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Draft lesson.

Dual and tensor products of locally module stacks

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

The tensor product of two module categories is governed by the tensor product of their algebras. This remains true when the local algebras vary and the changes of description are Morita equivalences. Reversing a transport bimodule gives the dual stack; tensoring two transports gives the product stack. These constructions also retain functors, adjoints and their coherence.

Work in the sheaf category \(\mathcal E\) of a small site, with a fixed commutative ring sheaf \(R\). All algebra scalar maps are central and unital, and all modules and bimodules have agreeing scalar actions. Fix universes for the categories and diagrams involved. Restrictions mean localization within this sheaf category, interpreted on element sites. A locally module stack is an \(R\)-linear stack locally equivalent to \(\mathsf{Mod}(A)\) for a sheaf of \(R\)-algebras; the algebra may change from one local model to another.

We use Bimodules and adjoints of module-stack functors for tensor reconstruction and all transformations. Noncommutative gluing by invertible bimodules supplies presentations, coherent inverse transports, functor descent and refinements. The slice tensor and internal Hom are constructed in Tensor and Hom for twisted module stacks, Section 1.

1. Commuting actions and balanced comparisons

A module over \(A\otimes_R B\) is precisely an \(R\)-module with commuting left \(A\)- and \(B\)-actions. The algebra acts by \((a\otimes b)m=a(bm)\). Conversely its two inclusions give these actions, and their commutation follows from \((a\otimes1)(1\otimes b)=(1\otimes b)(a\otimes1)\). The two constructions are inverse on objects and maps, also on every localization. Similarly an \((A,B)\)-bimodule is a module over \(A\otimes_RB^{\mathrm{op}}\).

For an \((A,B)\)-bimodule \(L\), write \(L^{\mathrm{op}}\) for the \((B^{\mathrm{op}},A^{\mathrm{op}})\)-bimodule with the same underlying \(R\)-module and actions \[ b^{\mathrm{op}}\ell^{\mathrm{op}}a^{\mathrm{op}} =(a\ell b)^{\mathrm{op}}. \tag{1.1} \] For a \((C,B)\)-bimodule \(K\) and a \((B,A)\)-bimodule \(L\), there is a reversal isomorphism \[ L^{\mathrm{op}}\otimes_{B^{\mathrm{op}}}K^{\mathrm{op}} \simeq(K\otimes_B L)^{\mathrm{op}}, \qquad \ell^{\mathrm{op}}\otimes k^{\mathrm{op}} \longmapsto(k\otimes\ell)^{\mathrm{op}}, \tag{1.2} \] between \((A^{\mathrm{op}},C^{\mathrm{op}})\)-bimodules. Balancing becomes \(kb\otimes\ell=k\otimes b\ell\); reversing again gives the inverse map.

We also need the interchange isomorphism \[ (L\otimes_R K)\otimes_{B\otimes_R D}(L'\otimes_R K') \simeq (L\otimes_B L')\otimes_R(K\otimes_D K'), \tag{1.3} \] where \(L\) is an \((A,B)\)-bimodule, \(L'\) a \((B,C)\)-bimodule, \(K\) an \((E,D)\)-bimodule, and \(K'\) a \((D,F)\)-bimodule. It sends \((\ell\otimes k)\otimes(\ell'\otimes k')\) to \((\ell\otimes\ell')\otimes(k\otimes k')\). The \(B\)- and \(D\)-relations become exactly the two relations on the right. The reverse rule respects those relations and the scalar relation, so defines its inverse. These maps are linear for \(A\otimes_RE\) on the left and \(C\otimes_RF\) on the right. The sectionwise rules and their inverses sheafify and commute with localization. No flatness assumption enters this proof.

2. The stack of functors with a stack right adjoint

For the general internal functor stack, cartesian currying and two-fibre products, see Functor stacks, cartesian currying and two-fibre products, Sections 1–4. Write \(\mathsf H(\mathsf M,\mathsf N)\) for the stack whose objects over a base are \(R\)-linear functors between the restricted stacks which admit a right adjoint as a stack functor. Its arrows are all stack natural transformations; choosing an adjoint is unnecessary.

