Ind tensor products and compact duals

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

A tensor product of two formal filtered unions uses an independent index for each input. Extending the product is only the first step: its associator, unit maps, braiding and functor comparisons must extend on every morphism as well. The same construction explains a finiteness constraint on duality. In an Ind category a one-sided dual forces compactness; the converse needs the corresponding duals on the original objects.

Fix universes and choice. The original categories \(C,D\) are small, and all indexing categories are small and filtered. A filtered category is nonempty and every finite diagram in it admits a cocone. Write \(y:C\to\operatorname{Ind}(C)\) for the constant embedding. Here compact means that the ordinary set-valued functor \(\operatorname{Hom}(X,-)\) preserves small filtered colimits. This is also called finite presentation.

Ind-objects through their elements, Sections 1–3 proves the full Hom formula, pointwise filtered colimits, compactness of each \(yc\), and extension of a functor into any category with filtered colimits. Filtered stages and finite limits, Sections 1 and 4 proves the finite equality witnesses and the filtered Set comparison. Change an index without changing universal maps, Section 1 proves the cofinality criterion used below. Retracts and stabilization of formal objects, Sections 1–3 proves that Ind categories split idempotents and that their compact objects are precisely retracts of constants. We reuse those proofs.

For tensor conventions use Strict monoidal models, Sections 1–3 and Tensor duality, Sections 2–3. Module examples use Hom, tensor and the exactness of module limits for balanced tensor and filtered module colimits. There is no implicit interchange of tensor factors. The arguments require neither an additive structure nor finite colimits in \(C\).

1. Natural transformations between filtered extensions

Let \(E\) be locally small with small filtered colimits. Write \(\operatorname{Fun}_{\mathrm f}(\operatorname{Ind}(C),E)\) for functors preserving those colimits, with all natural transformations.

Lemma 1.1. Restriction along \(y\) is an equivalence

Proof. Essential surjectivity is exactly the complete extension theorem in the first prerequisite. We prove the assertion on all transformations. Let \(L,M\) preserve filtered colimits and let \(\alpha_c:L(yc)\to M(yc)\) be natural in \(c\). For a presentation \(X=\operatorname{colim}_i yc_i\), define \(\alpha_X\) by the cocone

where \(q_i:yc_i\to X\) are the given structure maps. Its compatibility follows from the naturality of \(\alpha_c\). The colimit comparison for \(L\) supplies one map \(L(X)\to M(X)\).

Every map \(h:yc\to X\) factors through some \(q_i\), by compactness of \(yc\). Naturality on \(C\) and (1.2) consequently give

for every such \(h\), not just the chosen structure maps. Conversely these equations determine \(\alpha_X\), since they include all the \(q_i\). This characterization proves independence of the presentation.

For \(f:X\to Y\), test naturality on every \(L(q_i)\). Apply (1.3) first to \(q_i\) and then to \(fq_i\). It gives

The colimit property proves the desired naturality. Any natural transformation extending \(\alpha_c\) must satisfy (1.3), so it is this one. Restriction is therefore full and faithful. In particular the transformation sets here are small: they are the natural-transformation sets on the small category \(C\). Restriction preserves identities and composition; its full faithfulness implies that an extended transformation is invertible exactly when its restriction is. \(\square\)

2. Product categories and independent indices

Lemma 2.1. Projection induces an equivalence

It carries a filtered diagram of pairs to the pair of its projected filtered colimits, and preserves filtered colimits.

Proof. Extend the two projections by the unary theorem. Consider two presentations by pairs, indexed by \(I,J\). The full Hom formula identifies their Hom set with

The first interchange is the finite-product/filtered-colimit theorem for sets; the second is the product universal property for limits. Each comparison is induced by the two projections. The last expression is precisely the Hom set between the two projected pairs of Ind objects. Thus the functor is fully faithful, including its action on all maps.

Given \(X=\operatorname{colim}_{i\in I}yc_i\) and \(Y=\operatorname{colim}_{j\in J}yd_j\), use the diagram \((c_i,d_j)\) on \(I\times J\). This product category is small and filtered: choose a common receiver, then an equalizer, in each of its two coordinates. The colimit of the first projected diagram is \(X\), and that of the second is \(Y\). Indeed colimits of a constant diagram over a nonempty connected category recover its value, so iterating the two colimit universal properties gives these identifications. This proves essential surjectivity. The construction by the unary theorem preserves filtered colimits in both projections, hence in the product category. An inverse equivalence also preserves them. If one original category is empty there are no objects on either side; the assertion still holds. \(\square\)

Lemma 2.2. For a functor \(H:A\times B\to E\), where all three categories have filtered colimits, preservation separately in the two variables is equivalent to preservation of filtered diagrams in the product category.

