Draft lesson.

Noncommutative gluing by invertible bimodules

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

Local descriptions of a module stack may use different algebras. On overlaps, those algebras need not be isomorphic: a Morita bimodule is enough to identify their module categories. Associative composition of these bimodules gives gluing data, and also controls functors between the resulting stacks.

Let $\mathcal E$ be the sheaf category of a small site, $R$ a commutative ring sheaf, and $U\to1$ an epimorphism. All scalar maps for algebras are central and unital. Products and pullbacks below are in $\mathcal E$, with their canonical associativity comparisons understood. A coordinate $i$ means a generalized coordinate of the cover, not a point. All modules on slices are interpreted on element sites.

We use Bimodules and adjoints of module-stack functors for bimodule reconstruction, transformations and composition. Effective descent of module sheaves is proved in Twisted modules from coherent gluing, Lemma 9.1. The latter lesson's central line cocycles are a special case of the construction here.

1. Coherent Morita transport and its units

Let $A$ be a sheaf of possibly noncommutative $R$-algebras on $U$. Write $A_i$ for its pullback to coordinate $i$. On $U^2$, let $L_{ij}$ be an invertible $(A_i,A_j)$-bimodule. Invertibility means that a bimodule in the reverse direction has tensor products isomorphic to the two regular bimodules; equivalently, tensoring gives a linear equivalence of module stacks. Suppose we have bimodule isomorphisms $\mu_{ijk}:L_{ij}\otimes_{A_j}L_{jk}\xrightarrow{\sim}L_{ik} \tag{1.1}$ on $U^3$, satisfying $\mu_{ikl}(\mu_{ijk}\otimes1) =\mu_{ijl}(1\otimes\mu_{jkl}) \tag{1.2}$ on $U^4$. Both sides have source $L_{ij}\otimes_{A_j}L_{jk}\otimes_{A_k}L_{kl}$ and target $L_{il}$.

Lemma 1.1. These data have unique diagonal isomorphisms $u_i:A_i\to L_{ii}$ which are two-sided units. In addition, $L_{ji}$ is a coherent inverse to $L_{ij}$.

Proof. On the diagonal put $D=L_{ii}$, an invertible $(A_i,A_i)$-bimodule, and $\mu=\mu_{iii}$. Left tensoring by $D$ is fully faithful. Hence there is a unique bimodule map $u:A_i\to D$ such that $1_D\otimes u=\mu^{-1}$, under the right unit identification $D\otimes_{A_i}A_i=D$. It is invertible because tensoring by $D$ reflects isomorphisms. Thus $\mu(1\otimes u)=1_D$. Put $a=\mu(u\otimes1)$. Associativity, precomposed with $1_D\otimes u\otimes1_D$, gives $\mu=\mu(1_D\otimes a)$. Cancel the invertible $\mu$ and use faithfulness of tensoring by $D$; it follows that $a=1_D$. This is the other unit equation. Any other right unit gives the same map under fully faithful tensoring by $D$, proving uniqueness. The construction therefore commutes with restrictions.

Under these diagonal identifications, $\mu_{iij}(u_i\otimes1)$ is an invertible endomorphism $a$ of $L_{ij}$. Equation (1.2) on $(i,i,i,j)$, inserting both diagonal units, gives $a^2=a$. Hence $a=1$. The equation on $(i,j,j,j)$ similarly gives $\mu_{ijj}(1\otimes u_j)=1$. These are the mixed unit equations.

Now $u_i^{-1}\mu_{iji}$ and $u_j^{-1}\mu_{jij}$ give the inverse identifications $L_{ij}\otimes_{A_j}L_{ji}\simeq A_i, \qquad L_{ji}\otimes_{A_i}L_{ij}\simeq A_j. \tag{1.3}$ Equation (1.2) on $(i,j,i,j)$, with the mixed units just proved, says that contracting either inverse pair in $L_{ij}\otimes L_{ji}\otimes L_{ij}$ gives the same map to $L_{ij}$. Together with the reversed-coordinate equation, these are the two triangle identities when the second contraction in (1.3) is inverted to form a coevaluation. Thus the supplied reverse bimodule is a coherent inverse. $\square$

2. Objects and effective descent

An object of $\mathsf T(A,L)(V)$ consists of an $A$-module $M$ on $U\times V$, and $A_i$-linear isomorphisms $\theta_{ij}:L_{ij}\otimes_{A_j}M_j\xrightarrow{\sim}M_i \tag{2.1}$ on $U^2\times V$ satisfying $\theta_{ij}(1\otimes\theta_{jk}) =\theta_{ik}(\mu_{ijk}\otimes1). \tag{2.2}$ A morphism $a:M\to M'$ is $A$-linear and satisfies $a_i\theta_{ij}=\theta'_{ij}(1\otimes a_j)$. Pullback in $V$ defines restriction functors.

