from its definition. Hence currying and its inverse restrict to the asserted bijection. Their section formulas show naturality in all arguments and compatibility with restriction. \(\square\)

6. Duals and the finiteness condition

For any \(M\in\mathsf T_L\), define \(M^\vee=\underline{\operatorname{Hom}}(M,R)\), where \(R\) is neutral. Thus \(M^\vee\in\mathsf T_{L^{-1}}\), and evaluation \(M^\vee\otimes M\to R\) always exists.

Theorem 6.1. The following are equivalent:

1. The underlying module of \(M\) on \(U\) is locally a retract of a finite free module.
2. There is a coevaluation \(R\to M\otimes M^\vee\) which, with evaluation, satisfies both triangle identities.
3. \(M\) has a dual object of inverse twist with evaluation and coevaluation satisfying both triangles.

When they hold, the canonical maps \[ \begin{gathered} M^\vee\otimes N\xrightarrow{\sim}\\ \underline{\operatorname{Hom}}(M,N),\\ \qquad M\xrightarrow{\sim}M^{\vee\vee} \end{gathered} \tag{6.1} \] are isomorphisms of twisted objects. Neither (6.1) nor coevaluation is asserted for arbitrary modules.

Proof. Under (1), (1.3) gives the first isomorphism locally. The map \(\alpha\otimes n\mapsto(m\mapsto\alpha(m)n)\) is equivariant by the evaluation and composition checks of Theorem 4.1, so is an isomorphism of twisted objects. Taking \(N=M\) identifies \(M\otimes M^\vee\) with the neutral endomorphism object, using symmetry. The identity endomorphism is equivariant. Its inverse image gives a global coevaluation, locally \(\sum_rm_r\otimes\alpha_r\). Lemma 1.2 proves both triangles, which remain valid after descent. This proves (2), and (2) implies (3).

Under (3), restrict its maps to \(U\). The inverse twist pairing identifies their targets and sources with the ordinary unit there. They give a dual pair of ordinary modules, so Lemma 1.2 supplies a local dual basis and hence (1). The same lemma identifies any such dual with \(M^\vee\).

For double duality, the canonical map sends \(m\) to \(\alpha\mapsto\alpha(m)\). With a local dual basis its inverse sends \(\xi\in M^{\vee\vee}\) to \(\sum_r\xi(\alpha_r)m_r\). One composite is (1.2). The other follows from \(\alpha=\sum_r\alpha(m_r)\alpha_r\), after applying \(\xi\). The maps are therefore inverse. Evaluation equivariance proves that the canonical map respects transitions; its inverse then does too. This proves the second isomorphism with its correct twist. \(\square\)

An invertible underlying module is included, even on a site where it has no local frame. Finite locally free modules are included as a special case. On a general ringed site the retract formulation is the one used here; no identification with local freeness is needed.

7. Reading the formulas in frames

On an open cover where the two line cocycles have frames, let their scalar factors be \(c_{ijk}\) and \(d_{ijk}\), and object transitions be \(t_{ij}\) and \(u_{ij}\). The product object has transition \(t_{ij}\otimes u_{ij}\) and factor \(c_{ijk}d_{ijk}\). The Hom transition is \[ f\longmapsto u_{ij}ft_{ij}^{-1}, \tag{7.1} \] and its successive composite equals \[ \begin{gathered} u_{ij}u_{jk}f(t_{ij}t_{jk})^{-1}\\ =d_{ijk}c_{ijk}^{-1}u_{ik}ft_{ik}^{-1}. \end{gathered} \tag{7.2} \] Commutativity of the scalar action justifies moving the two units past \(f\). The dual transition is \((t_{ij}^{-1})^*\), with factor \(c_{ijk}^{-1}\). In a finite free frame, its evaluation and the tensor corresponding to the identity matrix are the evaluation and coevaluation of Section 6. This is a specialization of the line formulas, not an assumption in their proofs.

8. Graded exercises with complete solutions

Exercise 8.1 (foundation). In the scalar description, let \(M,N,P\) have factors \(c,d,e\). Compute the twists of their tensor and Hom objects. Check the factor of composition and prove its overlap equation directly.

Solution. Tensor has factor \(cd\). The three Hom objects from \(M\) to \(N\), from \(N\) to \(P\), and from \(M\) to \(P\) have factors \(d/c,e/d,e/c\), by (7.2). The product of the first two factors is \(e/c\). Write the object transitions as \(t,u,v\). If \(f:M_j\to N_j\) and \(g:N_j\to P_j\), their transported composite is \[ (v_{ij}gu_{ij}^{-1})(u_{ij}ft_{ij}^{-1}) =v_{ij}(gf)t_{ij}^{-1}. \] This is the transport of the composite, so composition is equivariant. Evaluating the transported \(f\) on \(t_{ij}m\) gives \(u_{ij}f(m)\), which also directly checks evaluation. Identity maps transport to identity maps, and threefold composition agrees by associativity of module maps.

Exercise 8.2 (calculation). Let \(G=\{1,s\}\), \(s^2=1\), and use the normalized real scalar group cocycle with \(c(s,s)=-1\) and all other values one. A \(c\)-projective representation satisfies \(T_gT_h=c(g,h)T_{gh}\). Verify the cocycle equation, classify possible finite real dimensions, and compute the tensor square and endomorphisms of the two-dimensional example.