Proof. A constant diagram over a filtered category has its value as colimit. Thus joint preservation applied to diagrams constant in one coordinate gives separate preservation.

Conversely suppose preservation is separate, and take a filtered diagram \((X_i,Y_i)\). The diagonal \(I\to I\times I\) is cofinal. To check this directly, the comma category for \((i,j)\) consists of common receivers \(k\), with specified arrows from \(i,j\). It is nonempty. To connect two choices, receive their two targets and then equalize the two resulting arrows out of \(i\), and the two out of \(j\), in succession. The resulting common receiver receives both choices compatibly. The comma category is therefore connected. The stated cofinality criterion applies.

Separate preservation and iterated colimit universal properties now give

These are the canonical comparisons: both send a diagonal structure map to the indicated pair of structure maps. This proves joint preservation. \(\square\)

Iterate the two lemmas for any positive finite number of factors. Lemma 1.1 then gives

where \(\mathrm{sf}\) means separate filtered-colimit preservation. This is an equivalence on functors and all transformations. Formula (2.2) supplies the product equivalence; Lemma 2.2 supplies the equivalence of continuity conditions. For zero factors both product categories are terminal and restriction is identity. This covers the nullary unit data too.

3. The monoidal structure on the Ind category

Let \((C,\otimes,\mathbf1)\) be a small monoidal category, with its specified associator and unitors.

Theorem 3.1. There is a monoidal structure \((\operatorname{Ind}(C),\widehat\otimes,y\mathbf1)\), separately preserving filtered colimits, for which \(y\) is strong monoidal. For independent presentations it is given by

The structure is unique with these continuity and specified restriction properties, up to the unique comparison respecting them. A braiding or symmetry on \(C\) extends with the same property.

Proof. Apply (2.4) with target \(\operatorname{Ind}(C)\) to the functor \((c,d)\mapsto y(c\otimes d)\). This constructs the product on all objects and morphisms and proves (3.1), its independence of presentations, and separate continuity. It also supplies the specified tensor comparison for \(y\).

Both parenthesized three-variable product functors preserve filtered colimits in each variable: apply separate continuity to each occurrence of the product. Full faithfulness in (2.4) extends the associator of \(C\) to one transformation between them. Its inverse extends as well, and the two composites are identities because that is true on the original triples. The same unary argument extends both unitors, with unit \(y\mathbf1\).

The two routes around the pentagon are transformations between the same separately continuous four-variable functors. They agree on original quadruples by the pentagon in \(C\), so (2.4) makes them equal everywhere. The triangle follows by its two-variable instance. The normalization on original objects is exactly the strong-monoidal coherence of \(y\). Thus these are all the required monoidal axioms.

If a second product has the stated properties, (2.4) supplies its unique normalized invertible comparison with this one. The associator and unitor equations for that comparison hold by restriction, hence everywhere. This proves uniqueness as a monoidal comparison, not merely objectwise existence of isomorphisms.

A braiding extends using the two-variable full faithfulness. Its inverse extends too. Each hexagon is an equality of transformations of three-variable functors, so follows by restriction. When the original braiding is symmetric, its double interchange is identity on original pairs and therefore on all pairs. No braiding is asserted when none was specified. \(\square\)

For a diagram with a common filtered index, the diagonal argument in Lemma 2.2 can replace \(I\times I\) by \(I\) in (3.1). Independent indices remain the construction for two unrelated presentations.

4. Strong monoidal functors and all their transformations

Let \(E\) be locally small and monoidal, with filtered colimits preserved separately by its tensor product.

Theorem 4.1. Restriction along \(y\) gives an equivalence between the category of strong monoidal filtered-colimit-preserving functors \(\operatorname{Ind}(C)\to E\), with monoidal natural transformations, and the category of strong monoidal functors \(C\to E\), with monoidal natural transformations.

