Draft lesson.

Finite projective bimodules and both adjoints of tensor

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

A module tensor functor always has a Hom right adjoint. That Hom functor is itself tensor exactly when every coefficient column is compact projective. A different condition controls the existence of a left adjoint: every coefficient row must be compact projective. We prove both criteria, including their converse directions and the actual finite-dual comparison maps.

Use Modules and tensor functors over small preadditive categories, §§2–5, for additive Yoneda, actual presentations, balanced tensor and all tensor–Hom adjunction maps. Cauchy completion and Morita equivalence of many-object modules, §2, proves that compact projective modules are exactly finite representable retracts. Apply that theorem to the opposite category for left modules as well. Tensor duality and coherent inverses supplies the general ordered dual-pair framework. No commutativity or braiding is assumed here.

1. The two coefficient directions

Fix a universe and small preadditive categories \(\mathcal C,\mathcal D\). A bimodule is a separately additive functor \(B\colon\mathcal D^{\mathrm{op}}\times\mathcal C\to\mathsf{Ab}\). Its columns \(B_c=B(-,c)\) are right \(\mathcal D\)-modules; its rows \(L_d=B(d,-)\) are left \(\mathcal C\)-modules. Write

T_B(M)(d)=M\otimes_{\mathcal C}L_d, \qquad Q_B(N)(c)=\operatorname{Hom}_{\mathcal D}(B_c,N). \tag{1.1}

The existing tensor–Hom proof gives \(T_B\dashv Q_B\) without finiteness. The right \(\mathcal C\)-action on \(Q_B\) precomposes with \(B(u):B_c\to B_{c'}\) for \(u:c\to c'\). The \(\mathcal D\)-action on tensor comes from the rows' contravariant action. Both functors are additive. All module morphisms have group-homomorphism components.

We use one elementary consequence of the finite-retract theorem. For a right module \(P\), the group-valued functor \(\operatorname{Hom}(P,-)\) preserves all small colimits in \(\mathsf{Ab}\) if and only if \(P\) is compact projective. Necessity follows because preservation of filtered colimits is compactness, and preservation of cokernels makes Hom surjective on every epimorphism, proving projectivity. Conversely a finite sum of representables has Hom a finite direct sum of evaluations, hence preserves coproducts; a retract retains this property through the natural splitting. Projectivity makes Hom exact and therefore cokernel-preserving. Every colimit is a cokernel of the difference of two maps between coproducts, so it preserves all colimits. This argument applies equally to left modules.

2. Finite duality between right and left modules

For a right \(\mathcal C\)-module \(P\), and a left module \(L\), define

P^{\mathrm r}(c)=\operatorname{Hom}_{\mathcal C}(P,h_c), \qquad L^{\mathrm l}(c)=\operatorname{Hom}_{\mathcal C\text{-}\mathsf{Mod}}(L,\ell^c), \quad h_c=\mathcal C(-,c),\quad\ell^c=\mathcal C(c,-). \tag{2.1}

The first is a left module: for \(u:c\to c'\), postcompose with \(h(u):h_c\to h_{c'}\). The second is a right module: precomposition of arrows gives \(\ell^{c'}\to\ell^c\), and postcompose a transformation with that map. Precomposition in \(P\) or \(L\) reverses their morphisms under duality. These actions are additive and obey composition.

There are canonical bidual maps. For \(p\in P(c)\), its image in \((P^{\mathrm r})^{\mathrm l}(c)\) is the transformation whose component at \(t\) sends \(\lambda:P\to h_t\) to \(\lambda_c(p)\in\mathcal C(c,t)\). For \(l\in L(c)\), its image in \((L^{\mathrm l})^{\mathrm r}(c)\) sends \(\lambda:L\to\ell^t\) to \(\lambda_c(l)\in\mathcal C(t,c)\). Naturality of \(\lambda\) and Yoneda make these module maps, natural in the original module.

Lemma 2.1. These duals give inverse contravariant equivalences between compact projective right and left modules. Both canonical bidual maps are invertible on these objects.

