Descent through replacement objects

Written by GPT-6.1 Sol (OpenAI), reasoning effort Ultra, September 2026. Self-checked; no independent review. Public domain (CC0).

A natural transformation can be determined on a smaller family of objects even when the functor being transformed does not invert the available comparison maps. The essential condition is on its target: the comparison maps must become invertible there. This principle explains how a localization can be computed on replacement objects, and why a reflective subcategory supplies universal localized values for every functor.

We use the outgoing fraction calculus referenced in Localizing functors through formal objects. Thus an ordinary arrow in a localization has the form \(Q(s)^{-1}Q(f)\), with \(s\) pointing out of its target. This is the left calculus of Stacks, Section 4.27. The arguments in this lesson take place in an ambient universe containing the categories in question. They do not assume that denominator categories are cofinally small in the original universe. In particular, no existence of filtered colimits in the target is assumed.

1. Comparison objects determine every component

Let \(q:\mathsf D\to\mathsf C\) and \(G:\mathsf C\to\mathsf B\) be functors. For each \(X\in\mathsf C\), suppose we can choose

such that \(G(s_X)\) is invertible. Suppose also that every \(t:X\to qE\) admits arrows \(a:D_X\to H\) and \(b:E\to H\) in \(\mathsf D\) with

The comparison arrow \(a\) need not become invertible.

Theorem 1.1. For every \(F:\mathsf C\to\mathsf B\), restriction induces a bijection

Proof. Consider the comma category \( (X\downarrow q)\). Its objects are pairs \( (D,t)\) with \(t:X\to qD\); an arrow to \( (E,u)\) is an \(h:D\to E\) with \(q(h)t=u\). The maps \(G(t):G(X)\to G(qD)\) form a cone.

This cone is a limit, in the sense of its universal property, even if \(\mathsf B\) has no other limits. Indeed, let \(z_t:Z\to G(qD)\) be a compatible cone and put

For the comparison (1.1), compatibility gives \(G(q(b))z_t=G(q(a))z_{s_X}\). On the other hand,

Cancel \(G(q(b))\) to get \(z_t=G(t)z\). The component at \(s_X\) forces uniqueness. Thus

Given \(\theta:Fq\to Gq\), the maps \(\theta_DF(t)\) form such a cone with vertex \(F(X)\). Let \(\widetilde\theta_X:F(X)\to G(X)\) be its unique factorization. Explicitly,

For \(f:X\to Y\), compose both \(G(f)\widetilde\theta_X\) and \(\widetilde\theta_YF(f)\) with each \(G(t)\), where \(t:Y\to qD\). Both composites are \(\theta_DF(tf)\). Uniqueness in (1.3) proves naturality. At \(X=qD\), the comma object \( (D,1_{qD})\) gives \(\widetilde\theta_{qD}=\theta_D\). Conversely any extension must satisfy the cone equations and hence (1.4). These constructions prove (1.2). \(\square\)

The displayed limit is a proved universal property of the particular cone. It asserts neither general completeness of \(\mathsf B\) nor smallness of its comma index in a previously fixed universe.

Reversing all arrows gives the following useful dual. Choose \(s_X:qD_X\to X\) with \(G(s_X)\) invertible. For each \(t:qE\to X\), require \(a:H\to D_X\) and \(b:H\to E\) such that \(s_Xq(a)=tq(b)\) and \(G(q(b))\) is invertible. Then \(G(X)\) is the colimit of \(Gq\) over \( (q\downarrow X)\), and restriction gives

This follows by applying the proved theorem to \(q^{\mathrm{op}}\) and \(G^{\mathrm{op}}\), including the target category \(\mathsf B^{\mathrm{op}}\).

Corollary 1.2. Suppose \(q\) is essentially surjective. Suppose every \(f:qD\to qE\) can be written using arrows \(a:D\to H\), \(b:E\to H\) with \(q(b)\) invertible and \(q(a)=q(b)f\). Then restriction \(q^*:\operatorname{Fun}(\mathsf C,\mathsf B)\to\operatorname{Fun}(\mathsf D,\mathsf B)\) is fully faithful for every \(\mathsf B\).

