One-sided fractions and saturation

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

Some ordinary arrows become invertible after localization even though they were not among the selected denominators. Fractions detect exactly which ones. The same calculus also controls comma categories, zero objects, and the faithfulness of a functor after the arrows it inverts have been localized.

Throughout, \(S\) contains isomorphisms, is closed under composition, and satisfies the outgoing square and cancellation axioms. Let \(Q:\mathsf C\to\mathsf L=\mathsf C[S^{-1}]\) be its localization. We retain the fraction construction from Stacks, Section 4.27: this reference calls our outgoing calculus left. In particular, every map \(QX\to QY\) is \(Q(s)^{-1}Q(f)\), where \(f:X\to Y'\) and \(s:Y\to Y'\) with \(s\in S\). The other interface we will repeatedly use is

for ordinary parallel arrows \(f,g: X\rightrightarrows Y\), with \(t\) pointing out of \(Y\). Denominator categories may be interpreted in an ambient universe; no cofinal-smallness assumption is made here.

1. Two successive composites detect invertibility

Theorem 1.1. For \(f:X\to Y\), the localized map \(Q(f)\) is invertible if and only if there are \(g:Y\to Z\) and \(h:Z\to W\) such that

Proof. Write an inverse of \(Q(f)\) as \(Q(s)^{-1}Q(a)\), where \(s:X\to X'\) lies in \(S\) and \(a:Y\to X'\). Then \(Q(af)=Q(s)\). By (0.1), there is \(t:X'\to Z\) in \(S\) with \(taf=ts\). Put \(g=ta\); its composite \(gf=ts\) is in \(S\). Since \(Q(g)Q(f)\) and \(Q(f)\) are invertible, \(Q(g)\) is invertible. Apply the argument just proved to \(g\): it supplies \(h:Z\to W\) with \(hg\in S\).

Conversely, write \(A=Q(f)\), \(B=Q(g)\), \(C=Q(h)\). If \(BA\) and \(CB\) are invertible, then \(A(BA)^{-1}\) is a right inverse of \(B\), while \( (CB)^{-1}C\) is a left inverse. A left inverse and a right inverse of the same arrow agree: if \(lB=1\) and \(Br=1\), then \(l=lBr=r\). Thus \(B\) is invertible, and so is \(A=B^{-1}(BA)\). \(\square\)

Call an outgoing system saturated if \(gf,hg\in S\) always imply \(f\in S\). Theorem 1.1 proves that this is equivalent to

This formulation also shows that all three of \(f,g,h\) in (1.1) have invertible localized images, although the definition only asks for \(f\in S\).

Theorem 1.2. The family

is the smallest saturated outgoing multiplicative system containing \(S\). The canonical functor \(\mathsf C[S^{-1}]\to\mathsf C[\widehat S^{-1}]\) is an equivalence.

Proof. Identities, isomorphisms and compositions lie in \(\widehat S\). For \(s:X\to Y\) in \(\widehat S\) and \(f:X\to Z\), express the localized map \(Q(f)Q(s)^{-1}:QY\to QZ\) as \(Q(t)^{-1}Q(g)\), with \(t:Z\to W\) in \(S\) and \(g:Y\to W\). Then \(Q(gs)=Q(tf)\). Choose \(v:W\to W'\) in \(S\) with \(vgs=vtf\). The arrows \(vg\) and \(vt\in S\subseteq\widehat S\) supply the required outgoing square.

For cancellation, if \(f,g: X\rightrightarrows Y\) agree after precomposition by an arrow of \(\widehat S\), invert its image to obtain \(Qf=Qg\). Equation (0.1) supplies a further arrow of \(S\), and hence of \(\widehat S\), which equalizes them. Thus \(\widehat S\) is an outgoing system. It is saturated: if \(gf,hg\in\widehat S\), the inverse argument in Theorem 1.1 makes \(Qf\) invertible. Any saturated outgoing system containing \(S\) contains every \(f\in\widehat S\), because Theorem 1.1 gives witnesses with both composites in \(S\). This proves minimality.