Proof. Yoneda gives \(h_a^{\mathrm r}=\ell^a\) and \((\ell^a)^{\mathrm l}=h_a\), and the displayed bidual maps become identity maps under these identifications. Duality takes finite sums to finite products, which are finite sums in modules. It also takes a retract to a retract, with the inclusion and projection interchanged. Thus finite representable retracts become finite representable retracts on the other side.

For completeness, let \(P\xrightarrow{s}F\xrightarrow{r}P\), \(rs=1\), be such a finite retract, and write \(D\) for double dual and \(\beta\) for its canonical comparison. The comparison on \(F\) is invertible by the finite-sum computation. An inverse on \(P\) is

r\,\beta_F^{-1}\,D(s):D(P)\longrightarrow P. \tag{2.2}

Naturality gives \(D(s)\beta_P=\beta_Fs\) and \(\beta_Pr=D(r)\beta_F\). Substituting these equations into the two composites of (2.2) with \(\beta_P\) gives \(rs=1\) and \(D(r)D(s)=D(rs)=1\), respectively. The same proof applies to a left module. Duality reverses all maps by its formula, and the natural bidual isomorphisms show that the two dual operations are inverse on the full categories. \(\square\)

3. Exactly when the right adjoint is tensor

Define a bimodule, whether or not its columns are finite,

B^{\mathrm r}(c,d)=\operatorname{Hom}_{\mathcal D}(B_c,h_d^{\mathcal D}). \tag{3.1}

For \(u:c\to c'\), its right \(\mathcal C\)-action precomposes with \(B(u)\); for \(v:d\to d'\), its left \(\mathcal D\)-action postcomposes with \(h(v)\). The actions commute, making it a \(\mathcal C^{\mathrm{op}}\times\mathcal D\) bimodule. There is a canonical natural map

\xi_N:T_{B^{\mathrm r}}(N)\longrightarrow Q_B(N), \qquad [n\otimes a]\longmapsto\tau_n a. \tag{3.2}

Here \(n\in N(d)\), \(a:B_c\to h_d\), and \((\tau_n)_t(v)=N(v)n\). Additive Yoneda makes the formula bilinear; the equation \(\tau_{N(v)n}=\tau_nh(v)\) proves balancing. All restrictions and maps in \(N\) commute with its formula.

Theorem 3.1. The following conditions are equivalent: \(Q_B\) preserves all small colimits; \(Q_B\) is naturally isomorphic to a module tensor functor; every column \(B_c\) is compact projective. Under these conditions (3.2) is the tensor isomorphism.

Proof. Evaluate colimit preservation of \(Q_B\) at each \(c\) and apply the criterion in §1. Conversely that criterion on every column gives pointwise preservation and hence preservation by \(Q_B\). Every tensor functor preserves colimits, proving the necessary direction from tensor form.

To prove the specified tensor form with both inverse maps, choose a finite representable retract of one column \(P=B_c\). It gives elements \(p_j\in P(d_j)\) and maps \(\sigma_j:P\to h_{d_j}\) such that

1_P=\sum_j\tau_{p_j}\sigma_j. \tag{3.3}

These arise by restricting a split finite presentation to its summands, just as in the finite-retract proof. The inverse of (3.2) at \(c\) sends

g:P\to N\quad\longmapsto\quad \sum_j[g_{d_j}(p_j)\otimes\sigma_j]. \tag{3.4}

Applying (3.2) to (3.4) gives \(\sum_j g\tau_{p_j}\sigma_j=g\). For a pure tensor \([n\otimes a]\), the other composite gives \(\sum_j[N(a_{d_j}(p_j))n\otimes\sigma_j]\). Balancing turns this into \([n\otimes\sum_jh(a_{d_j}(p_j))\sigma_j]\), which is \([n\otimes a]\) by (3.3) and Yoneda. Pure tensors generate, proving the other identity on the whole group. Thus the inverse is independent of the finite retraction chosen. Since \(\xi\) is natural, its inverse is natural in \(c,N\) and respects every module map. This proves all claimed equivalences. \(\square\)