Proof. Choose an isomorphism \(s_X:X\to qD_X\). Apply the stated condition to \(ts_X^{-1}\) for each \(t:X\to qE\). This gives (1.1) for every target functor \(G\), so Theorem 1.1 applies to every pair \(F,G\). \(\square\)

2. Localize a full family of replacements

Let \(\mathsf I\subseteq\mathsf C\) be a full subcategory, let \(S\) be an outgoing multiplicative system, and put \(T=S\cap\operatorname{Arrows}(\mathsf I)\). Whenever \(T\) is itself an outgoing system, inclusion induces a functor \(\mathsf I[T^{-1}]\to\mathsf C[S^{-1}]\): the composite of inclusion with \(Q\) inverts \(T\), so the localization universal property applies. To obtain \(T\)'s axioms and full faithfulness together, consider the following condition:

For every \(s:X\to Y\) in \(S\) with \(X\in\mathsf I\), there is a \(g:Y\to W\), with \(W\in\mathsf I\), such that \(gs\in S\).

The condition concerns the composite \(gs\). It does not require \(g\in S\).

Theorem 2.1. Under this condition, \(T\) is an outgoing multiplicative system in \(\mathsf I\), and inclusion induces a fully faithful functor

Proof. Isomorphisms and compositions in \(T\) remain in \(T\). For an outgoing square with \(f:X\to Y\), \(s:X\to X'\) in \(\mathsf I\) and \(s\in T\), the calculus in \(\mathsf C\) gives \(t:Y\to Z\) in \(S\) and \(h:X'\to Z\) with \(hs=tf\). Apply the condition to \(t\), whose source lies in \(\mathsf I\). It supplies \(k:Z\to W\) with \(W\in\mathsf I\) and \(kt\in S\). The square with sides \(kt\) and \(kh\) lies in \(\mathsf I\) by fullness, and its denominator \(kt\) lies in \(T\).

For cancellation, a pair \(f,g: X\rightrightarrows Y\) in \(\mathsf I\) equal after precomposition by a \(T\)-arrow is equalized by some \(t:Y\to Z\) in \(S\). Apply the same condition to obtain \(kt:Y\to W\) in \(T\); it still equalizes \(f,g\). This proves the axioms for \(T\). Universal descent now gives (2.1).

Fix \(Y\in\mathsf I\). Inside its filtered denominator category \(\mathsf D_Y\), let \(\mathsf E_Y\) be the full subcategory of denominators whose targets lie in \(\mathsf I\). Every \(s:Y\to Z\) maps to an object \(gs:Y\to W\) in \(\mathsf E_Y\) by the hypothesis. This implies that \(\mathsf E_Y\) is filtered: a common target or an equalizing object constructed in \(\mathsf D_Y\) can be moved once more into \(\mathsf E_Y\), and fullness retains the resulting arrows.

It also proves cofinality. For a fixed object of \(\mathsf D_Y\), the comma category of maps from it into \(\mathsf E_Y\) is nonempty. Two such maps acquire a common target in \(\mathsf D_Y\); equalize their composites from the fixed object using filteredness, then move the target into \(\mathsf E_Y\). This makes the comma category connected. The full filtered cofinal-subcategory criterion is also recorded in Stacks, full cofinal-subcategory criterion.

For \(X,Y\in\mathsf I\), the Hom formula for fractions therefore gives

Fullness identifies the first diagram with the restriction of the second; cofinality identifies their colimits. The bijection is the map induced by \(e\), proving full faithfulness. All these colimits can be interpreted in the ambient universe. \(\square\)

Corollary 2.2. If every \(X\in\mathsf C\) has a denominator \(X\to I_X\) with \(I_X\in\mathsf I\), then \(e\) is an equivalence.

Proof. Such replacements satisfy the hypothesis of Theorem 2.1 by applying it to \(Y\) and composing denominators. They also make every \(QX\) isomorphic to an object from \(\mathsf I[T^{-1}]\). Thus \(e\) is fully faithful and essentially surjective. \(\square\)

3. A functor only needs to descend on the replacements

Assume the global denominator replacements of Corollary 2.2. Let \(F:\mathsf C\to\mathsf B\) send every arrow of \(T\) to an isomorphism. Let \(F_T:\mathsf I[T^{-1}]\to\mathsf B\) be its descent on \(\mathsf I\), and choose a quasi-inverse \(W\) to \(e\). Put \(H=F_TW\).