The diagonal transition is the identity after the identification $u_i$. Equation (2.2) there makes it an invertible idempotent; it is therefore the identity. This proves normalization instead of imposing a second unrelated choice of diagonal maps.

Proposition 2.1. These categories form an $R$-linear stack.

Proof. Given descent data over an epic covering of $V$, pull that cover back to $U\times V$. Effective descent glues the underlying $R$-module sheaves. The local action maps $A\otimes_R M\to M$ agree under the gluing identifications, so glue uniquely too. Unit and associativity are equalities of maps and hold after the covering; they therefore hold globally. This gives the glued $A$-module. Over $U^2\times V$ the $\theta$ maps glue by the equivariance equations of the descent isomorphisms. Their inverses glue as well. Their linearity and (2.2) hold after the cover and hence globally. This proves effective object descent. An equivariant morphism is a module map satisfying an equality of sheaf maps, so compatible local morphisms glue uniquely and stay equivariant. Morphisms thus form sheaves, and their operations are inherited from $R$-linear module maps. $\square$

3. The local module model

Suppose the pulled-back cover over $V$ has a section $s:V\to U$, and write $A_s=s^*A$. Define $\Phi_s(N)_x=L(x,s(v))\otimes_{A_s}N_v, \qquad E_s(M)=s^*M. \tag{3.1}$ The first is an $A_x$-module; its transition is multiplication of the two successive $L$ factors followed by the identity on $N$.

Theorem 3.1. These functors are inverse $R$-linear equivalences of stacks between $\mathsf{Mod}(A_s)$ and $\mathsf T(A,L)|_V$.

Proof. Equation (1.2) gives (2.2) for $\Phi_sN$. The diagonal unit gives $E_s\Phi_sN\simeq N$. For a twisted object, its transition $L(x,s(v))\otimes_{A_s}M_{s(v)}\to M_x$ gives $\Phi_sE_sM\simeq M$. Its compatibility with the transition from $y$ to $x$ is exactly (2.2) evaluated at $(x,y,s(v))$. These isomorphisms are natural, and every formula pulls back to the same formula on a further base. They give inverse equivalences as asserted. Sections exist after pulling the epic cover back to itself, using its diagonal section. Thus such module models exist locally everywhere. $\square$

For two sections $s,t$, the change from the $s$-model to the $t$-model is tensor with the $(A_t,A_s)$-bimodule $L(t,s)$. Three such changes are compared by $\mu$, and their fourfold compatibility is (1.2). This records the comparisons between the local models as well as each individual equivalence.

4. Reconstructing a locally module stack

Theorem 4.1. Every $R$-linear stack locally equivalent to module stacks over $R$-algebras has a presentation $\mathsf T(A,L)$ on a cover as above.

Proof. Choose a cover $U$ on which the stack $\mathsf S$ is described by $\mathsf{Mod}(A)$, with equivalences $\phi_i$ and inverse equivalences $\psi_i$. Choose coherent units and counits, as in the adjoint-equivalence proof in Section 6 of the bimodule-functor lesson. On $U^2$, the change of descriptions $F_{ij}=\phi_i\psi_j:\mathsf{Mod}(A_j)\longrightarrow\mathsf{Mod}(A_i)$ is a linear equivalence of stacks. The reconstruction theorem writes it as tensor with an invertible bimodule $L_{ij}$.

Contracting $\psi_j\phi_j$ with the inverse unit gives a comparison $F_{ij}F_{jk}\to F_{ik}$. The full faithfulness on transformations in that theorem identifies it with a unique bimodule isomorphism $\mu_{ijk}$ in (1.1). For four indices, the two contractions remove the same two intervening inverse-equivalence pairs. Naturality and interchange make their composites equal. Faithfulness on bimodule maps then gives (1.2). The diagonal comparisons similarly give the units of Lemma 1.1, since that lemma proves their uniqueness.