Both localization functors invert \(S\), and \(Q\) also inverts \(\widehat S\). Their universal properties therefore give functors in both directions. Their composites restrict to the original localization functors on \(\mathsf C\). Full faithfulness of restriction on natural transformations identifies those composites with the identities, proving the equivalence. \(\square\)

If \(S\) admits both incoming and outgoing calculi, the existing stronger Stacks saturation lemma applies. Its definition of saturation asks that \(gf,hg\in S\) imply the middle arrow \(g\in S\). Its complete statement and proof identify the saturation with the inverse image of the isomorphisms, and give the alternative test \(gf,fh\in S\) for suitable \(g,h\). Thus in the two-sided case that middle-arrow condition is equivalent to (1.2), and hence to our saturation condition. Theorem 1.1 supplies the successive-composite test with only the outgoing calculus; it does not assume the additional incoming axioms.

2. Fractions cover the localized comma category

For \(X\in\mathsf C\), let \(\mathsf D_X\) have objects \(s:X\to X_s\) in \(S\), with arbitrary arrows \(a:X_s\to X_t\) satisfying \(as=t\). This category is filtered by the retained fraction calculus. Let \(\mathsf E_X=(Q\downarrow QX)\): its objects are \( (Y,u)\) with \(u:QY\to QX\); an arrow \(b:(Y,u)\to(Z,v)\) is an ordinary \(b:Y\to Z\) satisfying \(vQb=u\). Define

An arrow \(a:s\to t\) goes to the same ordinary \(a\); the equality \(Q(t)^{-1}Q(a)=Q(s)^{-1}\) follows from \(as=t\).

Theorem 2.1. The functor \(\theta\) is cofinal, and \(\mathsf E_X\) is filtered.

Proof. Fix \( (Y,u)\in\mathsf E_X\). A fraction representation \(u=Q(s)^{-1}Q(b)\) gives a comma object \(b:(Y,u)\to\theta(s)\), so \( ((Y,u)\downarrow\theta)\) is nonempty.

For two such objects \( (s,b),(t,c)\), choose a common target \(r\) in \(\mathsf D_X\), with arrows \(a:s\to r\), \(d:t\to r\). The ordinary arrows \(ab,dc:Y\to X_r\) have the same localized image, namely \(Q(r)u\). Choose \(w:X_r\to Z\) in \(S\) with \(wab=wdc\) by (0.1). The denominator \(wr\) and that common composite give a common target in the comma category. For parallel arrows \(a,d:(s,b)\rightrightarrows(t,c)\), filteredness of \(\mathsf D_X\) supplies a further arrow equalizing \(a,d\); its composite with \(c\) supplies the corresponding comma object. Thus the comma category is filtered, in particular connected. The nonempty-connected comma criterion for cofinality proves the first assertion; see Stacks, cofinal functors.

Every object of \(\mathsf E_X\) maps into the image of \(\theta\). Two objects consequently have a common target by filteredness of \(\mathsf D_X\). If \(b,c:(Y,u)\rightrightarrows(Z,v)\) are parallel in \(\mathsf E_X\), choose \(a:(Z,v)\to\theta(s)\). Then

Choose \(w:X_s\to W\) in \(S\) with \(wab=wac\). The composite \(wa:(Z,v)\to\theta(ws)\) equalizes \(b,c\). Nonemptiness follows from \(\theta(1_X)\). This proves filteredness. \(\square\)

In the convention that a functor is right exact when all its target comma categories are filtered, Theorem 2.1 says that \(Q\) is right exact. It does not require \(\mathsf C\) to have finite colimits.

The existing Stacks, Lemma 4.27.9 says more precisely that the outgoing localization preserves every existing finite colimit. Its proof uses the filtered denominator Hom formula and commutation of filtered set colimits with finite limits; those are exactly the interfaces used here.

The existence statements can be separated. If \(\mathsf C\) has finite coproducts, their preserved images supply finite coproducts in \(\mathsf L\), because every localized object is \(QX\). If \(\mathsf C\) has coequalizers of all parallel pairs, then so does \(\mathsf L\): the common-denominator lemma in Section 4.27 writes a localized pair as \(Q(s)^{-1}Q(f)\), \(Q(s)^{-1}Q(g)\) with the same denominator. If \(c\) coequalizes \(f,g\) in \(\mathsf C\), then \(Q(c)Q(s)\) coequalizes the two localized arrows, by transporting the preserved coequalizer across \(Q(s)\). This does not require finite coproducts. Consequently, if \(\mathsf C\) has all finite colimits, so does \(\mathsf L\): finite coproducts and coequalizers construct them, as in Stacks, finite colimits. Here a coequalizer is the general categorical construction on two parallel arrows; no zero or additive structure is required.