Lemma 2.1. These objects and arrows form an \(R\)-linear stack.

Proof. Given local functors and invertible comparisons satisfying a cocycle, evaluate them on the restrictions of an input object. The target stack glues their outputs and their maps. Its uniqueness on maps makes identity, composition, linearity and the restriction comparisons hold after gluing: each is an equality already true on the cover. This constructs the glued stack functor. Transformations glue componentwise in the same way; their naturality and restriction equations are local equalities.

Choose local stack right adjoints. On an overlap, an isomorphism \(\alpha:F_i\to F_j\) induces the inverse-direction isomorphism \(G_j\to G_i\): for every \(X,Y\), use the adjunctions and precomposition with \(\alpha_X\) to identify \(\operatorname{Hom}(X,G_jY)\) with \(\operatorname{Hom}(X,G_iY)\), then take the represented map. Its inverse uses \(\alpha^{-1}\). Composition of these Hom bijections proves the overlap cocycle. Because the given adjunctions are stack adjunctions, these comparisons commute with every further restriction. The source stack therefore glues the adjoints to a stack functor \(G\). Units and counits correspond to identities under these same bijections, so agree on overlaps and glue. Their triangle identities hold on the cover and hence globally. Thus the glued functor lies in \(\mathsf H\). Additive and scalar operations on transformations are componentwise, completing the proof. \(\square\)

Separate adjoints of fibre functors are insufficient here; their restriction mates must also be invertible. The stack-adjoint requirement keeps this distinction explicit.

Theorem 2.2. Suppose \(\mathsf M=\mathsf T(A,L)\) and \(\mathsf N=\mathsf T(B,K)\) have presentations on a common epic cover. Then \(\mathsf H(\mathsf M,\mathsf N)\) is locally a module stack, with local algebra and transport \[ C_i=B_i\otimes_R A_i^{\mathrm{op}},\qquad J_{ij}=K_{ij}\otimes_R(L_{ji})^{\mathrm{op}}. \tag{2.1} \] Its multiplication uses that of \(K\) and the reversed multiplication of \(L\).

Proof. Locally, bimodule reconstruction identifies a functor with \(P_i\), a left \(B_i\)-, right \(A_i\)-module, and identifies every transformation with a bimodule map. In the gluing theorem its comparison is \[ K_{ij}\otimes_{B_j}P_j\simeq P_i\otimes_{A_i}L_{ij}. \tag{2.2} \] Tensor its right side with \(L_{ji}\) and contract the coherent inverse pair. This is equivalent to a transition \[ K_{ij}\otimes_{B_j}P_j\otimes_{A_j}L_{ji}\simeq P_i. \tag{2.3} \] The domain is \(J_{ij}\otimes_{C_j}P_j\): the comparison sends \((k\otimes\ell^{\mathrm{op}})\otimes p\) to \(k\otimes p\otimes\ell\). The \(B_j\)-relation balances the first two factors; the \(A_j^{\mathrm{op}}\)-relation balances the last two. Reversing that rule gives its inverse.

Recover (2.2) from (2.3) by inserting the inverse contraction \(A_j\to L_{ji}\otimes_{A_i}L_{ij}\), then applying (2.3) to its first three factors. The two coherent-inverse triangles prove that this recovery and the previous contraction are mutually inverse. The comparison cocycle becomes the twisted-object cocycle for \(J\). Explicitly, expanding two successive transitions applies the two \(K\) comparisons, then the two functor comparisons, then contracts the successive \(L\) inverse pairs. The functor comparison law replaces their composite by the single comparison after the \(K\) multiplication and the \(L\) multiplication. Associativity and the inverse-pair triangles combine the two contractions into the contraction for \(L_{ik}\otimes L_{ki}\). This is exactly the transition after multiplication in \(J\). Units follow from the same triangles. Conversely inserting the inverse contractions reverses this calculation, recovering the functor law.