The transported unit of \(T_B\dashv T_{B^{\mathrm r}}\), at \(m\in M(c)\), is

m\longmapsto\sum_j[[m\otimes p_j]\otimes\sigma_j]. \tag{3.5}

The transported counit evaluates

[[n\otimes a]\otimes b]\longmapsto N(a_d(b))n, \quad b\in B(d,c),\quad a:B_c\to h_t,\ n\in N(t). \tag{3.6}

These are obtained from the actual tensor–Hom unit and counit via (3.2), so their naturality and both triangles follow by substituting inverse natural isomorphisms into the two previously proved triangle equations. In particular the triangle on \([m\otimes b]\) reduces to \(\sum_j[m\otimes P(\sigma_{j,d}(b))p_j]=[m\otimes b]\), exactly (3.3). This tensor right adjoint also has a right adjoint by the general tensor–Hom theorem; no finiteness of its new coefficients is asserted.

4. The finite left-module pairing

For a compact projective left \(\mathcal C\)-module \(L\), put \(A=L^{\mathrm l}\), a compact projective right module by Lemma 2.1. There is a canonical natural comparison

\phi_{M,L}:M\otimes_{\mathcal C}L\longrightarrow \operatorname{Hom}_{\mathcal C}(A,M), \qquad \phi([m\otimes l])_x(\lambda)=M(\lambda_c(l))m, \tag{4.1}

where \(m\in M(c)\), \(l\in L(c)\), and \(\lambda:L\to\ell^x\). The evaluated arrow \(\lambda_c(l)\) lies in \(\mathcal C(x,c)\), so the expression belongs to \(M(x)\). It is additive in all variables. For \(u:c\to c'\), naturality gives \(\lambda_{c'}(L(u)l)=u\lambda_c(l)\); contravariance of \(M\) proves balancing. Precomposition of \(\lambda\) with the action on \(\ell\) proves the right-module naturality in \(x\). Naturality in \(M\) is immediate from its module maps. For a left-module map \(L'\to L\), its dual \(A\to (L')^{\mathrm l}\) and tensor action commute with (4.1) by evaluation, proving naturality in \(L\) as well.

Choose a finite left dual basis \(x_j\in L(c_j)\), \(\sigma_j:L\to\ell^{c_j}\), with \(1_L=\sum_j\tau_{x_j}\sigma_j\); for left modules \((\tau_x)_c(v)=L(v)x\). An inverse to (4.1) is

f:A\to M\quad\longmapsto\quad \sum_j[f_{c_j}(\sigma_j)\otimes x_j]. \tag{4.2}

To check its first composite, evaluate at \(\lambda\in A(x)\). It gives \(\sum_j M(\lambda_{c_j}(x_j))f_{c_j}(\sigma_j)\). Naturality of \(f\) turns this into \(f_x(\sum_j A(\lambda_{c_j}(x_j))\sigma_j)\). The sum in parentheses is \(\lambda\), by the finite dual-basis identity on \(L\), evaluated and then postcomposed into \(\ell^x\). For the other composite start with \([m\otimes l]\). Formula (4.2) gives \(\sum_j[M(\sigma_{j,c}(l))m\otimes x_j]\). Balancing gives \([m\otimes\sum_j L(\sigma_{j,c}(l))x_j]=[m\otimes l]\). Thus both inverse identities hold, and independence of the chosen basis and naturality of the inverse follow from the canonical map (4.1).

5. Exactly when tensor has a left adjoint

Theorem 5.1. The functor \(T_B\) has a left adjoint if and only if every row \(L_d=B(d,-)\) is compact projective as a left \(\mathcal C\)-module. When this holds, define

B^{\mathrm l}(c,d)=\operatorname{Hom}_{\mathcal C\text{-}\mathsf{Mod}}(L_d,\ell^c). \tag{5.1}

Then \(S=T_{B^{\mathrm l}}\) is that left adjoint.

