Small orthogonality and accessible reflective localizations

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

An extension test can require a solution, or it can require exactly one solution. The second requirement constructs reflective subcategories. We give the small-set case of the classical Freyd–Kelly orthogonality theorem in a locally presentable category: pushouts supply extensions, coequalizers impose uniqueness, and a regular-length chain makes both conditions hold simultaneously. The resulting reflector is accessible and its local category is again locally presentable.

Work in fixed universes with small Hom sets, diagrams and coproducts. Use set-indexed choice and chosen colimit functors in an enlargement containing the category data. Let \(C\) be locally presentable and \(S=\{s\colon A_s\to B_s\}\) a small set of arrows. There is no assumption that its arrows are monic or epic. A local object is an object \(y\) for which each restriction map is bijective:

Here \(C(a,b)\) denotes the entire Hom set. Write \(D=S^\perp\) for the full subcategory of local objects, with inclusion \(i\colon D\to C\). We prove a reflection \(L\dashv i\). Its universal property for inverting \(S\) among colimit-preserving functors will be stated precisely; it need not be an ordinary localization at \(S\) for arbitrary functors.

Presentable probes and accessible adjoints, Sections 1–4 proves the set-colimit comparison, all small limits in \(C\), raising regular bounds and the individual ranks used below. Relative injectivity and small test families treats extension without uniqueness. Sieve local isomorphisms proves the separate sheaf-localization interface. The classical orthogonality setting is discussed in [AHS].

Choose one infinite regular \(\lambda\) for which \(C\) is locally \(\lambda\)-presentable and every \(A_s,B_s\) is \(\lambda\)-presentable. The linked rank and cardinal-raising theorem supplies this common bound because \(S\) is small. In particular maps out of each endpoint factor through a stage of any \(\lambda\)-filtered colimit, and equality of two represented maps is witnessed at a common later stage. The latter assertion follows from injectivity of the actual Hom comparison, not only its factorization half.

1. Add all extensions with one pushout

For \(x\in C\), let \(I_x\) consist of all pairs \((s,a)\) with \(s\in S\) and \(a\colon A_s\to x\). This is a small set. Form

The two maps out of \(U_x\) are the evaluations \(a\) into \(x\) and the sum of the maps \(s\) into \(V_x\). Let \(p_x\colon x\to P(x)\), and let \(j_{s,a}\colon B_s\to P(x)\) be the other component maps. The defining equations and universal property are

A map from \(P(x)\) is exactly a map from \(x\) and one map from each \(B_s\) satisfying these equations.

This is functorial. For \(h\colon x\to x'\), the function \((s,a)\mapsto(s,ha)\) gives maps of both coproducts using identity maps on their corresponding summands. The pushout gives the unique map \(P(h)\) satisfying

Identities and composites satisfy exactly these equations, so uniqueness proves both functor laws. It also proves naturality of \(p\).

2. Impose uniqueness with a coequalizer

Let \(Q_x\) consist of all triples \((s,u,v)\) with \(s\in S\), \(u,v\colon B_s\to P(x)\), and \(us=vs\). This too is a small set. Take the simultaneous coequalizer

where the two component maps at \((s,u,v)\) are \(u,v\). Thus \(q_xu=q_xv\) for every such pair, and maps out of \(K(x)\) are exactly the maps out of \(P(x)\) with all these equalities. Put

For \(h\colon x\to x'\), the triple \((s,u,v)\) is sent to \((s,P(h)u,P(h)v)\), since its two restrictions remain equal. These component maps, and the coequalizer universal property, give a unique \(K(h)\) with \(K(h)q_x=q_{x'}P(h)\). Its identity and composition equations follow after precomposition with the coequalizer, which is epic by universal uniqueness. Together with (1.4), this proves that \(K\) is a functor and \(k\colon1_C\to K\) is natural.