Multiplication in \(J\) is well typed by (1.2) and (1.3): on its second factor it sends \(\ell_{ji}^{\mathrm{op}}\otimes\ell_{kj}^{\mathrm{op}}\) to \(\mu_{kji}(\ell_{kj}\otimes\ell_{ji})^{\mathrm{op}}\). Its four-coordinate equation is the original one with coordinates reversed. Invertibility follows by tensoring the coherent inverses of its two factors and using (1.3). Thus \(J\) is genuine invertible-bimodule gluing data. The gluing theorem identifies its objects and all arrows with the functors and transformations above, including their actual stack adjoints. Lemma 2.1 makes this an equivalence of stacks. \(\square\)

3. Duality of locally module stacks

Define \[ \mathsf D\mathsf M=\mathsf H(\mathsf M,\mathsf{Mod}(R)). \tag{3.1} \] Theorem 2.2 gives \[ \mathsf D\mathsf{Mod}(A)\simeq\mathsf{Mod}(A^{\mathrm{op}}), \qquad A_i^D=A_i^{\mathrm{op}},\quad L_{ij}^D=(L_{ji})^{\mathrm{op}}. \tag{3.2} \] The dual stack is thus built from right modules and reversed transports. This operation does not mean merely reversing the arrows of the original module category.

A stack-adjoint functor \(F:\mathsf M\to\mathsf N\) gives \(\mathsf DF:\mathsf D\mathsf N\to\mathsf D\mathsf M\) by precomposition. This also has a stack right adjoint. Indeed locally \(F=P\otimes_A-\), with \(P\) an \((B,A)\)-bimodule, and precomposition sends a right \(B\)-module \(Q\) to \(Q\otimes_BP\). As a functor on left opposite-algebra modules, this is tensor with \(P^{\mathrm{op}}\), an \((A^{\mathrm{op}},B^{\mathrm{op}})\)-bimodule, by (1.2). It therefore has the tensor–Hom stack adjoint. These local stack adjoints glue by Lemma 2.1. A transformation \(\alpha:F\to F'\) is sent to \(\eta\alpha:\eta F\to\eta F'\); its component maps are the same underlying bimodule maps. Precomposition preserves the reversed order of composition and the identity comparisons.

Proposition 3.1. Evaluation gives a natural equivalence \[ \mathsf M\xrightarrow{\sim}\mathsf D\mathsf D\mathsf M, \qquad X\longmapsto(\eta\longmapsto\eta(X)). \tag{3.3} \]

Proof. In a local \(A\)-model, \(\eta\) is represented by a right \(A\)-module \(Q\). Its value on \(X\) is \(Q\otimes_AX\), equivalently \(X\otimes_{A^{\mathrm{op}}}Q\). For fixed \(X\) this is a stack-adjoint tensor functor, so evaluation defines an object of the double dual. Sending \(X\) to that functor is exactly the bimodule reconstruction equivalence for \((A^{\mathrm{op}})^{\mathrm{op}}=A\). In particular it is fully faithful on all maps and essentially surjective locally. Evaluation commutes with the model changes because it uses the actual functor value, and the transport formula in (3.2) reverses twice to \(L_{ij}\). The local inverse equivalences therefore agree on overlaps and descend; their inverse comparisons are the local reconstruction comparisons and glue too. This proves (3.3). Naturality at \(F\) is the equality \((\eta F)(X)=\eta(FX)\), and on transformations it is equality of the corresponding component maps. \(\square\)

No finite-generation or module-reflexivity hypothesis on \(X\) is needed in this stack biduality statement.

4. The internal tensor product and external objects

Define \[ \mathsf M\boxtimes_R\mathsf N=\mathsf H(\mathsf D\mathsf M,\mathsf N). \tag{4.1} \] Combining Theorem 2.2 with (3.2) and flipping the two algebra factors gives local data \[ A_i\otimes_RB_i,\qquad L_{ij}\otimes_RK_{ij}, \qquad\mu\otimes\nu \tag{4.2} \] with the interchange (1.3) understood in its multiplication. Consequently \[ \mathsf{Mod}(A)\boxtimes_R\mathsf{Mod}(B) \simeq\mathsf{Mod}(A\otimes_RB). \tag{4.3} \] These are tensor products over the fixed \(R\); no derived tensor products occur here.