For $S\in\mathsf S(V)$, put $M_i=\phi_i(S|_U)$. Its transition is the map from $\phi_i\psi_j\phi_j(S)$ to $\phi_i(S)$ obtained by the same inverse unit. The comparison law gives (2.2). This constructs an $R$-linear stack functor $\mathsf S\to\mathsf T(A,L)$.

Conversely, for a twisted object $M$, put $S_i=\psi_i(M_i)$ on the cover. The map $\theta_{ij}$, transported by $\psi_i$, and the equivalence comparison $\psi_i\phi_i\psi_j\simeq\psi_j$, give an isomorphism $S_j\to S_i$. Equation (2.2) and the definition of $\mu$ give its cocycle equation; the diagonal unit makes it normalized. The stack axiom for $\mathsf S$ glues it to an object over $V$. Equivariant maps give exactly compatible maps between these local objects and therefore glue uniquely. The equivalence units and counits give inverse comparisons for the two constructions after the cover. Their overlap equations are the same transitions just used; hence they descend to inverse natural isomorphisms globally. All operations commute with base restriction. This proves the equivalence of stacks and the presentation theorem. $\square$

5. Functors, their adjoints and their transformations

Let $(A,L,\mu)$ and $(B,K,\nu)$ be data on the same cover. A coherent bimodule family from the first to the second has $(B_i,A_i)$-bimodules $P_i$, and isomorphisms $\gamma_{ij}:K_{ij}\otimes_{B_j}P_j \xrightarrow{\sim}P_i\otimes_{A_i}L_{ij} \tag{5.1}$ compatible with the diagonal units and satisfying $\gamma_{ik}(\nu_{ijk}\otimes1) =(1\otimes\mu_{ijk})(\gamma_{ij}\otimes1)(1\otimes\gamma_{jk}). \tag{5.2}$ Both sides take $K_{ij}\otimes_{B_j}K_{jk}\otimes_{B_k}P_k$ to $P_i\otimes_{A_i}L_{ik}$.

Theorem 5.1. Such families classify $R$-linear functors $\mathsf T(A,L)\to\mathsf T(B,K)$ with a right adjoint as a stack functor. All transformations are bimodule maps $\tau_i:P_i\to P'_i$ satisfying $(\tau_i\otimes1)\gamma_{ij} =\gamma'_{ij}(1\otimes\tau_j). \tag{5.3}$ The functor is an equivalence exactly when its local bimodules are invertible.

Proof. Define the functor's object locally by $P_i\otimes_{A_i}M_i$. Its transition first applies $\gamma_{ij}\otimes1$, then $1\otimes\theta^M_{ij}$. Equation (5.2) followed by (2.2) proves its triple equation; their unit equations prove its diagonal equation. Tensoring an equivariant map is equivariant by the same formula. Restriction compatibility is built into the sheaf maps, so this is a linear functor of stacks.

We construct its stack right adjoint, rather than assuming that local adjoints automatically glue. On an overlap put $E_L=L_{ij}\otimes_{A_j}-$ and $E_K=K_{ij}\otimes_{B_j}-$, with their coherent inverses from Lemma 1.1. Locally put $F_i=P_i\otimes_{A_i}-$ and $H_i=[P_i,-]_{B_i}$, the latter an $A_i$-module by its right action. Equation (5.1) is an isomorphism $E_KF_j\simeq F_iE_L$. For test modules $X$ over $A_i$ and $N$ over $B_j$, it gives natural bijections $\begin{aligned} \operatorname{Hom}_{A_i}(X,E_LH_jN) &\simeq\operatorname{Hom}_{A_j}(E_L^{-1}X,H_jN)\\ &\simeq\operatorname{Hom}_{B_j}(F_jE_L^{-1}X,N)\\ &\simeq\operatorname{Hom}_{B_i}(E_KF_jE_L^{-1}X,E_KN)\\ &\simeq\operatorname{Hom}_{B_i}(F_iX,E_KN)\\ &\simeq\operatorname{Hom}_{A_i}(X,H_iE_KN). \end{aligned} \tag{5.4}$ The fourth bijection uses the inverse of the indicated $\gamma$ comparison by precomposition. The represented objects are therefore canonically isomorphic: $E_LH_jN\simeq H_iE_KN$. Compose with $H_i(\theta^N_{ij})$ to give the transition for the modules $H_iN_i$.