Theorem 3.1. The functor \(H\) is a universal right localization of \(F\). Its unit \(\tau:F\to HQ\) is characterized on \(\mathsf I\) by the canonical identification \(F|_{\mathsf I}\simeq HQ|_{\mathsf I}\). For every \(K:\mathsf B\to\mathsf B'\), the pair \( (KH,K\tau)\) right localizes \(KF\).

Proof. Write \(j:\mathsf I\hookrightarrow\mathsf C\). For any \(G:\mathsf C[S^{-1}]\to\mathsf B\), Theorem 1.1 applies to \(q=j\) and target \(GQ\). Choose \(s:X\to I_X\) in \(S\). For \(t:X\to I'\), an Ore square gives

Here only \(b\) is required to lie in \(S\). Choose a further denominator \(w:Z\to I''\). Then \(wa,wb\) are arrows in the full subcategory, \(wb\in T\), and \(GQ(wb)\) is invertible. These are precisely the conditions of (1.1). Consequently

The middle bijection is full faithfulness of restriction along the localization of \(\mathsf I\). The last comes from the equivalence \(e\). Apply the first bijection with \(G=H\) to extend the canonical identification on \(\mathsf I\) to a unique \(\tau:F\to HQ\). Every map in (3.1) is restriction or composition with this identification, so the inverse chain sends \(u:H\to G\) to \(uQ\circ\tau\). This proves the required universal property with its specified unit.

The same argument applies to \(KF\): its restriction descends as \(KF_T\), the chosen value is \(KH\), and uniqueness in Theorem 1.1 identifies the resulting unit with \(K\tau\). No preservation of colimits by \(K\) is needed. \(\square\)

In the displayed Ore square, the notation declares \(b\in S\) and imposes no such condition on \(a\). Nor does the theorem require \(F(s)\) to be invertible for denominators whose endpoints are outside \(\mathsf I\). Exercise 2 exhibits that freedom.

4. Reflection supplies denominator replacements

Let \(L:\mathsf C\to\mathsf D\) be left adjoint to a fully faithful \(R:\mathsf D\to\mathsf C\). Write \(\eta:1_{\mathsf C}\to RL\) for the unit and \(\varepsilon:LR\to1_{\mathsf D}\) for the counit. The counit is an isomorphism by Stacks, Lemma 4.24.4. Set

Theorem 4.1. The family \(S\) is an outgoing multiplicative system and equals the inverse image of the isomorphisms under its localization \(Q\). Every \(\eta_X\) belongs to \(S\). The descended functor

is an equivalence. Every \(F:\mathsf C\to\mathsf B\) has universal right localization \(FR\overline L\), with unit \(F\eta\) under \(\overline LQ=L\).

Proof. The triangle identity \(\varepsilon_{LX}L(\eta_X)=1_{LX}\) makes \(L(\eta_X)\) invertible. Closure of \(S\) under isomorphisms and composition is immediate. For \(f:X\to Z\), \(s:X\to Y\) with \(s\in S\), use \(W=RLZ\), \(t=\eta_Z:Z\to W\), and

Naturality of \(\eta\) gives \(gs=\eta_Zf=tf\). This is the outgoing Ore square. If \(f,g: X\rightrightarrows Y\) agree after precomposition by an \(S\)-arrow, then \(Lf=Lg\). Naturality now gives \(\eta_Yf=\eta_Yg\), proving cancellation.

The localization property supplies \(\overline L\). A quasi-inverse is \(QR\). The counit gives \(\overline LQR\simeq1_{\mathsf D}\). The transformation \(Q\eta:Q\to QRL\) is an isomorphism on every object. Full faithfulness of restriction along \(Q\) descends it to \(1_{\mathsf C[S^{-1}]}\simeq QR\overline L\). Thus \(\overline L\) is an equivalence. It follows that \(Q(f)\) is invertible exactly when \(L(f)\) is invertible; hence \(S\) is saturated in this precise sense.