There is an external object operation \((X,Y)\mapsto X\boxtimes Y\). On a common cover it is \(X_i\otimes_RY_i\), with commuting actions \((a\otimes b)(x\otimes y)=ax\otimes by\). Its transition is the interchange map followed by the transitions for \(X\) and \(Y\). Their triple and unit equations prove its own equations. Tensoring maps defines its arrows, compatibly with restriction. Refining the cover restricts all these same formulas, so the refinement equivalences give a well-defined operation on the stacks.

For fixed \(X\), this is a stack-adjoint functor in \(Y\). Locally its bimodule is \(X\otimes_RB\), a left \(A\otimes_RB\)-, right \(B\)-module. The gluing transitions of \(X\) supply its functor comparisons; the gluing theorem therefore supplies an actual stack right adjoint. The same argument applies in the other variable. This allows arbitrary modules \(X,Y\), including nonflat ones.

5. Closed currying

Theorem 5.1. For locally module stacks there is a natural equivalence of stacks \[ \mathsf H(\mathsf M\boxtimes_R\mathsf N,\mathsf P) \simeq\mathsf H(\mathsf M,\mathsf H(\mathsf N,\mathsf P)). \tag{5.1} \] It sends \(F\) to \(X\mapsto(Y\mapsto F(X\boxtimes Y))\).

Proof. On a common cover let the three algebras be \(A,B,C\). The left side is the category of \((C,A\otimes_RB)\)-bimodules \(T\). On the right, the inner stack is \(\mathsf{Mod}(C\otimes_RB^{\mathrm{op}})\); reconstruction for the outer functor gives \((C\otimes_RB^{\mathrm{op}},A)\)-bimodules. Both descriptions are exactly an \(R\)-module with commuting left \(C\)-action and right \(A\)- and \(B\)-actions. Both have precisely the maps preserving these three actions.

The displayed currying functor identifies them by the balanced isomorphism \[ T\otimes_{A\otimes_RB}(X\otimes_RY) \simeq (T\otimes_AX)\otimes_BY, \quad t\otimes(x\otimes y)\longmapsto(t\otimes x)\otimes y. \tag{5.2} \] The right \(B\)-action on \(T\otimes_AX\) is \((t\otimes x)b=(tb)\otimes x\). It is well-defined because the two right actions commute. The \(A\)- and \(B\)-relations on either side of (5.2) become those on the other; the reverse rule is its inverse. Thus currying really gives the functors indicated, not just two abstractly isomorphic module categories. The outer functor \(X\mapsto T\otimes_AX\) and each inner functor have stack right adjoints by reconstruction. On maps, both descriptions send a three-action map \(h\) to the transformation induced by \(h\), so the identification is fully faithful and essentially surjective.

Formula (5.2) commutes with all transport comparisons: their scalar interchange moves the same factors, and their algebra balancing removes the same actions on \(T,X,Y\). Thus the local currying comparisons agree on overlaps and define the stated global functor. Conversely an object of the right side gives local three-action modules. The overlap isomorphisms of its functors are, by full faithfulness of reconstruction, precisely isomorphisms of those modules satisfying the left-side transport law. They descend to a left-side functor. Its local tensor adjoints glue by Lemma 2.1. The inverse comparisons are (5.2) and its inverse, and they agree on overlaps by the same calculation. Transformations descend uniquely componentwise. This proves the global equivalence, naturality in all three stacks and compatibility with restriction. \(\square\)

6. Coherence and scalar extension

The unit for \(\boxtimes_R\) is \(\mathsf{Mod}(R)\). The unit comparisons, symmetry and associator are induced in local models by \[ R\otimes_RA\simeq A,\qquad A\otimes_RB\simeq B\otimes_RA, \qquad (A\otimes_RB)\otimes_RC\simeq A\otimes_R(B\otimes_RC). \tag{6.1} \] The same maps on transport bimodules respect their multiplications by (1.3), so give global equivalences. All coherence diagrams commute: on local pure tensors, the pentagon changes only parentheses, the two unit triangles delete the same unit factor, and the hexagons perform the same permutation of three factors. Reversing the symmetry twice is identity. Pure tensors locally generate; equality there proves equality of the sheaf maps, hence of the global comparisons. For stack-adjoint functors represented by \(P,Q\), the product is represented by \(P\otimes_RQ\). Equation (1.3) identifies the product of composites with the composite of products, and the same ordered-factor calculation proves its interchange and unit laws.