Proof. Start with a strong monoidal \(F:C\to E\). Use comparisons in the direction

Lemma 1.1 gives its filtered extension \(L\). Both functors \(L(X)\otimes L(Y)\) and \(L(X\widehat\otimes Y)\) preserve filtered colimits separately. Formula (2.4) extends the comparisons \(J_{c,d}\), and their inverses, to a natural isomorphism \(J_{X,Y}\). The unit map is \(j_0\), followed by the specified identification \(F(\mathbf1)\simeq L(y\mathbf1)\).

For presentations as in (3.1), the tensor comparison has the actual formula

It takes each pair of original structure maps to its prescribed strong comparison. The associator and both unit equations for \(L\) are equalities of transformations between separately continuous functors. They hold on the original objects by the strong-monoidal equations for \(F\), so (2.4) proves them everywhere. This constructs a strong monoidal extension.

Given a monoidal transformation \(\eta:F\to G\), Lemma 1.1 gives its unique underlying natural extension. The equation comparing its tensor component with the two strong comparisons is an equality of two transformations of separately continuous two-variable functors. It holds on original pairs, hence everywhere. The unit equation is the original nullary equation. Thus the extension is monoidal. Conversely restriction of any monoidal transformation is monoidal, and Lemma 1.1 supplies uniqueness. This proves full faithfulness.

Any strong continuous functor on \(\operatorname{Ind}(C)\) is the underlying extension of its restriction. The unique normalized underlying isomorphism is monoidal by the same restriction test. Hence restriction is essentially surjective and the theorem follows. If the given categories and functor are braided, the braided equation also follows on all objects by the two-variable restriction test. \(\square\)

In particular, a strong monoidal functor between small monoidal categories induces a strong monoidal functor between their Ind categories. The theorem asserts filtered continuity. It imposes no finite-limit or exactness conclusion on that extension.

5. Duals, compactness and retracts

We specify the direction of a left dual \(D\) of \(X\) by maps

with the two triangle identities. A right dual has the reversed order. An object with both is called dualizable here. There is no symmetry hypothesis.

Proposition 5.1. In any locally small monoidal category with filtered colimits preserved separately by tensor and with compact unit, an object with a left dual is compact. The corresponding statement holds for a right dual.

Proof. The duality adjunction, with its actual maps from the tensor-duality prerequisite, is

It takes \(a\) to \((a\otimes1_D)b\). Its inverse sends \(z:\mathbf1\to Y\otimes D\) to the composite \(X\to Y\otimes D\otimes X\to Y\), using \(z\) and \(e\). The triangle identities prove the two inverse equations, and these formulas prove naturality in \(Y\).

For a filtered diagram \(Y_i\), tensoring with \(D\) preserves its colimit. Compactness of \(\mathbf1\) then identifies the right side of (5.2), at \(\operatorname{colim}_iY_i\), with \(\operatorname{colim}_i\operatorname{Hom}(\mathbf1,Y_i\otimes D)\). Naturality identifies this bijection with the canonical compactness comparison for \(X\). A right dual gives the same argument with \(D\otimes Y\) in place of \(Y\otimes D\). \(\square\)

The unit \(y\mathbf1\) is compact in \(\operatorname{Ind}(C)\) by the Ind prerequisite. Thus either direction of duality in that category forces compactness.

Theorem 5.2. Suppose every object of \(C\) has a left dual. Then an object of \(\operatorname{Ind}(C)\) has a left dual if and only if it is compact, if and only if it is a retract of \(yc\) for some \(c\). The right-dual statement holds under the right-dual hypothesis on \(C\). If \(C\) is rigid in both directions, the compact objects are exactly the dualizable objects.

Proof. Compact objects are precisely retracts of constants by the cited complete retract theorem. Proposition 5.1 proves the implication from a left dual to compactness. A strong monoidal embedding transports an original dual, its evaluation, coevaluation and both triangle identities, so each constant \(yc\) has its indicated left dual.

It remains to prove that the relevant dual survives a retract; we give the maps. Let \(X\) be a retract of \(V=yc\), with

Let \(D\) be a left dual of \(V\), with evaluation \(e\) and coevaluation \(b\). Duality transposes \(a\) to an idempotent \(a^\vee:D\to D\); reversing composition still sends \(a^2=a\) to \((a^\vee)^2=a^\vee\). The evaluation and coevaluation naturality equations are

They follow from the two-variable duality adjunction, or directly by inserting its evaluation–coevaluation formula for \(a^\vee\); its triangle identities cancel the adjacent evaluation and coevaluation. These are the ordinary dual morphism equations proved in the duality prerequisite.