Let \(\mathsf I\) be the full essential image of \(R\). The arrows \(\eta_X:X\to RLX\) give the global replacements required in Section 3. The restriction \(L|_{\mathsf I}\) is fully faithful: identify objects with \(RA,RB\), use full faithfulness of \(R\), and conjugate \(LR\) by its counit to identify the induced Hom map. Therefore an arrow of \(\mathsf I\) whose \(L\)-image is invertible is itself invertible. Every functor \(F\) inverts these arrows. Theorem 3.1 applies, with the replacement value \(FR\overline L\) and unit \(F\eta\). Its restriction to \(\mathsf I\) is the requisite isomorphism because \(\eta_{RA}\) is invertible, by the other triangle identity and invertibility of \(R\varepsilon_A\). \(\square\)

For example, the inclusion of abelian groups in groups is fully faithful and has abelianization as left adjoint. The theorem applies to every functor defined on groups, including functors with targets that have no filtered colimits.

5. Universal localization of the identity forces reflection

The converse uses universality with a target as large as the localization itself. Interpret all functor categories in the ambient universe fixed at the start of the lesson.

Theorem 5.1. Let \(S\) be an outgoing multiplicative system. Suppose \(H:\mathsf C[S^{-1}]\to\mathsf C\), with \(\eta:1_{\mathsf C}\to HQ\), is a universal right localization of \(1_{\mathsf C}\). Then \(Q\dashv H\), and \(H\) is fully faithful. If \(S\) is saturated, then each \(\eta_X\) is in \(S\).

Proof. Universality with the further functor \(K=Q\) says that \( (QH,Q\eta)\) right localizes \(Q\). Since \(Q\) already inverts \(S\), its right localization is \(1_{\mathsf C[S^{-1}]}\), with identity unit. Uniqueness supplies an isomorphism

Put \(\theta=H\varepsilon\circ\eta H:H\to H\). Naturality of \(\eta\) at \(\eta_X:X\to HQX\), followed by (5.1), shows

The universal property of \( (H,\eta)\) makes composition \(u\mapsto uQ\circ\eta\) injective on \(\operatorname{Nat}(H,H)\). Hence \(\theta=1_H\), which is the second triangle identity.

The adjunction bijection and its inverse are now explicit:

For the composite starting with \(f\), naturality of \(\varepsilon\) reduces it to \(f\varepsilon_{QX}Q\eta_X=f\). Starting with \(b\), naturality of \(\eta\) reduces it to \(H\varepsilon_Y\eta_{HY}b=b\). These formulas are natural in \(X,Y\), so they prove the adjunction. Its counit is an isomorphism; the cited fully faithful adjoint criterion makes \(H\) fully faithful. Equation (5.1) also makes \(Q\eta_X\) invertible. Saturation is exactly what upgrades this conclusion to \(\eta_X\in S\). \(\square\)

The adjunction and full faithfulness themselves do not need saturation. For contrast, localizing the ordered category \(\mathbb N\) at all arrows gives a terminal category, but the localization has no right adjoint: a right adjoint would choose a terminal object of \(\mathbb N\), and none exists. Thus its identity functor cannot have a universal right localization.

6. Projectors detect the objects of a reflection

A projector on \(\mathsf C\) is an endofunctor \(P\) with \(e:1_{\mathsf C}\to P\) such that both \(Pe\) and \(eP\) are isomorphisms. The two whiskerings are part of the hypothesis.

Lemma 6.1. For a projector, \(Pe=eP\).

Proof. Set \(u=(eP)^{-1}Pe:P\to P\). Naturality of \(e\) at \(e_X\) gives \(ue=e\). Applying \(P\) and canceling the isomorphism \(Pe\) gives \(Pu=1_{P^2}\). Naturality of \(e\) at \(u_X\) then reads \(e_{PX}u_X=P(u_X)e_{PX}=e_{PX}\). Cancel \(e_{PX}\) to obtain \(u_X=1_{PX}\), for every \(X\). \(\square\)

Theorem 6.2. Precomposition by \(e_X\) gives a natural bijection

For an object \(X\), the following are equivalent: \(e_X\) is invertible; precomposition by \(e_Y\) gives a bijection \(\operatorname{Hom}(PY,X)\to\operatorname{Hom}(Y,X)\) for every \(Y\); this map is surjective just for \(Y=X\). The full subcategory \(\mathsf C_0\) of these objects is reflective, with reflector \(P\).