There is also an equivalence \[ \mathsf D(\mathsf M\boxtimes_R\mathsf N) \simeq\mathsf D\mathsf M\boxtimes_R\mathsf D\mathsf N. \tag{6.2} \] Locally it is \((A\otimes_RB)^{\mathrm{op}}\simeq A^{\mathrm{op}}\otimes_RB^{\mathrm{op}}\). The reverse transports on both sides are \((L_{ji}\otimes_RK_{ji})^{\mathrm{op}}\), identified with the tensor of the two reverse transports. Reversal and interchange identify their multiplications, so these local equivalences descend. Their associativity and symmetry compatibility follows from the same pure-factor calculations.

Let \(S\) now be a commutative \(R\)-algebra sheaf. An object of \(\mathsf{Mod}(S;\mathsf M)\) is an object of \(\mathsf M\) with an \(R\)-algebra map \(S\to\underline{\operatorname{End}}(X)\); its maps preserve that action. These objects and maps form a stack since actions and their laws descend. Commuting actions in Section 1 give an equivalence \[ \mathsf{Mod}(S)\boxtimes_R\mathsf M\simeq\mathsf{Mod}(S;\mathsf M). \tag{6.3} \] In local models both sides have algebra \(S\otimes_RA_i\) and transport \(S\otimes_RL_{ij}\). Interchange and the unit of \(S\) identify their inverse transports and their cocycle; the object comparisons just retain the commuting \(S\)- and \(A_i\)-actions. They therefore glue to (6.3), including all maps.

7. The two adjoints of forgetting a scalar action

The forgetful functor \(U:\mathsf{Mod}(S;\mathsf M)\to\mathsf M\) has both adjoints: \[ S\otimes_R-\ \dashv\ U\ \dashv\ [S,-]_R. \tag{7.1} \] Here the two outer operations are formed in local module models and then descended. In the rightmost object the local algebra acts on the values of a map, and \(S\) acts by \((s\cdot f)(t)=f(ts)\).

Proof. In an \(A\)-model the free object is \(S\otimes_RX\), with \(A\) acting on \(X\) and \(S\) on its first factor. An \(A\)-linear map \(q:X\to UN\) extends uniquely to the \(S\otimes_RA\)-linear map \(s\otimes x\mapsto s q(x)\); restriction to \(1\otimes x\) is its inverse. This proves the left adjunction.

For an \(A\)-linear map \(q:UN\to X\), put \(\widehat q(n)(s)=q(sn)\). It is \(A\)-linear in \(n\), since the \(S\)- and \(A\)-actions on \(N\) commute. Also \(\widehat q(s_0n)(s)=q(ss_0n)=(s_0\widehat q(n))(s)\). Thus it defines an \(S\otimes_RA\)-linear map \(N\to[S,X]_R\). Conversely evaluate such a map at \(1\) to get \(q\). The formulas are inverse: \(\widehat q(n)(1)=q(n)\), while \(S\)-linearity gives \(h(sn)(1)=h(n)(s)\). This proves the right adjunction. Both bijections are natural and commute with slice restriction. Their triangles follow by extending then restricting, or by evaluating then recovering the same functions.

These local constructions descend along a Morita transport \(E=L_{ij}\otimes_{A_j}-\). Interchange gives \(E(S\otimes_RX)\simeq S\otimes_REX\). The right-adjoint comparison is invertible as well: the inverse equivalence to \(E\), the preceding comparison, and the tensor–Hom adjunction give, naturally in a test module \(Z\), \[ \begin{aligned} \operatorname{Hom}_{A_i}(Z,E[S,X]_R) &\simeq\operatorname{Hom}_{A_j}(E^{-1}Z,[S,X]_R)\\ &\simeq\operatorname{Hom}_{A_j}(S\otimes_RE^{-1}Z,X)\\ &\simeq\operatorname{Hom}_{A_i}(S\otimes_RZ,EX)\\ &\simeq\operatorname{Hom}_{A_i}(Z,[S,EX]_R). \end{aligned} \tag{7.2} \] The represented isomorphism commutes with the \(S\)-action, since precomposition by multiplication on \(S\) is the same in all four bijections. Composing two such chains is the chain for the composite transport; the inverse-equivalence triangles cancel the intervening pair. Hence it respects the transport cocycle and units. The free and cofree objects, and their displayed Hom bijections, therefore descend. Their units, counits and triangles descend as in Lemma 2.1. This proves (7.1) for the stacks, without flatness or finiteness of \(S\). \(\square\)

