Internal Hom in Ind categories and flat module obstructions

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

A tensor product on a filtered completion need not have an internal Hom. For an original tensor factor the original adjunction extends directly. For an arbitrary filtered union of factors, the right adjoint requires an inverse limit. We construct it when the original small category has finite colimits, keep the order of the tensor factors explicit, and characterize when the resulting internal Hom preserves filtered colimits. A square-zero ring gives a concrete obstruction when the needed limits are absent.

Fix universes and choice. All original categories and indexing categories are small. Filtered categories are nonempty and every finite diagram in them admits a cocone. Write \(D=\operatorname{Ind}(C)\), with constant embedding \(y\), and write \(\widehat\otimes\) for its tensor. Compact means preservation of small filtered colimits by the ordinary set-valued functor \(\operatorname{Hom}(X,-)\).

Use Ind tensor products and compact duals, Sections 1–5 for the tensor on all Ind morphisms, separate filtered continuity, all monoidal coherence maps, compact constants and the compact-dual theorem. Ind-objects through their elements, Sections 1–3 supplies the full Hom formula and filtered extensions. Limits and colimits of formal objects, Proposition 1.1 supplies the recognition of Ind presheaves by finite-colimit tests. Retracts and stabilization of formal objects, Section 3 proves the compact-retract criterion. All of these are complete course proofs.

The module arguments use Hom, tensor and the exactness of module limits for the balanced tensor universal property and filtered module colimits, and Flat modules as an Ind completion of finite free modules for the full flat-module realization. Formal colimits and compact presentations, Sections 1–2 gives the realization criterion used in the exercises. Basic references are [Schapira] and [EGNO], listed at the end; the proofs below use the stated course interfaces.

1. Right adjoints for original tensor factors

Let \(C\) be small and monoidal. Assume that it is right closed in the following specified sense: for every \(c\), the functor \(a\mapsto a\otimes c\) has a right adjoint \([c,-]_C\). Thus

naturally in all three entries. The contravariant action in \(c\) is the unique map obtained by transposing precomposition with \(a\otimes f\). The adjunction therefore supplies the indicated bifunctor, not just objects representing separate Hom sets. No braiding or additive structure is assumed.

Lemma 1.1. Tensoring by \(yc\) on the right in \(D\) has right adjoint

This right adjoint preserves filtered colimits, and for \(Y=\operatorname{colim}_j yd_j\) its value is \(\operatorname{colim}_j y[c,d_j]_C\).

Proof. The unary extension theorem constructs \(G_c\), with the displayed value, on all objects and maps, and proves its filtered continuity. For \(A=\operatorname{colim}_a ya_a\), the tensor and full Hom formulas give

Every comparison uses the actual structure maps of the presentations and the original adjunction. It is therefore natural on original stage maps. A map from a constant into a filtered presentation factors through a stage; equality of two factors is witnessed at a later stage. Testing each source structure map gives naturality for every Ind morphism. Equivalently this is the all-transformation extension property proved in the tensor prerequisite. Consequently (1.3) is an adjunction on the actual functors. Its construction is independent of the presentations by the full Hom formula.

For \(f:c\to c'\), transposition in (1.1) gives \([c',-]_C\to[c,-]_C\), hence \(G_{c'}\to G_c\). Under (1.3) this is precisely precomposition with \(1_A\widehat\otimes yf\). This also proves its naturality in \(Y\) and the contravariant identity and composition laws. \(\square\)

This lemma does not require finite colimits in \(C\). It constructs right adjoints for constant tensor factors; it does not yet construct one for every \(X\in D\).

2. Completeness and the internal-Hom formula

From now through Section 4 assume also that \(C\) has finite colimits.

Lemma 2.1. The category \(D\) has all small limits, computed in its presheaf category.