The same calculus clears a commuting square, not just a pair of arrows. We retain the full Stacks, Lemma 4.27.10: given ordinary \(f:X\to Y\), \(f':X'\to Y'\) and localized \(a:QX\to QX'\), \(b:QY\to QY'\) with \(Q(f')a=bQ(f)\), it supplies ordinary \(g:X\to X''\), \(h:Y\to Y''\), \(f'':X''\to Y''\), denominators \(s:X'\to X''\), \(t:Y'\to Y''\), and equalities

with \(a=Q(s)^{-1}Q(g)\) and \(b=Q(t)^{-1}Q(h)\). Its direction agrees with our outgoing convention, so no dualization or triangulated hypothesis is needed.

3. A zero object survives localization

Assume \(\mathsf C\) has a zero object \(0\). Write \(0_{XY}:X\to Y\) for its zero morphism. No additivity is assumed.

Theorem 3.1. The object \(Q0\) is a zero object of \(\mathsf L\). Moreover,

Proof. The fraction formula gives \(\operatorname{Hom}_{\mathsf L}(Q0,QX)\) as a filtered colimit of singletons, since \(\operatorname{Hom}_{\mathsf C}(0,X_s)\) is a singleton. Hence \(Q0\) is initial.

A map \(QX\to Q0\) has a representative \( (f:X\to Y,s:0\to Y)\), with \(s\in S\). The arrows \(1_Y,0_{YY}\) agree after precomposition by \(s\), because there is only one map \(0\to Y\). Cancellation supplies \(t:Y\to Z\) in \(S\) with \(t=t0_{YY}=0_{YZ}\). Thus \(tf=0_{XZ}\) and \(ts=0_{0Z}\). The refined roof has zero numerator and denominator \(ts\in S\). It represents \(Q(0_{X0})\): both \(ts\circ0_{X0}\) and \(tf\) are the unique zero map \(X\to Z\). Every map \(QX\to Q0\) is therefore the same; this proves terminality. Ordinary zero morphisms localize to zero morphisms, since their factorization through \(0\) remains a factorization through \(Q0\).

If \(QX\) is zero, then \(Q(1_X)=Q(0_{XX})\). Equation (0.1) gives \(t:X\to Y\) in \(S\) with \(t=t0_{XX}=0_{XY}\), proving the forward implication. Conversely, if \(0_{XY}\in S\), its image is an invertible zero morphism. Composing it and its inverse in both orders shows that both \(1_{QX}\) and \(1_{QY}\) are zero. In a category with a zero object, an object whose identity is zero has exactly one map to and from every object. Thus \(QX\) is zero. \(\square\)

Saturation is unnecessary in this theorem. In particular, the zero denominator may have a target other than \(X\).

4. Finite colimits make descent faithful

Let \(F:\mathsf C\to\mathsf B\). Assume \(\mathsf C\) has finite colimits and that \(F\) preserves them. The target need only have the particular colimits represented by these images. Define

Theorem 4.1. The family \(S_F\) is a saturated outgoing multiplicative system. Its descended functor \(\overline F:\mathsf C[S_F^{-1}]\to\mathsf B\) is faithful.

Proof. Closure under isomorphisms and composition follows from applying \(F\). Given \(f:X\to Z\) and \(s:X\to Y\) in \(S_F\), form their pushout in \(\mathsf C\). The pushout leg \(t:Z\to Y\amalg_X Z\) becomes an isomorphism under \(F\), because the pushout of the isomorphism \(F(s)\) is an isomorphism. This proves the outgoing square axiom.

If \(f,g: X\rightrightarrows Y\) become equal after precomposition by an \(S_F\)-arrow, then \(Ff=Fg\). Let \(c:Y\to Z\) be their coequalizer. Its image is a coequalizer of an equal pair, hence an isomorphism: the identity of \(FY\) is such a coequalizer, and the universal property uniquely identifies the two coequalizers. Therefore \(c\in S_F\), which proves cancellation.

If \(gf,hg\in S_F\), apply the inverse argument of Theorem 1.1 in \(\mathsf B\) to conclude that \(Ff\) is invertible. Thus \(S_F\) is saturated.