Idempotents split in \(\operatorname{Ind}(C)\). Choose \(j:D'\to D\), \(q:D\to D'\) with \(jq=a^\vee\), \(qj=1_{D'}\), and define

The first triangle composite for \(X\) is

The first equality inserts \(jq=a^\vee\); the second uses (5.4); the last uses the triangle for \(V\), \(ai=i\), and \(pi=1_X\). In the other triangle the inserted projector is \(ip=a\). Move it across the evaluation by the first equation of (5.4). The composite becomes

Here the middle equality is the triangle for \(D\), and \(a^\vee j=jqj=j\). Thus (5.5) supplies a genuine left dual of \(X\). The reversed-order construction proves the right-dual assertion. Applying the two directions separately proves the rigid case. \(\square\)

When \(C\) is idempotent complete, each retract of \(yc\) is itself represented by an object of \(C\): fullness identifies the idempotent with one on \(c\), and its splitting is preserved by \(y\). Thus in the rigid idempotent-complete case the original objects are, up to isomorphism, exactly the dualizable Ind objects. Without rigidity a compact object need not have a dual.

6. Four exercises with complete solutions

Exercise 1 — finite sets and the rigidity boundary

Take a small skeleton of finite sets, with cartesian product. Identify its monoidal Ind category and classify its compact and dualizable objects. Explain the role of the diagonal index.

Solution. Every small set is the filtered union of its finite subsets. A map from a finite set into a filtered colimit of sets factors at one common stage, by choosing stages for finitely many images. Equality of two such maps holds at a common later stage, by imposing finitely many equality witnesses. Finite sets are therefore compact. The full realization criterion in Formal colimits and compact presentations, Sections 1–2 now identifies this Ind category with sets, on all maps, not just on objects.

Cartesian product of sets preserves filtered colimits in each variable: any pair consists of two entries, and any equality is witnessed at a later stage. Formula (3.1) therefore realizes as the ordinary cartesian product, with its actual pair maps. For two chains of initial segments of the natural numbers, a pair of integers lies in one product stage after taking the larger bound. This is exactly the diagonal cofinality comparison of Lemma 2.2, not an identification of two unrelated chosen indices.

If a set is compact, its identity factors through one member of its finite-subset union. That forces every element into that finite member. Hence the compact sets are exactly the finite sets.

For a cartesian dual pair \(X,D\), evaluation to the one-point unit is unique. Coevaluation chooses \((x_0,d_0)\in X\times D\). The triangle on \(X\) is consequently the constant map \(x\mapsto x_0\); for it to be identity, \(X\) must have exactly one element. The other triangle makes \(D\) a singleton too. Conversely the singleton unit is its own dual. Empty sets cannot occur because there is no coevaluation into the empty product. Thus only singleton sets are dualizable. In particular a two-element set is compact without a dual: the original finite-set category is not rigid, as required by Theorem 5.2.

Exercise 2 — the tensor and duals of flat modules

Let \(R\) be a commutative unital ring and let \(C\) be the category of finite rank free \(R\)-modules. Identify its monoidal Ind category, its compact objects, and its dualizable objects. Give a flat nondualizable example over the integers.

Solution. The complete theorem in Flat modules as an Ind completion of finite free modules, Sections 1–4 identifies \(\operatorname{Ind}(C)\) with flat \(R\)-modules, with the usual underlying filtered colimits and all module maps. A tensor product of two flat modules is flat: tensoring a module injection successively with the two flat modules remains injective, and associativity identifies that composite with tensoring by their product. The tensor functor preserves filtered colimits by its balanced universal property. These are the exact hypotheses of Theorem 3.1.

On finite free modules the ordinary product has \(R^m\otimes_R R^n\simeq R^{mn}\), with basis indexed by ordered pairs. Therefore (3.1) identifies the extended tensor with the usual tensor on flats. The comparison is the map from the original balanced pair of structure maps. Associators and unitors agree too: both are determined by the same ordered pure tensors.