8. Graded exercises with complete solutions

Exercise 8.1 (foundation). Take \(R=\mathbb Z\) on a point and \(A=B=\mathbb Z/2\mathbb Z\). Verify the local product algebra and explain why the tensor comparisons above do not imply exactness of tensoring over \(R\).

Solution. Multiplication gives \(A\otimes_{\mathbb Z}B\simeq\mathbb Z/2\mathbb Z\), with inverse \(\bar x\mapsto\bar x\otimes\bar1\). Indeed \(\bar x\otimes\bar y=\overline{xy}\otimes\bar1\) by scalar balancing, and multiplication followed by this inverse fixes every pure tensor. Thus the product stack is \(\mathsf{Mod}(\mathbb Z/2\mathbb Z)\). On a one-label cover both regular transports are invertible; their product is the regular bimodule of the displayed product algebra. Its inverse contraction is multiplication over that algebra, independently of flatness over \(\mathbb Z\).

The injection \(\mathbb Z\xrightarrow{2}\mathbb Z\) becomes the zero map \(\mathbb Z/2\mathbb Z\to\mathbb Z/2\mathbb Z\) after tensoring with \(A\). That map is not injective. Tensoring with \(A\) is therefore not exact. The earlier comparisons were proved from balancing and explicit inverse maps; they assert isomorphisms between specified tensor expressions, not preservation of all injections.

Exercise 8.2 (calculation). Let \(A=M_2(k)\), \(B=M_3(k)\), \(X=k^2\), \(Y=k^3\), and \(T=k^{1\times2}\otimes_k k^{1\times3}\), with its two right matrix actions. Compute the functor represented by \(T\) on \(X\boxtimes Y\) and its curried version. Check both inverse evaluation maps.

Solution. The two actions on \(T\) affect different factors, so commute even though each matrix algebra is noncommutative. Interchange and evaluation give \[ T\otimes_{A\otimes_kB}(X\otimes_kY) \simeq(k^{1\times2}\otimes_A k^2)\otimes_k (k^{1\times3}\otimes_B k^3)\simeq k. \] For size \(n\), row–column multiplication is balanced and sends \(e_i^*\otimes e_j\) to \(\delta_{ij}\). Balancing with \(E_{ii}\) makes the tensor zero for \(i\ne j\). For equal indices, balancing with \(E_{1i}\) gives \(e_i^*\otimes e_i=e_1^*\otimes e_1\). Thus \(c\mapsto c e_1^*\otimes e_1\) is its inverse on a spanning set, and evaluation of that tensor is \(c\). This proves both inverse comparisons without a dimension shortcut.

Currying first forms \(T\otimes_AX\simeq k^{1\times3}\) as a right \(B\)-module, then tensors it with \(Y\) to give \(k\). The uncurried scalar is \((u x)(v y)\) for \((u\otimes v)\otimes(x\otimes y)\); the curried scalar is the same expression. Formula (5.2) and its reverse identify the two balanced tensors before evaluation. Hence this calculation respects both actions and the actual currying comparison.

Exercise 8.3 (obstruction). Explain why stack biduality does not imply that \(\mathsf M\boxtimes_R\mathsf D\mathsf M\) is equivalent to \(\mathsf{Mod}(R)\) by the regular-bimodule functor. Compare \(A=k[t]\) with \(A=M_2(k)\). Compute that functor's right adjoint.