These transitions obey the triple equation. To check it, apply (5.4) twice to any test module. Successive precompositions use the two successive $\gamma$ comparisons. Equation (5.2) identifies their composite with the single comparison after $\nu$ and $\mu$. The inverse-equivalence contractions remove the intermediate transport pairs by their triangles. Equation (2.2) for $N$ then gives exactly the single transported Hom map. Equality for every test module makes the represented transition maps equal. The diagonal equation follows from the units and the adjunction triangles. Thus these local Hom modules form an actual source-twisted object. All the bijections commute with restriction, so this construction is a stack functor. Local units and counits are compatible on overlaps by (5.4): they are the transposes of the identity maps under these same bijections. They consequently glue, and their triangles hold locally and therefore globally. This constructs the stack right adjoint.

Conversely, use the local equivalences of Section 3 to express a given stack-adjoint functor as functors between $\mathsf{Mod}(A_i)$ and $\mathsf{Mod}(B_i)$. The bimodule reconstruction theorem gives $P_i$. Compatibility with the change of local models gives (5.1), since all transformations between the local tensor functors are uniquely bimodule maps. Three changes give (5.2), and identity changes give the unit condition. The local identifications with tensor functors agree on overlaps by these equations, so glue to an isomorphism with the constructed functor.

For transformations the same local full faithfulness gives exactly $\tau_i$; compatibility with restrictions is precisely (5.3). Conversely (5.3) makes $\tau_i\otimes1$ equivariant on every twisted object, so these components define the required transformation and uniquely determine it. This proves the classification on arrows as well as objects.

If every $P_i$ is invertible, each local tensor adjunction is an equivalence, so its unit and counit are isomorphisms locally. Their glued maps are therefore isomorphisms, proving a global equivalence. Conversely a global equivalence restricts to local equivalences; bimodule reconstruction then makes each $P_i$ invertible. $\square$

Composition of a family $P$ with a family $Q$ uses $Q_i\otimes_{B_i}P_i$. Its comparison first moves a target transport past $Q_j$, then moves the intermediate transport past $P_j$. Substituting the two equations (5.2) proves that of the composite; balanced tensor associativity gives the comparison for three families. Transformations compose by their component maps and tensor products, with interchange on pure tensors. Thus the classification also retains the composition data.

6. Refining the cover

Let $a:V\to U$ be a map such that $V\to1$ is epic. Pull $A,L,\mu$ back along $a,a^2,a^3$. Restriction gives an equivalence between the old and new twisted stacks. Surjectivity onto $U$ is unnecessary.

Here is its inverse on objects, including its descent. For $N$ on the new cover, put on $U\times V$ $Z(x,v)=L(x,a(v))\otimes_{A_{a(v)}}N_v. \tag{6.1}$ Its descent map from the $w$-copy to the $v$-copy first uses $\mu^{-1}$ to replace $L(x,a(w))$ by $L(x,a(v))\otimes L(a(v),a(w))$, then applies the transition of $N$ to the latter factor and $N_w$. Associativity and (2.2) prove the ordinary descent cocycle; the units prove its diagonal equation. All these are $A_x$-linear maps. Since $U\times V\to U$ is epic, the module descent of Section 2 glues them to an $A$-module on $U$.

The maps $L(x,y)\otimes_{A_y}Z(y,v)\to Z(x,v)$ supplied by $\mu$ commute with those descent maps: on the three line factors this is (1.2), followed by the transition of $N$. They descend to the old-cover twisted transition. It is invertible after the epic cover, hence globally, and its triple equation holds there and hence globally.

For an old-cover object $M$, its maps $L(x,a(v))\otimes M_{a(v)}\to M_x$ descend to a natural comparison between this inverse after restriction and $M$. For a new-cover object $N$, restrict (6.1) to $x=a(v)$, use the auxiliary label $v$, and contract the diagonal unit. This gives the comparison back to $N$. Comparing two auxiliary labels is exactly its descent equation, so these comparisons are morphisms of twisted objects and are inverse natural isomorphisms. Every formula pulls back over a further base. This proves the equivalence of stacks.

Successive restrictions agree under their canonical pullback identifications. The section formulas substitute the same composite map, so the identity and three-refinement laws hold. Inverse comparisons are forced by full faithfulness of restriction once specified after the cover. Any two covers have the common epic refinement $U\times V$. Thus the presentations and the functor classification can be compared on common covers with coherent further-refinement maps.