Finally, let two localized maps have the same \(\overline F\)-image. Use a common denominator to write them as \(Q(s)^{-1}Q(f)\), \(Q(s)^{-1}Q(g)\) with \(s:Y\to Y'\) in \(S_F\). Their images being equal means \(Ff=Fg\), since \(Fs\) is invertible. The preceding coequalizer construction supplies \(c\in S_F\) with \(cf=cg\). Their refined fractions agree, so the original localized maps agree. This is faithfulness. \(\square\)

The source convention for right exactness uses filtered comma categories, whereas many accounts define it by finite-colimit preservation. Here these agree because \(\mathsf C\) has finite colimits. To check the interface explicitly, suppose first every \( (F\downarrow U)\) is filtered. Its ordinary Hom formula is

The comparison sends \(a:X\to Y\) to \(uF(a)\). It is surjective by taking \( (X,v)\) and \(a=1_X\) for each \(v:FX\to U\). Every representative \(a\) equals this canonical representative for its image, via the comma arrow \(a:(X,uFa)\to(Y,u)\); hence it is also injective. If \(X\) is the colimit of a finite diagram, substitute its Hom universal property into (4.1) and commute the finite limit with the filtered set colimit. The resulting natural bijection is precisely the colimit property of \(FX\).

Conversely, if \(F\) preserves finite colimits, then \( (F\downarrow U)\) has finite colimits: form them in \(\mathsf C\) and use the compatible maps to \(U\) to obtain the induced comma arrow. In particular it has an initial object, binary coproducts, and coequalizers, so it is filtered. Thus Theorem 4.1 retains the full source hypothesis. The Hom calculation can be performed in the ambient universe and does not silently impose cofinal-smallness.

Faithfulness does not imply fullness. The free-vector-space example in Exercise 3 gives an explicit distinction.

5. Incoming denominators and the universal interface

An incoming system satisfies the outgoing axioms in \(\mathsf C^{\mathrm{op}}\). Explicitly, a map \(f:X\to Y\) and a denominator \(t:Y'\to Y\) can be completed to \(tg=fs\), with \(s:X'\to X\) in \(S\); equality after postcomposition by a denominator can be made into equality after precomposition by one. “Incoming” and “outgoing” describe directions, so they also keep track of the opposite naming conventions in the references.

Let \(\mathsf I_X\) consist of incoming denominators \(s:X_s\to X\). An arrow \(s\to t\) in \(\mathsf I_X\) is an arbitrary \(a:X_s\to X_t\) satisfying \(ta=s\). These arrows need not belong to \(S\). The category \(\mathsf I_X\) is cofiltered, by applying the filtered-denominator result to \(\mathsf C^{\mathrm{op}}\). Set \(\mathsf J_X=\mathsf I_X^{\mathrm{op}}\). Thus an arrow \(s\to t\) in \(\mathsf J_X\) is an ordinary \(a:X_t\to X_s\) with \(sa=t\).

The retained incoming construction is Stacks, Remark 4.27.15. Its Hom formula, with the indexing variance displayed, is

The transition map associated to \(a:X_t\to X_s\) sends \(f\) to \(fa\). A representative \(f:X_s\to Y\) gives \(Q(f)Q(s)^{-1}\). This explains why the opposite of the incoming denominator category occurs: Hom is contravariant in its first argument.

If both calculi hold, Stacks, Lemma 4.27.19 identifies their localizations canonically. Consequently there is also a natural bijection

Here \(s:X_s\to X\), \(t:Y\to Y_t\), and a representative \(f:X_s\to Y_t\) maps to \(Q(t)^{-1}Q(f)Q(s)^{-1}\).

To verify the two comparisons in (5.2), first take the outgoing colimit in \(t\). It is \(\operatorname{Hom}_{\mathsf L}(QX_s,QY)\). Precomposition with \(Q(s)^{-1}\) identifies this set with \(\operatorname{Hom}_{\mathsf L}(QX,QY)\). For \(a:X_t\to X_s\) with \(sa=t\), the identity

makes these identifications compatible with every incoming transition. The resulting diagram on the nonempty filtered category \(\mathsf J_X\) is therefore constant up to its specified natural isomorphism. Its colimit is that same Hom set. Interchanging the two set colimits gives the incoming comparison as well. Colimits over a product category agree with iterated colimits because compatible maps out of either construction are exactly compatible maps out of all pairs of indices. The two maps from the single-denominator Hom formulas to (5.2), obtained by inserting \(1_X\) or \(1_Y\), are thus bijections. All these constructions can be made in an ambient universe.

For clarity, the universal interface retained from Stacks, Lemma 4.27.8 concerns natural transformations as well as objects. For every target \(\mathsf B\), restriction along \(Q\) is an equivalence onto the full subcategory of functors \(\mathsf C\to\mathsf B\) which invert \(S\). The formula for a descended functor on an outgoing roof is \(F(s)^{-1}F(f)\); the full associative fraction construction is Stacks, Lemma 4.27.2.