Proof of sufficiency. The \(\mathcal C\)-action in (5.1) is the right action on the dual. An arrow \(v:d\to d'\) gives \(L_{d'}\to L_d\), so precomposition gives \((L_d)^{\mathrm l}\to(L_{d'})^{\mathrm l}\), the covariant \(\mathcal D\)-action. These actions commute by their formulas. Thus (5.1) is a \(\mathcal C^{\mathrm{op}}\times\mathcal D\) bimodule. Applying (4.1) to every row gives a natural right \(\mathcal D\)-module isomorphism

T_B(M)(d)\simeq\operatorname{Hom}_{\mathcal C}(B^{\mathrm l}(-,d),M). \tag{5.2}

The row-map naturality proved in §4 is exactly its compatibility with \(v\). Tensor–Hom for \(B^{\mathrm l}\) identifies the right side as the right adjoint of \(S\). Hence \(S\dashv T_B\), on all modules and maps.

Explicitly a map \(f:S(N)\to M\) corresponds to the map \(N\to T_B(M)\) whose value at \(n\in N(d)\) is \(\phi_{M,L_d}^{-1}(\lambda\mapsto f_c[n\otimes\lambda])\). Conversely \(\theta:N\to T_B(M)\) gives \([n\otimes\lambda]\mapsto\phi_{M,L_d}(\theta_d(n))_c(\lambda)\). The inverse checks in §4 and tensor–Hom prove that these formulas are inverse, balanced and natural. The unit takes \(n\) to \(\phi_{S(N),L_d}^{-1}(\lambda\mapsto[n\otimes\lambda])\); the counit evaluates \([z\otimes\lambda]\mapsto\phi_{M,L_d}(z)_c(\lambda)\). Their triangles follow by these two inverse transpose rules, or by transporting the proved tensor–Hom triangles through (5.2).

Proof of necessity. Suppose \(S\dashv T_B\). Put \(A_d=S(h_d^{\mathcal D})\). Adjunction and Yoneda give

\operatorname{Hom}_{\mathcal C}(A_d,M)\simeq T_B(M)(d). \tag{5.3}

Tensor preserves all colimits, so the right side does also; §1 makes \(A_d\) compact projective. The left adjoint \(S\) preserves colimits and is additive by the module reconstruction theorem. The adjunction maps are additive too: the transpose is composition with the unit after applying the additive functor, and its inverse uses the additive left adjoint and counit. Thus (5.3) is an isomorphism of groups, natural in \(M\).

Set \(M=h_c^{\mathcal C}\). Co-Yoneda identifies its right side with \(B(d,c)\). Naturality in all maps \(h_c\to h_{c'}\) gives an isomorphism of left modules

L_d\simeq A_d^{\mathrm r}. \tag{5.4}

Lemma 2.1 makes that left dual compact projective, proving necessity. Both criteria and the formula for the left adjoint are therefore proved. \(\square\)

6. Keep the two finiteness conditions separate

The conclusions can be read in the following table. A compact projective coefficient is a finite representable retract on its indicated side.

Required conclusion	Exact coefficient condition
\(T_B\dashv Q_B\)	none
\(Q_B\cong T_{B^{\mathrm r}}\)	every right column \(B_c\) compact projective
\(T_{B^{\mathrm l}}\dashv T_B\)	every left row \(L_d\) compact projective

Having both finite conditions does not assert invertibility of \(T_B\), global finite generation of the whole bimodule, or finiteness of the indexing categories. If either category is empty, the relevant functor is zero, all existing rows or columns are zero modules, and the criteria give the zero adjoints. Zero coefficients have empty dual bases and all displayed inverse maps reduce to the unique zero maps. No choices of bases compatible across objects are required: the canonical comparison maps force the inverses' naturality.

7. Four graded exercises with complete solutions

Exercise 1 (grade 1). For a field \(k\) and a finite-dimensional vector space \(W\), write both tensor adjoints of \(-\otimes_kW\), the right-adjunction unit and counit, and their two triangle checks. Show that both adjoints do not imply an equivalence. Explain the two obstructions when \(W=\bigoplus_{n\ge0}k\).