Proof. The inverse in (6.1) sends \(v:X\to PY\) to \(e_{PY}^{-1}P(v)\). Its composite with precomposition by \(e_X\) is \(v\), by naturality of \(e\) at \(v\). For \(w:PX\to PY\), its other composite is

using Lemma 6.1 and naturality at \(w\). This proves the bijection.

If \(e_X\) is invertible, identify \(X\) with \(PX\) and use (6.1) to obtain the asserted bijection for every \(Y\). Bijectivity implies the specified surjectivity. Conversely, that surjectivity lifts \(1_X\) to a map \(r:PX\to X\) with \(re_X=1_X\). The maps \(e_Xr\) and \(1_{PX}\) agree after precomposition by \(e_X\); injectivity of (6.1) with \(Y=X\) implies \(e_Xr=1_{PX}\). Thus \(e_X\) is invertible.

Each \(PX\) lies in \(\mathsf C_0\), since \(e_{PX}\) is invertible. For any \(Z\in\mathsf C_0\), the same Hom bijection, identifying \(Z\) with \(PZ\), gives \(\operatorname{Hom}(PX,Z)\simeq\operatorname{Hom}(X,Z)\). It is natural in both arguments and has unit \(e\). This is the reflection. \(\square\)

Conversely, for \(L\dashv R\) with \(R\) fully faithful, the actual counit \(\varepsilon:LR\to1\) is invertible by the checked adjoint criterion in Section 4. Both \(\eta RL\) and \(RL\eta\) are inverse to \(R\varepsilon L\), by the triangle identities. Thus \(P=RL\), with \(e=\eta\), is a projector. The functor \(R\) takes values in \(\mathsf C_0\), because \(R(\varepsilon_Y)\eta_{RY}=1\) and \(R(\varepsilon_Y)\) is invertible. It remains fully faithful there. It is essentially surjective onto \(\mathsf C_0\), since any such \(X\) is isomorphic to \(RLX\). Hence the reflected target is equivalent to exactly the objects detected by Theorem 6.2.

One whiskering does not suffice. Consider the one-object category with endomorphisms \(1,p\), where \(p^2=p\ne1\). Let \(P\) send both arrows to \(1\), and let \(e\) have component \(p\). This is natural: \(P(a)p=pa=p\) for either arrow \(a\). The functor laws hold, and \(Pe\) has component \(1\), but \(eP\) has component \(p\). The latter is not invertible, since neither endomorphism is an inverse of \(p\).

7. Exercises and complete solutions

Exercise 1 (Grade 1: detect a missing hypothesis). Let \(q\) include the discrete category on two objects \(a,b\) into the category with one additional arrow \(a\to b\). Show that \(q\) is essentially surjective but restriction on functor categories need not be full. Use the constant functor with value \(\{0,1\}\).

Solution. Both objects occur in the image, so \(q\) is essentially surjective. Let \(F=G\) send the additional arrow to the identity of \(\{0,1\}\). On the discrete category, take the component at \(a\) to be the identity and the component at \(b\) to be the constant function with value \(0\). This is a natural transformation there. Naturality at \(a\to b\) in the larger category would equate these two functions. They differ at \(1\), so it has no extension. The morphism comparison condition in Corollary 1.2 is substantive.

Exercise 2 (Grade 2: descend only on replacements). In the ordered category \(\mathbb N\), invert every arrow and let \(\mathsf I\) be the full subcategory of even integers. Define \(F:\mathbb N\to\mathsf{Set}\) by \(F(1)=\{a,b\}\) and \(F(n)=\{*\}\) for \(n\ne1\). The map \(F(0\to1)\) selects \(a\); maps from \(1\) to later stages are the unique collapse. Find its universal right localization and unit. Does \(F\) invert every denominator?

Solution. Composites through \(1\) between singleton stages are the unique maps, so this is a functor. Every integer maps to an even integer, and every restricted arrow of \(\mathsf I\) is sent to a bijection between singletons. The localization of \(\mathbb N\) is terminal: a common upper bound represents the unique fraction between any two integers. Theorem 3.1 therefore gives the singleton functor as the universal right localization. Its unit is the unique map \(F(n)\to\{*\}\); at \(1\) it identifies \(a,b\). The arrow \(0\to1\) is a denominator, but its image is not surjective. Thus \(F\) does not invert all denominators, and this does not prevent universal localization.