Lemma 2.1. For every local \(y\), restriction along \(k_x\) is a natural bijection

Proof. Given \(f\colon x\to y\), every \(fa\colon A_s\to y\) has a unique extension from \(B_s\). These extensions give the unique map \(g\colon P(x)\to y\) with \(gp_x=f\). For any triple in \(Q_x\), the maps \(gu,gv\) have equal restrictions along \(s\); injectivity in (1.1) makes them equal. Thus \(g\) descends uniquely across \(q_x\). Conversely restricting this factor returns \(f\). Uniqueness first in the pushout and then in the coequalizer proves the other inverse identity. Equations (1.4) and its coequalizer counterpart prove naturality in \(x\); postcomposition with maps between local objects proves naturality in \(y\). \(\square\)

3. A regular-length chain gives the reflector

For each \(x\), define \(x_\alpha\) for \(\alpha\le\lambda\), together with all transition maps, by

Use \(k_{x_\alpha}\) at a successor. At a limit the cocone determines the transitions, and their composition equations follow from its uniqueness. Naturality of \(k\) constructs maps between chains at successors; the colimit universal property constructs them at limits. Transfinite induction on these equations proves identity, composition and compatibility with every transition. Consequently

define a functor and a natural map. Regularity says that the ordinal index \(\{\alpha:\alpha<\lambda\}\) is \(\lambda\)-filtered: fewer than \(\lambda\) ordinals have a common bound below \(\lambda\). Thus every endpoint's presentability applies to this final colimit.

Theorem 3.1. Every \(T(x)\) is local.

Proof: existence. For \(a\colon A_s\to T(x)\), choose a factor \(a_\alpha\colon A_s\to x_\alpha\). Its pushout generator \(j_{s,a_\alpha}\colon B_s\to P(x_\alpha)\), followed by \(q_{x_\alpha}\) and the transition from \(x_{\alpha+1}\) to \(T(x)\), extends \(a\). Here \(\alpha+1<\lambda\), since the infinite cardinal \(\lambda\) is a limit ordinal.

Proof: uniqueness. Suppose \(u,v\colon B_s\to T(x)\) satisfy \(us=vs\). Presentability of \(B_s\) puts both maps at a common stage \(\alpha\). Injectivity of the Hom comparison for \(A_s\) puts equality of their restrictions at a later common stage \(\beta<\lambda\). Write the resulting maps \(B_s\to x_\beta\) as \(u_\beta,v_\beta\). Their composites with \(p_{x_\beta}\) are a pair in \(Q_{x_\beta}\), since their restrictions along \(s\) are equal. The next coequalizer makes their composites to \(x_{\beta+1}\) equal. They therefore give the same maps to \(T(x)\). This proves bijectivity in (1.1). \(\square\)

Theorem 3.2. Regard \(T(x)\) as an object \(L(x)\) of \(D\). Restriction along \(\eta_x\) is a natural bijection

Hence \(L\dashv i\), with the displayed actual unit.

Proof. Fix a local \(y\). A map \(x\to y\) extends uniquely along every successor step by Lemma 2.1. At a limit stage the already compatible family has exactly one colimit factor to \(y\). Transfinite induction gives a unique extension from \(x_\lambda\). Conversely a map from that colimit restricts to precisely this family, so both inverse identities hold. Naturality in \(x\) follows by induction from the chain maps; naturality in \(y\) follows by postcomposition and uniqueness. Since \(D\) is full, this is exactly (3.3). \(\square\)

Every other selection of the colimits gives the same reflector up to the unique natural isomorphism preserving the unit: extend one unit through the other by (3.3), in both directions. The composites are identities by the uniqueness clause. If \(S\) is empty, the pushout and coequalizer are identities, and the construction gives \(T\simeq1_C\).

4. The actual idempotent monad and its local objects

For a local \(y\), the identity extends to \(r\colon T(y)\to y\) with \(r\eta_y=1_y\). The two maps \(\eta_yr,1_{T(y)}\) agree after precomposition with \(\eta_y\); their target is local, so (3.3) makes them equal. Thus \(\eta_y\) is invertible.

Since \(T(x)\) is local, \(\eta_{T(x)}\) is invertible. Naturality at \(\eta_x\) and uniqueness of extensions into the local \(T^2(x)\) show

Define \(\mu_x\colon T^2(x)\to T(x)\) to be their common inverse. These maps are natural, by naturality of \(\eta\) and of its inverse on local objects. The two unit equations follow immediately. For associativity, naturality at \(\mu_x\) gives

The second equality uses the inverse defining \(\mu_{T(x)}\); the first uses \(T(\mu_x)\eta_{T^2(x)}=\eta_{T(x)}\mu_x\). Cancel the invertible \(\eta_{T^2(x)}\). This proves the monad associativity equation. The counit at \(y\in D\) is \(\eta_y^{-1}\), and its two adjunction triangles are precisely these unit inverse identities.