Every finite free module has both duals, with evaluation given by the coordinate pairing and coevaluation by the sum of basis vectors tensored with their dual basis. Both triangles fix every basis vector, hence are identity. By Theorem 5.2, compactness and dualizability in the flat category are therefore equivalent to being a retract of a finite rank free module. These retracts are precisely finitely generated projective modules: a retract of a finite free module is finitely generated and has the lifting property, while a finite generating surjection onto a projective module splits. Its dual is the corresponding retract of the dual finite free module; (5.5) gives its evaluation and coevaluation.

Over \(\mathbb Z\), the directed system with term \(\mathbb Z\) at \(n\ge1\), map \(n\to n+1\) multiplication by \(n+1\), and representative \(a\mapsto a/n!\), has colimit \(\mathbb Q\). Every rational has such a representative, and equality is witnessed after taking a common factorial denominator. It is flat because it is this filtered colimit of free modules. It is not compact: every group map \(\mathbb Q\to\mathbb Z\) is zero, since the image of each rational is divisible by every positive integer in \(\mathbb Z\). Thus its identity cannot factor through any free stage of that presentation. Theorem 5.2 says it is not dualizable.

Exercise 3 — why the unit must be compact

Let \(k\) be a field and let \(E=\prod_{n\ge1}\operatorname{Vect}_k\), with componentwise tensor product. Show that tensor preserves filtered colimits separately, but its dualizable unit is not compact.

Solution. Filtered colimits and tensor are computed coordinatewise. Vector tensor preserves colimits by the bilinear universal property, so it preserves these filtered colimits in both variables. The unit is \(U=(k,k,\ldots)\). Its componentwise unit evaluation and coevaluation make it its own dual: both triangle equations are identity in every coordinate and hence in \(E\).

For each finite subset \(S\) of the positive integers, let \(U_S\) have value \(k\) on \(S\) and zero elsewhere, with the evident inclusions for \(S\subseteq T\). These objects form a small filtered diagram whose colimit is \(U\), since every coordinate is eventually \(k\) with identity transitions. If \(U\) were compact, \(1_U\) would factor through some \(U_S\). Choose \(n\notin S\). In that coordinate the composite factors through zero, whereas \(1_k\ne0\). This is impossible. The unit is therefore not compact. Separate tensor continuity alone cannot give Proposition 5.1; the compact-unit assumption, which holds in the monoidal Ind construction, does essential work.

Exercise 4 — a strong extension need not preserve equalizers

For a field \(k\), let \(F\) send a finite set \(S\) to the vector space \(k[S]\) freely spanned by \(S\). Equip finite sets with cartesian product and vector spaces with their usual tensor. Construct its strong monoidal filtered extension to all sets, and show that it does not preserve one explicit equalizer.

Solution. The comparison \(k[S]\otimes k[T]\to k[S\times T]\) sends \(e_s\otimes e_t\) to \(e_{(s,t)}\). These are bases, so the map is invertible, with the indicated inverse on each basis vector. The unit comparison sends \(1\in k\) to the basis vector of the singleton. Naturality follows on basis vectors, and the associator and both unit equations send each ordered tuple to its same basis vector. This is a strong monoidal functor.

Under Exercise 1, its filtered extension is \(S\mapsto k[S]\) on all small sets: the free vectors have finite support, so both generation and every equality are captured in a finite-subset stage. Equivalently maps from \(k[S]\) to a vector space are exactly functions from \(S\) to its underlying set; this universal property proves the colimit comparison on all diagrams for which the set colimit is taken. Formula (4.2) gives the same ordered-pair tensor comparison on arbitrary sets. Theorem 4.1 supplies its full monoidal naturality.

Take \(S=\{0,1\}\), with the identity and the swap \(\sigma\). Their equalizer in sets is empty, whose free vector space is zero. The equalizer of \(1_{k[S]}\) and \(k[\sigma]\) is the nonzero line spanned by \(e_0+e_1\): a vector \(a e_0+b e_1\) is fixed exactly when \(a=b\), over any field, including characteristic two. Thus the image of the set equalizer does not equal the vector-space equalizer. Strong monoidality and filtered continuity do not imply finite-limit preservation.

7. Further reading

• Pierre Schapira, An Introduction to Categories and Homological Algebra, Section 2.7, for formal Ind objects. The complete functor-extension proof used here is in the linked course prerequisite.
• Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik, Tensor Categories, author-hosted final version, Chapter 2, for monoidal functors and the two directions of duality.