Solution. If any argument in \(c(g,h)c(gh,k)=c(h,k)c(g,hk)\) is one, normalization makes both sides equal to the remaining factor. For \((s,s,s)\), both sides are \(-1\). Thus it is a cocycle. The representation consists of \(T_1=1\) and \(J=T_s\) with \(J^2=-1\). For dimension \(r\), determinants give \((\det J)^2=(-1)^r\), so \(r\) is even. Conversely every even dimension has a block diagonal sum of \[ J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \] including the zero-dimensional case. The factor cannot be removed by a normalized real scalar cochain: that would require \(b_s^2=-1\).

On \(\mathbb R^2\otimes\mathbb R^2\), the transition is \(J\otimes J\), whose square is \(1\); its factor is the neutral cocycle \(c^2\). Its \(+1\) eigenspace has basis \[ e_1\otimes e_1+e_2\otimes e_2,\qquad e_1\otimes e_2-e_2\otimes e_1, \] and its \(-1\) eigenspace has basis \[ e_1\otimes e_1-e_2\otimes e_2,\qquad e_1\otimes e_2+e_2\otimes e_1. \] Applying \(J\) to each factor checks these signs. The four vectors form a basis, by adding and subtracting the displayed pairs. The neutral Hom action on matrices is \(A\mapsto JAJ^{-1}\). Its invariant matrices satisfy \(AJ=JA\); writing four entries and equating them gives exactly \(A=aI+bJ\). Thus the projective representation has endomorphism algebra \(\mathbb C\) as a real algebra. Its dual action is \((J^{-1})^{\mathsf t}=J\), so evaluation is invariant. These are group-cocycle calculations; they do not claim a nontrivial open-cover cocycle on a point.

Exercise 8.3 (obstruction). Over a field \(k\), let \(V=\bigoplus_{n\geq0}ke_n\) with neutral twist. Identify \(V^*\), describe the image of \(V^*\otimes V\to[V,V]\), and decide whether \(V\) has a dual pair.

Solution. A linear functional is determined by arbitrary values on the basis, because each input has finite support. Hence \(V^*=\prod_{n\geq0}k\). A finite sum \(\sum_r\lambda_r\otimes v_r\) gives the operator \(v\mapsto\sum_r\lambda_r(v)v_r\), whose image lies in the finite-dimensional span of the \(v_r\). Conversely a finite-rank operator has image with a finite basis \(v_1,\ldots,v_m\); its coordinate functions on that image, composed with the operator, are linear functionals \(\lambda_r\), giving the displayed tensor. Therefore the image is exactly the finite-rank operators. The identity has infinite-dimensional image and is absent. If a dual pair existed, its finite coevaluation expression and the triangle would write every \(v\) as a linear combination of finitely many fixed vectors. This contradicts the infinite independent basis. Internal Hom and evaluation exist, but coevaluation and the duality isomorphism do not.

Exercise 8.4 (synthesis). Suppose line changes send twists \(L,B\) to \(L',B'\), with underlying invertible modules \(P,Q\) on \(U\). Describe the induced line change on Hom objects, and show compatibility with evaluation. Use the conventions \[ L'_{ij}\otimes P_j\simeq P_i\otimes L_{ij}, \qquad B'_{ij}\otimes Q_j\simeq Q_i\otimes B_{ij}. \]

Solution. Invert the first line equation in the tensor group of invertible modules to obtain \[ (L'_{ij})^{-1}\otimes P_j^{-1} \simeq P_i^{-1}\otimes L_{ij}^{-1}. \] Tensor it with the second equation. This gives a line change from \(BL^{-1}\) to \(B'(L')^{-1}\) with underlying module \(Q\otimes P^{-1}\). Its triple equation is the product of the second original triple equation and the inverse of the first. Inverse multiplication is (2.2), so all contraction factors cancel in this product; the diagonal equation follows in the same way.

The canonical object comparison is (1.4): \[ \begin{gathered} Q\otimes P^{-1}\otimes\underline{\operatorname{Hom}}(M,N)\\ \simeq\underline{\operatorname{Hom}}(P\otimes M,Q\otimes N). \end{gathered} \] To check its overlap equation, insert the two line-change comparisons in the transported Hom map. The \(P\) comparison occurs once forward and once inverse, by (4.3), and cancels; the \(Q\) comparison remains forward. The result is exactly the product line change just constructed. Equivalently (1.5) verifies this equation after evaluation on every local \(p\otimes m\), which determines a Hom map. On a generating tensor \(q\otimes\lambda\otimes f\otimes p\otimes m\), evaluation on either side is \(\lambda(p)q\otimes f(m)\). Hence the comparison also respects evaluation. Since pure tensors locally generate, both equalities hold globally, with no frames assumed for \(P\) or \(Q\).

9. Further reading

• Andrea D'Agnolo and Pietro Polesello, Stacks of twisted modules and integral transforms, open preprint, Section 3.5, for products and opposite twisting data. Their operations on stacks also include functors between module stacks; those stack operations should be distinguished from the object-level internal Hom constructed here.
• The Stacks Project, Internal Hom on sites, Section 18.27, for the restricted-Hom definition and tensor–Hom adjunction.
• The Stacks Project, Duals, Section 17.18, for the finite-retract criterion on ringed spaces.

The line cocycles, their units, and the effectiveness of ordinary module descent used above are established in Twisted modules from coherent gluing. The compatibility of inverse identifications used in Lemma 1.2 is established in Tensor duality and coherent inverses.