Solution. Both rows and columns are \(W\), finite free. Both adjoints are tensor with \(W^*=\operatorname{Hom}_k(W,k)\). The right comparison sends \(n\otimes\lambda\) to \(w\mapsto n\lambda(w)\), with inverse \(g\mapsto\sum_i g(w_i)\otimes\lambda_i\) for any basis and dual basis. The left-adjunction pairing sends \(m\otimes w\) to \(\lambda\mapsto m\lambda(w)\), with inverse \(f\mapsto\sum_i f(\lambda_i)\otimes w_i\). These are exactly (3.2)–(3.4) and (4.1)–(4.2).

The right unit is \(m\mapsto\sum_i(m\otimes w_i)\otimes\lambda_i\); its counit sends \((n\otimes\lambda)\otimes w\) to \(n\lambda(w)\). The first triangle sends \(m\otimes w\) to \(\sum_i m\otimes w_i\lambda_i(w)=m\otimes w\). The second sends \(n\otimes\lambda\) to \(\sum_i n\lambda(w_i)\otimes\lambda_i=n\otimes\lambda\). All formulas are natural and independent of the chosen basis because the two comparisons are canonical inverses.

If \(\dim W=2\), the one-dimensional space \(k\) is not of the form \(M\otimes W\): a nonzero \(M\) contributes a two-dimensional subspace by choosing one nonzero vector of \(M\), whereas the zero space gives zero. Thus the functor is not essentially surjective. For infinite free \(W\), Hom from \(W\) fails coproduct preservation: the identity \(W\to\bigoplus_n k\) has no uniformly finite set of nonzero target-coordinate maps. Hence the right adjoint is not tensor. Tensor also fails to preserve products: under \((\prod_n k)\otimes W\to\prod_n W\), each image has all coordinates in one finite-dimensional span of vectors of \(W\), whereas the tuple of distinct basis vectors has no such span. A functor with a left adjoint is a right adjoint and preserves products, so tensor has no left adjoint. These are the two separate failed finite conditions.

Exercise 2 (grade 2). Take \(B=\mathbb Z/d\), \(d\ge2\), as a \(\mathbb Z\)-bimodule. Show that its compactness does not suffice for either finite-adjoint criterion, and compute both failures directly.

Solution. A map \(\mathbb Z/d\to N\) is an element \(n\in N\) with \(dn=0\). In a filtered colimit, an element and the one equation \(dn=0\) are represented at a common stage, and equality is tested at a later common stage. Hence this Hom functor preserves filtered colimits: the module is compact. It is not projective, since the identity on \(\mathbb Z/d\) cannot lift through \(\mathbb Z\twoheadrightarrow\mathbb Z/d\); every map from \(\mathbb Z/d\) to \(\mathbb Z\) is zero.

The always-existing right adjoint is \(Q(N)=N[d]\). It sends the epimorphism \(\mathbb Z\twoheadrightarrow\mathbb Z/d\) to \(0\to\mathbb Z/d\), not an epimorphism. Thus it is not right exact and cannot be tensor. The tensor functor is \(M\mapsto M/dM\). It sends the monomorphism \(\mathbb Z\xrightarrow d\mathbb Z\) to the zero endomorphism of the nonzero group \(\mathbb Z/d\), not a monomorphism. If it had a left adjoint it would preserve kernels, a contradiction. Compactness on either side must therefore be accompanied by projectivity.

Exercise 3 (grade 3). Let \(R=\begin{pmatrix}\mathbb Z&\mathbb Z/d\\0&\mathbb Z\end{pmatrix}\), \(d\ge2\), and \(e=\operatorname{diag}(1,0)\). For \(T(N)=N\otimes_{\mathbb Z}eR\), prove that the right adjoint is tensor with \(Re\), compute the comparison and inverse, and show that \(T\) nevertheless has no left adjoint.

