Quotients by a factorization ideal

Written by GPT-6.1 Sol (OpenAI), reasoning effort Ultra, October 2026. Self-checked by GPT-6.1 Sol, the AI that wrote it; no independent review. Public domain (CC0).

Suppose some objects are to count as zero. In an additive category, every map passing through one of these objects must then count as zero too. Quotienting the groups of maps does exactly this. The resulting objects retain their names, but their isomorphisms can change: two objects become isomorphic when suitable discarded summands make them isomorphic already.

We assume familiarity with additive categories and finite biproducts. The splitting terminology comes from Retracts and stabilization of formal objects, Section 1. No formal systems are needed here. Basic references are The Stacks Project, Preadditive and additive categories and the Mathlib treatments of category quotients and their additive structure.

Throughout, \(\mathsf C\) is a locally small additive category and \(\mathsf N\) a full additive subcategory. Thus its zero and finite biproducts can be chosen in \(\mathsf C\). Neither category is required to have a small set of objects. Write \(gf\) for first \(f\), then \(g\). Closure under objects will mean closure up to isomorphism unless otherwise stated.

1. Making the discarded objects zero

For \(X,Y\in\mathsf C\), put

There is no set of all intermediate objects in this formula. The subset \(\mathcal I(X,Y)\) is nevertheless a set, since it lies in the set \(\operatorname{Hom}_{\mathsf C}(X,Y)\).

Lemma 1.1. Each \(\mathcal I(X,Y)\) is an additive subgroup. Composing a member on either side with any map in \(\mathsf C\) gives another member.

Proof. The zero map passes through the zero object of \(\mathsf N\). If \(vu\) passes through \(Z\), its negative is \(v(-u)\). For factorizations through \(Z\) and \(Z'\), the sum factors through their biproduct:

Precomposing changes \(u\); postcomposing changes \(v\). Neither changes the intermediate object. \(\square\)

Such a system of subgroups is a two-sided additive ideal. More generally, for any such ideal \(\mathcal J\), the relation \(f\sim g\) defined by \(f-g\in\mathcal J\) is an equivalence relation on each group of maps and respects composition and addition. This is the congruence interface used in Mathlib's quotient construction. For (1.1), Lemma 1.1 verifies that interface directly.

Proposition 1.2. There is an additive category \(\mathsf C/\mathcal I\) with the objects of \(\mathsf C\) and groups of maps

The functor \(Q:\mathsf C\to\mathsf C/\mathcal I\) is the identity on objects and sends \(f\) to its class \([f]\). It is full and additive, preserves finite biproducts, and sends every object of \(\mathsf N\) to a zero object. Exactly the maps in \(\mathcal I\) become zero.

Proof. Use the group quotients in (1.3), with identity \([1_X]\) and composition \([g][f]=[gf]\). Changing representatives by \(m\in\mathcal I(X,Y)\) and \(n\in\mathcal I(Y,T)\) changes the composite by

Associativity, the identity laws and bilinearity descend from those in \(\mathsf C\). Every class has a representative, proving fullness. The group quotient maps prove additivity and the final assertion.

For finite biproducts, retain the biproduct criterion of Stacks, Remark 12.3.6: inclusion and projection maps with the usual matrix identities specify a biproduct. Their identities survive application of \(Q\), so the same objects with the classes of those maps are biproducts. The zero object also remains zero, by the identity criterion of Stacks, Lemma 12.3.2. These two criteria give an additive category.

For \(Z\in\mathsf N\), \(1_Z\) factors through \(Z\), so \([1_Z]=0\). The same identity criterion makes \(QZ\) a zero object. \(\square\)

Killing objects by this additive ideal is a different construction from Serre quotients and local saturation. Here there is no assumed abelian structure or closure under subobjects and quotients. Exactness of \(Q\) is not part of Proposition 1.2; Exercise 3 supplies an explicit failure.

2. The universal property, including transformations

An additive functor kills \(\mathsf N\) when it sends every object of \(\mathsf N\) to a zero object.

Theorem 2.1. Let \(\mathsf A\) be an additive category. An additive functor \(F:\mathsf C\to\mathsf A\) kills \(\mathsf N\) if and only if it has an additive factorization

With the object choices of Proposition 1.2, \(\overline F\) is unique. Natural transformations between such functors descend uniquely too. Consequently precomposition with \(Q\) identifies the category of additive functors from \(\mathsf C/\mathcal I\) with the full subcategory of additive functors from \(\mathsf C\) that kill \(\mathsf N\).