Proof. Its presheaves are on the small category \(C\), so the category of elements of each is small. The recognition theorem cited above says that a presheaf belongs to \(D\) exactly when it sends finite colimits in \(C\) to limits of sets. A pointwise limit of such presheaves still does so: both orders of taking limits describe the same families of compatible entries. This includes the empty limit. Hence the presheaf limit belongs to \(D\). Since \(D\) is a full subcategory, the presheaf limiting cone has the same universal property against its objects, and is its limit there. \(\square\)

Theorem 2.2. For every \(X\in D\), the functor \(A\mapsto A\widehat\otimes X\) has a right adjoint \([X,-]_D\). For independent presentations \(X=\operatorname{colim}_{i\in I}yc_i\) and \(Y=\operatorname{colim}_{j\in J}yd_j\),

The formula is independent of the presentations and defines the internal-Hom bifunctor on arbitrary Ind morphisms.

Proof: representation. For a chosen presentation of \(X\), let

For an arrow \(i\to i'\), Lemma 1.1 gives the transition \(G_{c_{i'}}(Y)\to G_{c_i}(Y)\). These are the correct inverse-limit arrows. The limit exists by Lemma 2.1. For every \(A\), separate tensor continuity, the ordinary colimit universal property, and (1.3) give

These are natural in \(A,Y\), with the indicated transition maps, so they are the required adjunction. Uniqueness of representing objects gives a unique isomorphism between different presentation formulas respecting (2.3). Applying Lemma 1.1 to \(Y\) gives (2.1).

Proof: evaluation and all maps. Each projection \(H_X(Y)\to G_{c_i}(Y)\) transposes to a map \(H_X(Y)\widehat\otimes yc_i\to Y\). The inverse-limit compatibility says that these maps form a cocone over \(I\). Tensor continuity therefore gives one map

Under (2.3), the transpose of \(h:A\to[X,Y]_D\) is exactly \(\epsilon_{X,Y}(h\widehat\otimes1_X)\). This description characterizes evaluation without a presentation.

Given \(f:X'\to X\) and \(g:Y\to Y'\), define

by transposing \(g\epsilon_{X,Y}(1\widehat\otimes f)\). Evaluating both sides proves the identity and composition laws, since ordinary composition and tensor satisfy them and transposition is a bijection. It also proves naturality of evaluation and agreement with every map in the original presentation formulas. Thus no choice of a reindexing functor for an arbitrary Ind morphism is required.

For fixed \(X\), the unit at \(A\) is the transpose of \(1_{A\widehat\otimes X}\):

By its definition,

For the other triangle, transpose \([X,\epsilon_{X,Y}]\eta_{[X,Y]}\). Naturality of evaluation, followed by (2.7) at \(A=[X,Y]\), makes its transpose \(\epsilon_{X,Y}\). This is also the transpose of \(1_{[X,Y]}\), so the two maps are equal. Both adjunction triangles are proved for the actual maps (2.4) and (2.6). \(\square\)

Each \([X,-]_D\) preserves limits. Indeed, for a diagram \(Y_s\), maps from any \(A\) into its two proposed limiting values identify by (2.3) with the same compatible family of maps \(A\widehat\otimes X\to Y_s\). Yoneda identifies their canonical comparison with an isomorphism. Filtered colimits behave differently.

3. Which internal Homs preserve filtered colimits?

Theorem 3.1. Under the hypotheses of Section 2, the following conditions on \(X\in D\) are equivalent:

1. \(X\) is compact.
2. \(X\) is a retract of a constant \(yc\).
3. The functor \([X,-]_D\) preserves filtered colimits.

Proof. The compact-retract theorem in the prerequisites gives the first equivalence. For a constant \(yc\), Lemma 1.1 identifies its internal-Hom functor with \(G_c\), which preserves filtered colimits.

If \(i:X\to yc\), \(p:yc\to X\) split a retract, contravariance gives

with composite identity. The filtered-colimit comparison for the first functor is therefore a retract of that for the middle one, by naturality. A retract of an isomorphism is an isomorphism: if \(f\) is a retract of an invertible \(g\), with source inclusion/retraction \(u,v\) and target inclusion/retraction \(s,t\), its inverse is \(vg^{-1}s\), as the two commuting squares and \(vu=1,\ ts=1\) show. Hence the third condition follows.