A \(T\)-algebra \(a\colon T(x)\to x\) satisfies \(a\eta_x=1_x\). This alone makes \(\eta_xa=1_{T(x)}\): compare these two maps into the local \(T(x)\) after \(\eta_x\), then apply (3.3). Thus \(x\) is isomorphic to a local object and is local itself, as its Hom tests are transported by that isomorphism. Its algebra map is forced to be \(\eta_x^{-1}\). Conversely, for a local \(x\), (4.1) shows that this inverse satisfies \(aT(a)=a\mu_x\), so it is an algebra. Every map between local objects respects these inverses by naturality of \(\eta\). Thus the entire algebra category is the full local category, including every morphism.

5. Limits, accessibility and local presentability

Proposition 5.1. The inclusion creates all small limits and all small \(\lambda\)-filtered colimits. The category \(D\) has all small colimits, with

The endofunctor \(T=iL\) preserves \(\lambda\)-filtered colimits.

Proof. A limit in \(C\) of local objects is local: applying either represented Hom functor \(C(B_s,-)\) or \(C(A_s,-)\) gives the corresponding limit of sets. The comparison in (1.1) is the limit of bijections and is bijective. This includes the empty limit. Fullness then supplies the same universal property in \(D\).

For a \(\lambda\)-filtered diagram of local objects, presentability of both endpoints makes (1.1) for its \(C\)-colimit the colimit of their bijective tests. It is again bijective, with the actual colimit comparison. The colimit object is therefore local, and fullness gives its colimit property in \(D\). This proves creation and preservation by inclusion.

For an arbitrary diagram in \(D\), adjunction gives maps from the right side of (5.1) to \(y\) as compatible families of maps from each \(d_j\) to \(y\). Hence (5.1), with its induced legs, is its colimit. The same Hom calculation says \(L\) preserves every small colimit. Composing this preservation with that of the inclusion for \(\lambda\)-filtered diagrams proves the assertion about \(T\). Its comparison is the one with the original colimit legs, since both functors use those very cocones. \(\square\)

Theorem 5.2. The category \(D\) is locally \(\lambda\)-presentable.

Proof. Let \(B\) be the small full presenting category of \(\lambda\)-presentable objects of \(C\), supplied by Presentable probes, Section 2. Each \(L(b)\) is \(\lambda\)-presentable in \(D\). Indeed for a \(\lambda\)-filtered diagram of local objects, adjunction and the created colimit identify its Hom comparison with the invertible comparison for \(C(b,-)\). Every local \(d\), as an object of \(C\), has a \(\lambda\)-filtered presentation \(d=\operatorname{colim}_j b_j\). Applying the colimit-preserving \(L\), and using \(L(i(d))\simeq d\), gives

The small family \(L(B)\) therefore consists of presentable objects and supplies every object by such a filtered colimit. Proposition 5.1 supplied all small colimits. These are exactly the defining conditions. \(\square\)

No exactness or finite-limit preservation of the reflector follows from these assertions. Exercise 3 gives a reflector here which fails to preserve a kernel.

6. The universal property for colimit-preserving functors

First \(L\) inverts each \(s\in S\). For every \(y\in D\), restriction along \(L(s)\) identifies by adjunction with the bijection (1.1). A map \(f\colon a\to b\) for which all maps \(D(b,y)\to D(a,y)\) are bijective is invertible: take \(y=a\) to obtain a left inverse \(r\), and use injectivity at \(y=b\) to deduce \(fr=1_b\) from their equal composites with \(f\). Applying this test proves the claim.

Let \(E\) be any locally small cocomplete category, and let \(F\colon C\to E\) preserve all small colimits and send every \(s\in S\) to an isomorphism.

Lemma 6.1. Every \(F(\eta_x)\) is invertible.

Proof. The image of \(p_x\) is a pushout of a coproduct of the isomorphisms \(F(s)\). The coproduct of their inverse maps is inverse to that coproduct map, and a pushout of an isomorphism has inverse supplied by its pushout universal property. Thus \(F(p_x)\) is invertible.

For \((s,u,v)\in Q_x\), equality \(us=vs\) and invertibility of \(F(s)\) give \(F(u)=F(v)\). Consequently the two maps from \(F(R_x)\) to \(F(P(x))\) agree on every coproduct component, and agree. Their coequalizer \(F(q_x)\) is invertible: the identity of \(F(P(x))\) is also their coequalizer, and the universal comparisons give inverse maps. Hence \(F(k_x)\) is invertible.