The omitted descent checks use only this formula. If an arbitrary refinement \(a\) takes a roof \((f,s)\) to \((af,as)\), then \(F(a)\) is invertible because both \(F(s)\) and \(F(as)\) are invertible. Thus \(F(as)^{-1}F(af)=F(s)^{-1}F(f)\); common refinements give well-definedness. To compose \((f,s)\) and \((g,t)\), an Ore square gives \(hs=vg\), with \(v\in S\), and the composite roof is \((hf,vt)\). Its image is

This is the composite of the two prescribed images. Identity roofs map to identities, and every roof is \(Q(s)^{-1}Q(f)\), so the descended functor is unique when its restriction is fixed.

The natural-transformation interface is visible directly. Let \(F,G: \mathsf L\to\mathsf B\) be functors. If \(\alpha:FQ\to GQ\) is natural on ordinary arrows, then

for \(s:Y\to Y'\) in \(S\). Combining this equality with naturality for a numerator proves naturality for every fraction. The components are already specified on all objects of the fraction category, so extension is unique. No invertibility of the components of \(\alpha\) is required.

This also gives uniqueness for an abstract localization with the same universal interface. Two such localizations produce comparison functors in both directions by descending their localization functors. Their composites become identities after restriction; full faithfulness of restriction lifts those identifications to natural isomorphisms. They are inverse equivalences. Passing to opposite categories preserves the interface: a functor \(\mathsf C^{\mathrm{op}}\to\mathsf B\) is a functor \(\mathsf C\to\mathsf B^{\mathrm{op}}\), and a transformation reverses direction when interpreted between these latter functors. Apply the interface with target \(\mathsf B^{\mathrm{op}}\), then reverse back. Thus \(\mathsf L^{\mathrm{op}}\) localizes \(\mathsf C^{\mathrm{op}}\) at \(S^{\mathrm{op}}\).

Finally, closure under isomorphisms and composition makes \(S\) the arrow set of a wide subcategory, with all objects of \(\mathsf C\); that subcategory need not be full. If \(E:\mathsf C\to\mathsf C'\) is an equivalence with quasi-inverse \(R\), transport the system as

The two closure axioms are immediate. To transport an outgoing square, apply \(R\), use the square in \(\mathsf C\), and transport its target and arrows back using \(E\) and the equivalence isomorphisms. Full faithfulness makes the resulting square commute. Its selected denominator belongs to \(S'\), since its \(R\)-image differs from the chosen \(S\)-arrow only by conjugating with isomorphisms. For cancellation, transport the equalizing arrow in the same way. Incoming axioms follow by opposition. A different quasi-inverse yields the same arrow family, since naturally isomorphic functors send each arrow to isomorphically conjugate arrows and \(S\) contains all isomorphisms.

6. Exercises and complete solutions

Exercise 1 (Grade 1: compute saturation). Regard the additive monoid \(\mathbb N\) as a category with one object, and take \(S=2\mathbb N\). Show that \(S\) is an outgoing system. Find its saturation and localized endomorphism monoid.

Solution. The sole identity is \(0\), and sums of even integers are even. An Ore square with ordinary arrow \(f\) and even denominator \(s\) can use the same denominator \(s\) and arrow \(f\), because \(f+s=s+f\). Cancellation follows from cancellation in \(\mathbb N\); equal arrows can be left unchanged using denominator \(0\). For every \(n\), take \(f=g=h=n\). Both successive composites are \(2n\in S\), so Theorem 1.1 puts every \(n\) in the saturation.

A fraction has numerator \(f\in\mathbb N\), denominator \(s\in2\mathbb N\), and value \(f-s\in\mathbb Z\). Equal fractions have equal differences. Conversely equal differences mean \(f+s'=f'+s\); the common denominator \(s+s'\) and refinement arrows \(s',s\) give equivalent roofs. Every integer is such a difference: choose an even \(s\ge\max(0,-z)\) and take \(f=z+s\). Composition adds differences. The localized endomorphisms are therefore the additive group \(\mathbb Z\); for example \( -1=1-2\).

Exercise 2 (Grade 2: zero denominators without additivity of the theorem). Let \(\mathsf C\) be the skeleton of finite-dimensional vector spaces over a field \(k\), with objects \(k^n\). Let \(S\) consist of all invertible endomorphisms and all zero endomorphisms, with no arrows between distinct objects. Verify the outgoing axioms. Determine the localization and its saturation.