Conversely the tensor unit \(y\mathbf1\) is compact. The adjunction gives the natural comparison

If \([X,-]_D\) preserves filtered colimits, compactness of \(y\mathbf1\) and (3.2) show that \(D(X,-)\) does too, with its canonical structure maps. Thus \(X\) is compact. \(\square\)

Right closedness does not say that every object has a dual. For an object with a left dual \(X^\vee\), with evaluation \(X^\vee\widehat\otimes X\to\mathbf1\), duality identifies

The tensor-duality adjunction proves this on all \(A\)-valued Hom sets; its evaluation is \(1_Y\widehat\otimes\operatorname{ev}\). If the original category is rigid in both directions, the compact-dual theorem from the preceding lesson adds dualizability to the three equivalent conditions. Without that hypothesis, the finite-set example below has compact objects without duals.

4. Currying, composition and endomorphism objects

The order in an iterated internal Hom is forced by the tensor adjunction. There is a natural isomorphism

To prove it, test against any \(A\). Both sides represent, in order,

Apply the two adjunctions and the given associator. These natural bijections give (4.1), its inverse, and naturality in every variable by Yoneda. No interchange of \(X\) and \(Y\) occurs.

Define the composition morphism

by transposing the ordered composite

The identity \(y\mathbf1\to[X,X]_D\) is the transpose of the left unitor at \(X\). Associativity follows by tensoring a triple of internal-Hom factors with the last input and transposing. Both resulting maps are the same ordered sequence of three evaluations; the associators agree by monoidal coherence. Since transposition is injective, the two composition morphisms agree. The unit laws follow in the same way by evaluating and using the unitor identities and the definition of the transposed identity.

Consequently \([X,X]_D\) is a monoid object. The maps (4.3) also give the left action of \([Y,Y]_D\) and the right action of \([X,X]_D\) on \([X,Y]_D\). Their associativity, units and commuting-action equation are precisely the composition equations just proved. These are statements in a monoidal category, with no additive or symmetry assumption.

5. A flat-module obstruction when the limits are absent

Let \(k\) be a field, let \(V\) have basis \(v_0,v_1,\ldots\), and form the commutative square-zero algebra

Thus \(V^2=0\), and \(\operatorname{ann}_R(v_0)=V\): multiplication by \(v_0\) sends \((\lambda,v)\) to \(\lambda v_0\).

Proposition 5.1. The module \(M=\prod_{n\ge1}R\) is not flat.

Proof. The principal ideal \((v_0)\) is isomorphic to \(R/V\), by \(\bar r\mapsto rv_0\). For any module \(N\), the tensor universal property gives

The maps are \(\bar1\otimes n\mapsto\bar n\) and \(\bar n\mapsto\bar1\otimes n\); the relation by \(V\) is exactly what makes both well defined. Under this isomorphism, tensoring the inclusion \((v_0)\subset R\) with \(N\) is the map \(\bar n\mapsto v_0n\). If \(N\) is flat, this map is injective. Therefore flatness requires

Take \(m=(v_1,v_2,\ldots)\in M\). It is killed by \(v_0\). But it is not in \(VM\). A member of \(VM\) is a finite sum \(\sum_{\ell=1}^s w_\ell m_\ell\), with \(w_\ell\in V\) and \(m_\ell\in M\). Since \(V^2=0\), every coordinate of that sum lies in the same finite-dimensional \(k\)-span of \(w_1,\ldots,w_s\). The coordinates \(v_n\) of \(m\) cannot all lie in such a span, by linear independence. Thus \(m\notin VM\), contradicting (5.3). \(\square\)

Let \(C_R\) be a small skeleton of finite rank free \(R\)-modules, with the usual tensor. It is rigid and right closed: the coordinate duals and the matrices of maps identify \([R^a,R^b]\) with \(R^{ab}\), with the ordinary tensor–Hom adjunction on matrices. The flat-module prerequisite gives its full monoidal Ind realization as flat \(R\)-modules.

Corollary 5.2. This monoidal Ind category is not right closed. In particular smallness, original rigidity and original right closedness do not alone give the conclusion of Theorem 2.2.