7. The scalar specialization

If $A_i=R$ everywhere, invertible bimodules with agreeing scalar actions are precisely invertible $R$-modules. Sections 1–4 recover the central line-cocycle construction.

For line cocycles $L,K$, the family $P$ in (5.1) is an arbitrary $R$-module on the cover. Tensor (5.1) with $L_{ij}^{-1}$ and use line evaluation. It becomes a twisted transition $(KL^{-1})_{ij}\otimes P_j\xrightarrow{\sim}P_i. \tag{7.1}$ Conversely tensor (7.1) with $L_{ij}$ and use the inverse evaluation isomorphism to recover (5.1). The product and inverse cocycle equations identify (5.2) with the twisted-object equation. The same contraction identifies (5.3) with equivariance of maps. Consequently the category of stack-adjoint linear functors $\mathsf T_L\to\mathsf T_K$ is equivalent to $\mathsf T_{KL^{-1}}$, via $P\mapsto(P\otimes-)$. Its right adjoint is $\underline{\operatorname{Hom}}(P,-)$, with target twist $K(KL^{-1})^{-1}=L$, as proved in Tensor and Hom for twisted module stacks. The object $P$ need not be a line; a line is exactly the equivalence case here.

8. Graded exercises with complete solutions

Exercise 8.1 (foundation). Cover a point by labels zero and one. Let $A_0=k$, $A_1=M_2(k)$, and set $L_{ij}=\operatorname{Mat}_{n_i,n_j}(k)$, where $n_0=1,n_1=2$. Prove that matrix multiplication gives the required invertible bimodule cocycle. Describe its twisted category.

Solution. More generally, multiplication gives $\operatorname{Mat}_{r,s}(k)\otimes_{M_s(k)} \operatorname{Mat}_{s,t}(k)\simeq\operatorname{Mat}_{r,t}(k)$ for positive sizes. For elementary rectangular matrices, balancing makes $E_{ab}\otimes E_{cd}=0$ if $b\ne c$, by inserting the middle diagonal unit $E_{bb}$. If $b=c$, use $E_{ab}=E_{a1}E_{1b}$ and balance to obtain $E_{a1}\otimes E_{1d}$. Thus every tensor is a linear combination of these last tensors. Multiplication sends them to the basis $E_{ad}$; the map sending $E_{ad}$ back to $E_{a1}\otimes E_{1d}$ is its inverse. This also proves the inverse-bimodule relations by taking $t=r$ and reversing the sizes. Ordinary associativity of matrices proves (1.2), and the diagonal units are identity matrices.

The category is equivalent to $k$-modules by evaluation at label zero. Its inverse sends $W$ to $M_0=W$, $M_1=k^2\otimes_k W$, with the transitions given by matrix multiplication. Section 3 supplies their two inverse comparisons. The local algebras $k$ and $M_2(k)$ are not isomorphic: the second is noncommutative, whereas the first is commutative. Their module categories nevertheless glue through the displayed Morita bimodules.

Exercise 8.2 (calculation). For a fixed algebra $A$, choose units $g_{ij}$, with $g_{ii}=1$, and put $\varphi_{ij}(b)=g_{ij}bg_{ij}^{-1}$. Give $L_{ij}=A$ the right action $p\cdot b=p\varphi_{ij}(b)$. Find the multiplication and verify coherence without commuting the units. Calculate $a_{012}$ when $g_{01}=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $g_{12}=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, and $g_{02}=I$, over $\mathbb Q$.

Solution. Put $a_{ijk}=g_{ij}g_{jk}g_{ik}^{-1}$. Then $\varphi_{ij}\varphi_{jk}=\operatorname{Ad}(a_{ijk})\varphi_{ik}$, so $\mu_{ijk}(p\otimes q)=p\varphi_{ij}(q)a_{ijk}$ is balanced and right $A_k$-linear. Balancing uses $p\varphi_{ij}(b)\varphi_{ij}(q)=p\varphi_{ij}(bq)$. Right-linearity uses $\varphi_{ij}\varphi_{jk}(b)a_{ijk}=a_{ijk}\varphi_{ik}(b)$. The map $L_{ij}\to A$, $p\mapsto pg_{ij}$, is an ordinary bimodule isomorphism, since $p\varphi_{ij}(b)g_{ij}=pg_{ij}b$. In these identifications the proposed multiplication is ordinary multiplication: $p\varphi_{ij}(q)a_{ijk}g_{ik}=pg_{ij}qg_{jk}$. Hence it is an isomorphism and all $L_{ij}$ are invertible.