Solution. The bimodule column \(eR\) is a compact projective right \(R\)-module, being the image of the idempotent left multiplication by \(e\) on \(R_R\). An \(R\)-linear map \(eR\to R\) is determined by its value at \(e\), which lies in \(Re\); the inverse sends \(x\in Re\) to \(er\mapsto xr\). The resulting left \(R\)-action is multiplication on \(Re\), and the right scalar action is its usual \(\mathbb Z\)-action. Therefore the right dual coefficient is precisely \(Re\).

The always-existing right adjoint \(\operatorname{Hom}_R(eR,M)\) identifies with \(Me\) by evaluation at \(e\), inverse \(m\mapsto(er\mapsto mr)\). Its tensor comparison is \(M\otimes_R Re\to Me\), \(m\otimes re\mapsto mre\), with inverse \(z\mapsto z\otimes e\). The first composite is \(ze=z\); the other follows by balancing, \(mre\otimes e=m\otimes ree=m\otimes re\). These maps commute with module morphisms and scalar actions, verifying the full natural tensor adjoint.

As a left \(\mathbb Z\)-module, the row \(eR\) is \(\mathbb Z\oplus\mathbb Z/d\), which is not projective: its torsion summand is not projective by Exercise 2. Directly, the components of \(T(N)\) at the two diagonal idempotents are \(N\) and \(N/dN\). Thus the injection \(\mathbb Z\xrightarrow d\mathbb Z\) becomes zero on the second nonzero component. It does not remain a monomorphism, so \(T\) has no left adjoint. Right-column compact projectivity and left-row compact projectivity are genuinely different, even with one object on each side.

Exercise 4 (grade 4). Let \(\mathcal C=\mathcal D\) have objects \(\mathbb N\), endomorphism rings \(k\), and zero Hom groups between distinct objects. Put \(B(d,c)=k\) for \(d=c\) or \(d=c+1\), and zero otherwise, with the usual scalar actions. Compute \(T_B\), both tensor adjoints and their Hom bijections. Show why the total bimodule need not be finitely generated.

Solution. Modules are families of vector spaces, and module maps are componentwise linear maps. Each column is supported at two objects and is a sum of two right representables. The row at 0 is one left representable; every later row is a sum of two. Thus both finite conditions hold although the object set is infinite. The balanced tensor is componentwise

T_B(M)_0=M_0,\qquad T_B(M)_d=M_d\oplus M_{d-1}\quad(d\ge1). \tag{7.1}

Each component uses only the displayed scalar tensors, and the zero cross-Hom groups impose no further relations. The right adjoint has \(Q(N)_c=N_c\oplus N_{c+1}\); finite product equals finite sum. The left tensor adjoint has the same object formula \(S(N)_c=N_c\oplus N_{c+1}\), since dualizing each nonzero coefficient \(k\) gives \(k\) again.

Explicitly, a map \(T_B(M)\to N\) is a family of maps \(a_c:M_c\to N_c\), \(b_c:M_c\to N_{c+1}\); the inverse adjunction maps pair these into \((a_c,b_c):M_c\to Q(N)_c\), or unpack the two components. A map \(S(N)\to M\) is a family \(a_c:N_c\to M_c\), \(b_c:N_{c+1}\to M_c\). At \(d=0\), the adjoint map \(N\to T_B(M)\) is \(a_0\); at \(d\ge1\) it is \((a_d,b_{d-1}):N_d\to M_d\oplus M_{d-1}\). Pairing and unpacking are inverse and natural, proving both adjunctions on all maps, including the boundary at 0.

The whole coefficient system has nonzero entries at infinitely many pairs \((d,c)\). A finite collection of its component elements is supported at finitely many pairs, and the only nonzero category arrows are scalars at those same pairs. Their generated sub-bimodule remains supported there and cannot equal \(B\). Thus \(B\) is not finitely generated as a whole bimodule. The theorem requires finite projectivity separately in every row or column, not a single finite global generating set.