Proof. The countably generated free module \(X=R^{(\mathbb N)}\) is flat. Suppose an internal Hom \(H=[X,R]\) existed in the category of flat modules. Its adjunction, tested at the free rank-one module, would give a natural bijection

This is an \(R\)-linear bijection. Naturality at scalar endomorphisms of \(R\) gives scalar compatibility. Naturality at the two inclusions into \(R^2\) identifies pairs of elements, and at the diagonal \(R\to R^2\) gives addition; the zero map gives zero. On the tensor side these same maps act on \(X\oplus X\) and its diagonal. Thus the ordinary set-valued adjunction really yields the claimed module isomorphism. Proposition 5.1 says its target is not flat, whereas \(H\) would have to be flat. This contradiction proves the corollary. \(\square\)

The finite-colimit hypothesis in Section 2 is a sufficient way of ensuring the required inverse limits. No necessity assertion about finite colimits in every original presentation is made.

6. Four graded exercises with complete solutions

Exercise 1 (introductory) — sets, powers and finite inputs

Apply the theorem to finite sets with cartesian product. Identify \([X,Y]_D\) for arbitrary sets. Determine exactly when it preserves filtered colimits in \(Y\), and compare that condition with dualizability.

Solution. Finite sets have finite colimits and are right closed, with \([S,T]_C=T^S\). The preceding Ind tensor lesson identifies their monoidal Ind category with all sets and cartesian product. Currying a function \(A\times X\to Y\) gives its unique function \(A\to Y^X\), with inverse evaluation. These formulas are natural in all three sets, so uniqueness of right adjoints identifies \([X,Y]_D\) with the ordinary function set \(Y^X\), including its evaluation.

If \(X\) is finite, each function into a filtered colimit is represented at a common stage by choosing finitely many values. Equality of two functions is witnessed at a common later stage by imposing finitely many equality witnesses. Thus \((-)^X\) preserves filtered colimits. If \(X\) is infinite, use the filtered system of its finite subsets. Their colimit is \(X\), but the identity function \(X\to X\) factors through no finite member. Hence the power comparison is not onto. This also agrees with Theorem 3.1 because the compact sets are precisely the finite sets. The tensor prerequisite proves that the only dualizable sets are singletons. Therefore a two-element set gives a filtered-continuous internal-Hom functor without being dualizable; finite sets are right closed but not rigid.

Exercise 2 (intermediate) — all linear maps and finite-rank maps

Take finite-dimensional vector spaces over a field. Identify the internal Hom on its Ind category. For arbitrary vector spaces \(X,Y\), compute the image of the canonical map \(Y\otimes X^*\to[X,Y]\), and show why it need not be onto.

Solution. Every vector space is the directed union of its finite-dimensional subspaces. Finite-dimensional source maps factor through a common stage by a finite basis, and equality is witnessed by finitely many basis equations. The full compact realization criterion therefore identifies the Ind category with all vector spaces. Ordinary bilinear currying gives

The natural map in (6.1) has exactly the finite-rank maps as image. Each finite tensor has image in the span of its finitely many \(Y\)-coefficients. Conversely, for a finite-rank map choose a basis \(y_1,\ldots,y_s\) of its image; its coordinate functions \(\lambda_\ell:X\to k\) give the tensor \(\sum_\ell y_\ell\otimes\lambda_\ell\). The map is injective too: rewrite a tensor using linearly independent \(Y\)-coefficients. If its associated function is zero, every corresponding coordinate functional is zero.

For \(X=Y=k^{(\mathbb N)}\), the identity has infinite rank, so it is not in that image. It also does not come from any \(\operatorname{Hom}_k(X,Y_j)\) in the directed system of finite-dimensional subspaces \(Y_j\subseteq Y\). Thus \([X,-]\) fails filtered continuity for this \(X\). For finite-dimensional \(X\), the coordinate basis and its dual make the map (6.1) an isomorphism for every \(Y\). Compactness, dualizability and filtered continuity agree here because the original finite-dimensional category is rigid.

Exercise 3 (advanced) — the missing finite colimit

For the square-zero ring of Section 5, show directly that \(v_0,0:R\rightrightarrows R\) have no coequalizer in \(C_R\), even as an internal coequalizer there. Explain why merely observing that the ambient quotient is not free would not be a proof.

Solution. If a coequalizer \(Q\) existed in \(C_R\), it would be a finite rank free module \(R^r\). Its universal property with target \(R\) would identify

This identification is \(R\)-linear: it sends a map to its composite with the specified quotient arrow, and therefore preserves addition and scalar multiplication. The left side is \(R^r\). For \(r=0\) it is zero, whereas \(V\ne0\). For \(r>0\), multiplication by \(v_0\) on \(R^r\) is nonzero, for example on its first basis vector. Multiplication by \(v_0\) on \(V\) is zero because \(V^2=0\). No module isomorphism (6.2) is possible. Thus there is no such internal coequalizer.