Directly, associativity is the identity $a_{ijk}a_{ikl}=\varphi_{ij}(a_{jkl})a_{ijl}.$ Both sides equal $g_{ij}g_{jk}g_{kl}g_{il}^{-1}$, in exactly that order. No interchange of two units is used. In the numerical case $a_{012}=\begin{pmatrix}2&1\\1&1\end{pmatrix}$; reversing the order would instead give $\begin{pmatrix}1&1\\1&2\end{pmatrix}$. The bimodule isomorphisms above identify the entire cocycle with ordinary multiplication data. This example checks noncommutative coherence and is a neutral presentation, not a claim of a nontrivial class.

Exercise 8.3 (obstruction). In the scalar specialization, explain why an arbitrary $(KL^{-1})$-twisted module gives a stack-adjoint functor, but only an invertible one gives an equivalence. Give a nonequivalence even when both twists are neutral.

Solution. Contracting the line pair in (5.1) gives (7.1), with its triple equation; conversely inverse evaluation reconstructs (5.1). Thus any such $P$ is a coherent family and Theorem 5.1 supplies $P\otimes-$ and its actual stack right adjoint $\underline{\operatorname{Hom}}(P,-)$. No finite or flat condition was used. By that theorem the functor is an equivalence exactly when $P$ is locally tensor-invertible. An invertible local module carries an inverse under transport; its evaluation maps commute with that transport by the transition isomorphisms, so the resulting inverse family is coherent. Conversely a local equivalence reconstructs such an invertible module. This is the invertible twisted-object condition.

On a point over a field, take $P=k^n$, $n\geq2$, and neutral twists. The functor is $M\mapsto M^{\oplus n}$, and its right adjoint is $N\mapsto N^n$, by the tensor–Hom adjunction. It cannot have the one-dimensional space in its essential image. If $M^{\oplus n}$ were one-dimensional, its first summand would make $M$ at most one-dimensional. Then $M=0$ gives dimension zero and a one-dimensional $M$ gives dimension $n$, both impossible. Hence this is a stack-adjoint functor which is not an equivalence.

Exercise 8.4 (synthesis). Refine the cover of Exercise 8.1 by the single label one. Describe the inverse refinement on an $M_2(k)$-module $N$, and prove explicitly how its value at label zero determines $N$.

Solution. The singleton still covers the point, although its map to the old two-label cover is not surjective. The pulled-back algebra is $M_2(k)$. Formula (6.1) gives $M_0=k^{1\times2}\otimes_{M_2(k)}N, \qquad M_1=M_2(k)\otimes_{M_2(k)}N\simeq N.$ Identify a row $r$ with the matrix $\widetilde r$ whose first row is $r$ and second row is zero. The map $r\otimes n\mapsto\widetilde r n$ identifies $M_0$ with $E_{11}N$. Its inverse sends $x\in E_{11}N$ to $(1,0)\otimes x$. One composite is identity because $E_{11}x=x$. For the other, balance $(1,0)\otimes\widetilde r n=((1,0)\widetilde r)\otimes n=r\otimes n$. Thus both composites are identity.

The transition recovering label one is $k^2\otimes_k E_{11}N\xrightarrow{\sim}N, \qquad e_i\otimes x\longmapsto E_{i1}x.$ Its inverse sends $n$ to $\sum_{i=1}^2e_i\otimes E_{1i}n$, where each $E_{1i}n$ is in $E_{11}N$. One composite is identity since $\sum_iE_{i1}E_{1i}=I$; the other is identity since $E_{1i}E_{j1}=\delta_{ij}E_{11}$. These are the actual module comparisons for the inverse refinement. They exhibit equivalence through the remaining label, not an identification of the two algebra sheaves.

9. Further reading

Andrea D'Agnolo and Pietro Polesello, Stacks of twisted modules and integral transforms, open preprint, Sections 3.3–3.4, for invertible-bimodule twisting data, local models and refinement. Here the presentation uses the nerve of an epic cover in the sheaf category; arbitrary semisimplicial data have additional covering hypotheses.
Bimodules and adjoints of module-stack functors, for full reconstruction and its restriction-sensitive adjoint convention.
Twisted modules from coherent gluing, for element sites and effective module descent.
Tensor and Hom for twisted module stacks, for product and inverse line cocycles in the scalar specialization.