Solution. For a local model, put \(A^e=A\otimes_kA^{\mathrm{op}}\). The regular-bimodule functor is \(V\mapsto V\otimes_kA\), with \((a\otimes b^{\mathrm{op}})(v\otimes x)=v\otimes axb\). An \(A^e\)-linear map from it to a bimodule \(N\) is determined by the values on \(v\otimes1\). Such values commute with every element of \(A\), since the two actions on \(1\) agree. Conversely a \(k\)-linear map \(V\to Z_A(N)=\{n:an=na\text{ for all }a\}\) extends by \(v\otimes a\mapsto a f(v)\); centrality makes it right-linear as well as left-linear. These rules are inverse. The right adjoint is therefore \(Z_A(N)=\underline{\operatorname{Hom}}_{A^e}(A,N)\).

For \(A=k[t]\), the enveloping algebra is \(k[t,u]\), and \(t-u\) annihilates every \(V\otimes_kA\). It does not annihilate the regular \(k[t,u]\)-module: its value on \(1\) is the nonzero polynomial \(t-u\). Thus that module is outside the essential image, and the regular-bimodule functor is not an equivalence.

For \(A=M_2(k)\), the map \(A^e\to\operatorname{End}_k(A)\) sends \(a\otimes b^{\mathrm{op}}\) to \(x\mapsto axb\). It preserves multiplication because composition reverses the right multipliers. Moreover \(E_{ab}\otimes E_{cd}^{\mathrm{op}}\) sends \(E_{ij}\) to \(\delta_{bi}\delta_{jc}E_{ad}\). These sixteen maps are exactly the elementary maps between the four basis vectors of \(A\), so the algebra map is an isomorphism. The regular \(A\) is the standard four-dimensional module for this matrix algebra. Its inverse bimodule is the row of length four. The row–column contraction from Exercise 8.2 is one inverse comparison; the column–row contraction over \(k\) is the outer-product isomorphism to \(M_4(k)\), since \(e_i\otimes e_j^*\) maps to its basis matrix \(E_{ij}\). The two contraction composites on a row or column are identity, by \(E_{ij}e_l=\delta_{jl}e_i\) and \(e_l^*E_{ij}=\delta_{li}e_j^*\). These prove the Morita equivalence \(V\mapsto V\otimes_kA\) and its inverse. Its right adjoint is the displayed centre functor. In particular \(Z_A(A)=kI\): commuting with \(E_{11}\) kills off-diagonal entries, and commuting with \(E_{12}\) equates the two diagonal entries.

In both examples Proposition 3.1 still gives \(\mathsf{Mod}(A)\simeq\mathsf D\mathsf D\mathsf{Mod}(A)\). Its local algebra is \((A^{\mathrm{op}})^{\mathrm{op}}=A\); it makes no assertion that \(A^e\) is Morita equivalent to \(k\). This separates the two claims.

Exercise 8.4 (synthesis). Take \(R=\mathbb Z\), \(S=\mathbb Z/2\mathbb Z\), and the ordinary module stack on a point. Compute the free and cofree objects on \(\mathbb Z\). Show directly why the free object cannot be the right adjoint to forgetting the \(S\)-action.

Solution. The free object is \(S\otimes_{\mathbb Z}\mathbb Z\simeq S\). The cofree object is \(\operatorname{Hom}_{\mathbb Z}(S,\mathbb Z)=0\): the image of \(\bar1\) must be killed by two, whereas \(\mathbb Z\) has no nonzero such element. These are the two adjoints in (7.1), with their respective directions.

If the free functor were right adjoint to forgetting, taking \(N=S\) and \(X=\mathbb Z\) would give a bijection \(\operatorname{Hom}_{\mathbb Z}(UN,\mathbb Z)\simeq\operatorname{Hom}_S(N,S\otimes_{\mathbb Z}\mathbb Z)\). The left group is zero, while the right is \(\operatorname{Hom}_S(S,S)\simeq S\), which has two elements. No such bijection exists. For the actual right adjoint both sides are zero. For the actual left adjunction, \(\operatorname{Hom}_S(S,S)\simeq\operatorname{Hom}_{\mathbb Z}(\mathbb Z,US)\simeq S\), with the maps given by extension from the image of \(1\). These direct computations confirm both directions and distinguish tensoring from coinduction.

9. Further reading