Solution. In this skeleton all isomorphisms are endomorphisms. Products of listed endomorphisms are invertible or zero. Given \(f:k^n\to k^m\) and denominator \(s:k^n\to k^n\), if \(s\) is invertible use denominator \(1_{k^m}\) and comparison \(fs^{-1}\). If \(s=0\), use denominator \(0_{k^m}\) and zero comparison; both composites are zero. For cancellation, precomposition by an invertible denominator forces the parallel arrows to be equal. In the other case, postcomposition by the zero endomorphism of their target equalizes them. Thus the outgoing axioms hold.

Each zero endomorphism \(0:k^n\to k^n\) is in \(S\). Theorem 3.1 makes every localized object zero. Every localized Hom set is consequently a singleton, so the localization is equivalent to a terminal category. Every arrow then has an invertible localized image; the saturation is all arrows of \(\mathsf C\), much larger than the specified endomorphism family.

Exercise 3 (Grade 3: faithful descent can fail to be full). Let \(F:\mathsf{Set}\to\mathsf{Vect}_k\) be \(X\mapsto k^{(X)}\), the free vector space with basis \(X\). Prove the finite-colimit hypothesis of Theorem 4.1, find \(S_F\), and show that the descended functor is faithful but not full.

Solution. A linear map from \(k^{(X)}\) to \(V\) is uniquely determined by arbitrary images of the basis vectors, so \(\operatorname{Hom}_k(k^{(X)},V)\simeq\operatorname{Map}(X,UV)\), naturally in \(X,V\). Thus \(F\) is left adjoint to the underlying-set functor. It preserves colimits by Stacks, Lemma 4.24.5: equivalently, maps from \(F(\operatorname{colim}X_i)\) into \(V\) correspond to compatible maps from each \(FX_i\), by this adjunction. Sets have finite colimits, so the theorem applies.

For a function \(f:X\to Y\), an omitted element of \(Y\) gives a basis vector outside the image of \(Ff\). Distinct \(x,x'\) with the same image give the nonzero kernel vector \(e_x-e_{x'}\), in any characteristic. Hence invertibility of \(Ff\) forces \(f\) to be both surjective and injective. Conversely a bijection gives a basis isomorphism. Thus \(S_F\) is the set of bijections, and localization leaves the category of sets equivalent to itself. The descended functor is faithful by the theorem, also directly because its action on basis vectors determines a function. It is not full: a singleton has only its identity set endomorphism, whereas the associated one-dimensional vector space has the zero endomorphism as well as the identity. The zero linear map has no set-map preimage.

Exercise 4 (Grade 4: why saturation alone cannot prove faithfulness). Let \(\mathsf C\) be a one-object category given by a nontrivial group \(G\), and let \(F:\mathsf C\to\mathsf 1\) be the unique functor to a terminal category. Compute \(S_F\) and the localization. Show that \(F\) is not right exact in the comma sense, and explain which hypothesis of Theorem 4.1 is missing.

Solution. Every image arrow is the identity, so \(S_F\) consists of all arrows. These are already invertible. Their outgoing Ore squares can be filled using inverses, and cancellation holds by group cancellation. The localization is equivalent to \(\mathsf C\), since its identity functor already inverts all arrows and satisfies the localization property. The descended functor still identifies distinct group elements, so it is not faithful.

The unique target comma category \( (F\downarrow *)\) is \(\mathsf C\) itself. If \(g\ne h\) are two parallel group arrows, there is no arrow \(t\) with \(tg=th\); multiplication by \(t^{-1}\) would imply \(g=h\). Thus this comma category is not filtered, so \(F\) is not right exact. Moreover \(\mathsf C\) has no initial object: its only object has more than one endomorphism. It therefore does not have all finite colimits. The right-exactness and finite-colimit hypotheses cannot be replaced by saturation alone.

References

Stacks, Section 4.27 supplies outgoing fractions, common denominators, equality and universal descent. We retain its finite-colimit preservation, commuting-square refinement and two-sided saturation results with the hypotheses stated above. The cofinal comma criterion is Stacks, Section 4.17; finite colimit construction is Stacks, Lemma 4.14.12. No canonical text is copied.

Localization by a multiplicative system is treated in Pierre Schapira, An Introduction to Categories and Homological Algebra, lecture notes, version of 1 March 2026, Section 3.1. Right exactness can also be defined through comma categories; the argument around (4.1) verifies its equivalence with the finite-colimit hypothesis used in this lesson.