Induct along (3.1). Successor transitions become isomorphisms. At a nonzero limit stage, all the maps from \(F(x)\) to the earlier stages are compatible isomorphisms. Via these maps the diagram is constant, and its ordinal index is nonempty and connected. Cocones from a constant object over such an index are exactly single maps from that object, since every two components are equated along a zigzag. Thus its colimit is the same object. Colimit preservation by \(F\) proves invertibility of the map from \(F(x)\) at the limit as well. At \(\lambda\) this is \(F(\eta_x)\). Naturality is inherited from \(\eta\). \(\square\)

Theorem 6.2. Composition with \(L\) is an equivalence

where the right side consists of the colimit-preserving functors inverting \(S\), with all natural transformations.

Proof. If \(H\colon D\to E\) preserves colimits, \(HL\) does so and inverts \(S\). For \(F\) on the right, put \(\overline F=Fi\). This restriction preserves colimits in \(D\): writing \(c=\operatorname{colim}_C i(d_j)\), formula (5.1) and Lemma 6.1 give

These are the comparisons with the specified colimit legs, by naturality of \(F(\eta)\). Moreover \(F(\eta)\) is a natural isomorphism \(F\to\overline F L\). In the other direction the counit gives \(HLi\simeq H\). Restriction and composition act on every natural transformation by their actual components. The two just specified natural isomorphisms commute with those transformations by naturality at \(\eta\) and at the counit. They supply inverse equivalences, proving (6.1) on objects and all maps. \(\square\)

For arbitrary functors, the analogous factorization holds when all the units \(\eta_x\) are inverted: the same restriction \(Fi\) and natural isomorphism \(F(\eta)\) give it, with uniqueness controlled by the unit. Inverting \(S\) alone is weaker for arbitrary functors. The colimit hypothesis in Theorem 6.2 is what makes those particular functors invert every unit. Exercise 4 shows that it cannot simply be discarded.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: roots need the whole countable chain)

In abelian groups take \(S=\{n\colon\mathbb Z\to\mathbb Z\}\), where \(n\ge2\). Identify its local objects and reflector. Starting at \(\mathbb Z\), compute the steps \(K\) explicitly and explain why the first finite step is insufficient.

Solution. Evaluation at \(1\) identifies (1.1) with multiplication by \(n\) on the target group. The local objects are exactly the groups on which this multiplication is invertible. Its inverse then defines a unique action of \(\mathbb Z[1/n]\): multiplication by \(a/n^k\) is multiplication by \(a\) followed by the \(k\)-fold inverse. Equality of fraction representatives gives equality of the operators after multiplying by the common invertible power of \(n\). Thus the local category is the category of \(\mathbb Z[1/n]\)-modules, with all underlying group maps between them.

For any group \(A\), the direct limit of the sequence

has invertible multiplication by \(n\): its inverse sends a represented element \([a,k]\) to \([a,k+1]\), since \([na,k+1]=[a,k]\). Any homomorphism \(f\colon A\to y\) into a local group extends uniquely by \([a,k]\mapsto n^{-k}f(a)\). The transition relation proves this is well defined, and every represented element forces that value. This identifies the reflector \(A[1/n]\), equivalently \(\mathbb Z[1/n]\otimes_{\mathbb Z}A\) by the ordinary balanced tensor universal property. The unit's kernel consists precisely of the elements killed by some power of \(n\), by eventual equality with zero in this direct limit.

For \(A=\mathbb Z\), the first pushout has generators \(z,x_a\) for all integers \(a\), with relations \(nx_a=az\). Its coequalizer kills all \(n\)-torsion: pairs of maps out of \(B_s=\mathbb Z\) have equal restrictions exactly when their values differ by such a torsion element. In the quotient, \(x_a=ax_1\), since their difference is killed by \(n\), and \(z=nx_1\). Thus \(K(\mathbb Z)\) is generated by \(x_1\). The map to \((1/n)\mathbb Z\) sending \(z\) to \(1\) and \(x_a\) to \(a/n\) proves it is infinite cyclic and identifies the unit with the inclusion \(\mathbb Z\subset(1/n)\mathbb Z\).

The same computation applies at each later finite stage, giving \((1/n^k)\mathbb Z\subset(1/n^{k+1})\mathbb Z\). None is local: \(1/n^k\) has no \(n\)-th root in that stage. Their countable union is \(\mathbb Z[1/n]\), which is local. Here the finitely presentable endpoints allow \(\lambda=\omega\), and the countable chain supplies exactly the required reflection.

Exercise 2 (standard: existence without uniqueness gives the wrong category)

In abelian groups take the test \(0\to\mathbb Z\). Compute the local category and one full step of the construction. Explain what goes wrong if the coequalizer is omitted.