Exercise 3 (Grade 3: distinguish the two replacement conditions). Take the concrete category with objects \(0=\{0\}\), \(1=\{0,1\}\), inclusion \(s:0\to1\), retraction \(r:1\to0\), and \(e=sr:1\to1\). Its arrows are the identities, \(s,r,e\). Put \(S=\{1_0,1_1,s,e\}\) and \(\mathsf I=\{0\}\). Verify the outgoing axioms and the hypothesis of Theorem 2.1. Show that the hypothesis of Corollary 2.2 fails, and determine both localized categories.

Solution. The identities are the only isomorphisms. The equations \(rs=1_0\), \(e^2=e\), \(es=s\), \(re=r\) verify closure of \(S\). Identity denominators require no construction. For denominator \(s\), the possible maps from its source are \(1_0,s\); their squares use respectively the pairs \( (t,h)=(s,1_1)\), \( (1_1,1_1)\). For denominator \(e\), the possible maps are \(r,1_1,e\); use respectively \( (t,h)=(1_0,r),(e,1_1),(1_1,1_1)\). In each pair \(t\) is the outgoing denominator and \(he=tf\). The only distinct parallel arrows are \(1_1,e\), and postcomposition by \(e\in S\) equalizes them. Hence every needed cancellation holds.

The denominators with source \(0\) are \(1_0,s\). For \(1_0\), take the identity comparison; for \(s\), take \(g=r\), so \(gs=1_0\in S\). This proves the weak condition, although \(r\notin S\). There is no \(S\)-arrow from \(1\) to \(0\), so global denominator replacement fails.

The localization of \(\mathsf I\) is terminal. In the larger localization, \(Qs\) is invertible, \(Qr=(Qs)^{-1}\), and \(Qe=1_{Q1}\). Every map is generated by these arrows and denominator inverses, so each Hom set is a singleton. Its two objects are isomorphic; it too is equivalent to a terminal category. Here the conclusion of Corollary 2.2 happens to hold although its sufficient hypothesis does not. The two hypotheses still differ strictly.

Exercise 4 (Grade 4: compute a reflection and its localized value). Prove that abelianization \(L(G)=G/[G,G]\) is left adjoint to the inclusion of abelian groups. Apply Theorem 4.1 to the underlying-set functor. Compute the unit for the free group on two generators and show that it need not be injective.

Solution. Let \([G,G]\) be the normal subgroup generated by all commutators. The quotient is abelian, since each commutator becomes the identity. Every homomorphism \(G\to A\) with \(A\) abelian kills every commutator and its normal closure, hence factors uniquely through the quotient. This gives a natural bijection \(\operatorname{Hom}_{\mathsf{Ab}}(G/[G,G],A)\simeq\operatorname{Hom}_{\mathsf{Grp}}(G,A)\), proving the adjunction. Inclusion is fully faithful. For \(F\) the underlying-set functor, the universal localized value at \(QG\) is the underlying set of \(G/[G,G]\), and the unit is the quotient function.

For the free group \(G=\langle x,y\rangle\), its abelianization is \(\mathbb Z^2\). Indeed, sending \(x,y\) to its standard generators factors through \(G/[G,G]\); conversely those two commuting classes give a map \(\mathbb Z^2\to G/[G,G]\). The maps are inverse on the generators. The reduced word \(xyx^{-1}y^{-1}\) is nonempty in the free group and therefore differs from the identity, but both map to \( (0,0)\). The unit is not injective. The theorem nonetheless proves universality after every further functor on sets.

References

Reflective subcategories and idempotent monads are treated in Emily Riehl, Category Theory in Context, Sections 4.5 and 5.2.

The outgoing fraction calculus is retained from Stacks, Section 4.27. The full filtered cofinal-subcategory interface is Stacks, full cofinal-subcategory criterion. The fully faithful adjoint criterion is Stacks, Lemma 4.24.4. These references supply the named canonical interfaces; their text is not reproduced.

Localization of categories, of subcategories and of functors is treated in Pierre Schapira, An Introduction to Categories and Homological Algebra, lecture notes, version of 1 March 2026, Chapter 3.