The ambient cokernel \(R/(v_0)\) not belonging to \(C_R\) would only show failure of closure under that ambient operation. A full subcategory can have a different internal universal object. The Hom test above rules out every possible internal finite free object, so it proves the required absence of a finite colimit. This supplies a concrete failure of the hypothesis that ensured completeness in Lemma 2.1.

Exercise 4 (advanced) — the order of bimodule evaluation

Let \(R\) be a finite-dimensional algebra over a field \(k\), and let \(C\) be a small monoidal model of finite-dimensional \(R\)-bimodules with equal left and right scalar action of \(k\). Identify its Ind category and its internal Hom. Give evaluation and composition explicitly. For \(R=M_2(k)\), compare the internal endomorphism object of the tensor unit with its ordinary bimodule endomorphisms.

Solution. These bimodules are left modules over the finite-dimensional algebra \(R^e=R\otimes_kR^{\mathrm{op}}\). Finite direct sums and cokernels stay finite-dimensional, and the ordinary finite-colimit construction by sums and a coequalizer proves that \(C\) has finite colimits. Its tensor product over \(R\) stays finite-dimensional.

Every \(R^e\)-module is the filtered union of its finitely generated submodules, which are finite-dimensional because \(R^e\) is. A finite-dimensional module has a finite generating surjection from \((R^e)^n\). Its kernel is finite-dimensional and hence finitely generated, so this is a finite presentation. Generator and relation witnesses at a common filtered stage prove compactness; equality of maps is witnessed on finitely many generators. The compact realization criterion identifies \(\operatorname{Ind}(C)\) with all these \(R\)-bimodules, on every map.

For bimodules \(X,Y\), put \(H=\operatorname{Hom}_{-R}(X,Y)\), the right \(R\)-linear maps, with actions

The two actions commute, satisfy their unit and associativity laws, and preserve right linearity. In the finite-dimensional case \(H\) is a subspace of \(\operatorname{Hom}_k(X,Y)\), so it belongs to \(C\). The inverse currying formulas are

Balancing says that the image of \(ab\) is the image of \(a\) with its right action by \(b\) from (6.3). The outer left action is respected because \(\phi\) is left linear; the image of \(a\) is right linear because \(\phi\) is right linear in \(x\). Conversely these three properties make \(a\otimes x\mapsto f(a)(x)\) a well-defined bimodule map. The formulas are inverse and natural. They prove right closedness of \(C\), and identify its Ind internal Hom with this same \(H\) on arbitrary bimodules.

Evaluation is \(f\otimes x\mapsto f(x)\). Composition is

sending \(g\otimes f\) to \(g\circ f\). It is balanced because \((g b)(f(x))=g(bf(x))=g((bf)(x))\). This is the order in (4.4), with no swap of factors.

For \(X=Y=R\), every right linear map is left multiplication \(L_a\), so the internal endomorphism object is the regular bimodule \(R\), with multiplication \(L_aL_b=L_{ab}\). Its identity morphism from the tensor unit sends \(b\) to \(L_b\). Ordinary bimodule maps \(R\to R\), however, are determined by a central element: \(h(r)=rh(1)=h(1)r\). Their ring is \(Z(R)\). For \(R=M_2(k)\), commuting with \(E_{11}\) forces a central matrix to be diagonal, and commuting with \(E_{12}\) forces its two diagonal entries to agree. Thus the ordinary endomorphism ring is \(k\), while the internal object is \(M_2(k)\). Its ordered multiplication distinguishes \(E_{12}E_{21}=E_{11}\) from \(E_{21}E_{12}=E_{22}\). Internal Hom is not the set of global morphisms.

7. References

• [Schapira] Pierre Schapira, An Introduction to Categories and Homological Algebra, Section 2.7, for the general Ind construction.
• [EGNO] Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik, Tensor Categories, author-hosted final version, Chapter 2, for tensor adjunctions and duality.