Solution. The restriction map in (1.1) is \(A\to\{*\}\). It is bijective exactly when \(A=0\), so the local category has only the zero group and the reflector is zero. In contrast it is surjective for every group: the zero homomorphism always extends the unique map from \(0\). Mere test-injectivity imposes no condition here.

There is one existence problem, since \(C(0,A)\) is a singleton. Its pushout is \(P(A)=A\oplus\mathbb Z\). Every pair of maps \(\mathbb Z\rightrightarrows P(A)\) has equal restrictions on \(0\). Comparing the map with value \(w\) at \(1\) to the zero map, for every \(w\in P(A)\), shows that the simultaneous coequalizer kills every element. Thus \(K(A)=0\), as required.

If only the pushout were iterated, the chain from \(0\) would instead give free abelian groups \(\mathbb Z^k\) and their direct sum at \(\omega\). That group satisfies the existence condition, as every group does, but is not local and is not the reflection. The uniqueness phase is substantive.

Exercise 3 (advanced: an accessible reflector need not preserve kernels)

In abelian groups use the quotient test \(s\colon\mathbb Z\to\mathbb Z/n\mathbb Z\), \(n\ge2\). Identify the reflector, check its universal map, and exhibit a kernel it fails to preserve.

Solution. The restriction map identifies with the inclusion \(A[n]\hookrightarrow A\). It is bijective exactly when \(nA=0\). For every such group \(y\), a homomorphism \(A\to y\) kills \(nA\), and factors uniquely through the quotient. Thus \(L(A)=A/nA\), with its ordinary quotient unit; maps of groups induce the quotient maps functorially. This also identifies the local category with all modules over \(\mathbb Z/n\mathbb Z\).

Multiplication by \(n\) on \(\mathbb Z\) has zero kernel. Applying \(L\) gives the zero endomorphism of \(\mathbb Z/n\mathbb Z\), whose kernel is the whole nonzero group. The canonical comparison from the reflection of the original kernel is therefore \(0\to\mathbb Z/n\mathbb Z\), not an isomorphism. Equivalently the monomorphism \(n\colon\mathbb Z\to\mathbb Z\) is sent to a nonmonomorphism. Both endpoints of the test are finitely presented groups, so the general construction is accessible at \(\omega\), yet it is not left exact. The inclusion of the local category still creates limits; that is a different assertion.

Exercise 4 (synthesis: the ordinary localization claim would be false)

For \(C=\operatorname{Set}\) and \(S=\{\varnothing\to1\}\), identify the reflection. Construct a functor from sets to abelian groups which inverts this one test but does not invert all reflection units. Explain the role of colimit preservation.

Solution. A local set \(y\) has \(y\to1\) bijective, so it is a singleton. The reflection sends every set to \(1\), with the unique unit map \(X\to1\). This is a left adjoint to the full category of singleton sets: a map from any set to a singleton is unique, including when the set is empty.

Define \(F(X)\) to be the group of finitely supported families of integers \((a_x)_{x\in X}\) with total sum zero. For a function \(u\colon X\to Y\), put

Only finitely many terms are nonzero. These families retain total sum zero. Partitioning a finite support into fibres proves \(F(vu)=F(v)F(u)\), and singleton fibres for the identity prove its identity law. Thus \(F\) is an ordinary functor.

Both \(F(\varnothing)\) and \(F(1)\) are zero, so it sends the test to an isomorphism. For a two-element set, \(F(2)\simeq\mathbb Z\) by \(a\mapsto(a,-a)\). The unit \(2\to1\) is therefore sent to the noninvertible map \(\mathbb Z\to0\). Consequently \(F\) cannot factor, even up to natural isomorphism, through this reflection: every such factor sends its units to isomorphisms.

It does not preserve coproducts: the coproduct of \(F(1)\) with itself is zero, while \(F(1\amalg1)\simeq\mathbb Z\). This is exactly the missing assumption in Theorem 6.2. Its universal property concerns colimit-preserving functors; arbitrary inversion of \(S\) alone does not give the claimed factorization.

References

[AHS] Jiří Adámek, Michel Hébert and Lurdes Sousa, The Orthogonal Subcategory Problem and the Small Object Argument, Coimbra preprint 07–40. Freely accessible departmental preprint. The introduction distinguishes orthogonality from injectivity; its pushout, coequalizer and transfinite-composition discussion gives broader context. The proofs here treat a small set of tests and include the actual reflector maps, accessibility and the stated colimit-preserving universal property.