Proof. If \(F\) kills \(\mathsf N\), every factorization (1.1) goes through a zero object after applying \(F\); its image is zero. Since \(F\) is additive, \(f-g\in\mathcal I\) implies \(Ff=Fg\). Define

This is well-defined, additive, and respects identities and composition. Conversely, \(Q\) kills \(\mathsf N\), and an additive functor preserves zero objects, so (2.1) implies that \(F\) kills \(\mathsf N\). The zero preservation used here is Stacks, Lemma 12.3.7.

Every object and every map of the quotient comes from \(Q\). Thus (2.1) determines all values of \(\overline F\), proving uniqueness. For a natural transformation \(\eta:F\to G\), retain the same components \(\eta_X\). Naturality for \([f]\) is precisely naturality for a representative \(f\), so they give a transformation \(\overline F\to\overline G\). All transformations between the descended functors are determined by these components, and composition and identities are unchanged. \(\square\)

This is a universal property with a specified factorization. If equivalent models of the quotient are chosen, the comparison is unique when required to respect that factorization.

3. Splitting a projector that passes through a discarded object

Recall that an idempotent \(e:Z\to Z\) splits if there are maps \(a:Z_0\to Z\) and \(b:Z\to Z_0\) with \(ab=e\) and \(ba=1_{Z_0}\). The subcategory \(\mathsf N\) is idempotent complete when every one of its idempotents splits in \(\mathsf N\).

The following lemma lets us use this assumption without asking for all projectors in \(\mathsf C\) to split.

Lemma 3.1. Suppose \(\mathsf N\) is idempotent complete. If an idempotent \(q:V\to V\) belongs to \(\mathcal I(V,V)\), then it splits through an object of \(\mathsf N\).

Proof. Choose \(q=vu\) with \(u:V\to W\), \(v:W\to V\), \(W\in\mathsf N\). Put \(t=uv\). The identity \(q^2=q\) gives

Hence \(p=t^2\) satisfies

Fullness puts this endomorphism of \(W\) in \(\mathsf N\). Split it as \(p=ab\), \(ba=1_{W_0}\), with \(W_0\in\mathsf N\). Since \(pa=a\) and \(bp=b\), (3.2) gives \(ta=a\) and \(bt=b\). Define

Then

These are the required splitting identities. They also imply \(qj=j\) and \(rq=r\). \(\square\)

In particular \(QV\) is zero exactly when \(1_V\) factors through an object of \(\mathsf N\). Under the hypothesis of Lemma 3.1, this means exactly that \(V\) is isomorphic to an object of \(\mathsf N\). Indeed, the splitting of \(1_V\) gives inverse maps \(V\leftrightarrows W_0\). The reverse implication follows from Proposition 1.2.

Without idempotent completeness, objects that are retracts of objects of \(\mathsf N\) still become zero, but they need not belong to \(\mathsf N\). Exercise 2 exploits this distinction.

4. An isomorphism after adding discarded summands

Call \(X,Y\) stably isomorphic relative to \(\mathsf N\) when some \(Z_1,Z_2\in\mathsf N\) give an isomorphism

Theorem 4.1. If \(\mathsf N\) is idempotent complete, then \(QX\simeq QY\) if and only if (4.1) holds. No idempotent completeness of \(\mathsf C\) is assumed.

Proof. An isomorphism (4.1) gives an isomorphism in the quotient, where both extra summands are zero. This proves one direction.

For the other, choose representatives \(f:X\to Y\), \(g:Y\to X\) of inverse isomorphisms in the quotient. In particular, write

The difference in (4.2) need not be zero. Make it zero by adjoining its intermediate object. On \(V=Y\oplus Z\), set

Now \(GF=gf-ab=1_X\). Thus

The quotient identifies \(QV\) with \(QY\), through the projection and inclusion of the first summand. Under this identification \(QF=[f]\) and \(QG=[g]\), because all maps to or from \(QZ\) are zero. They are inverse. Therefore \(Qq=0\), so \(q\in\mathcal I(V,V)\).

Lemma 3.1 gives \(j:W_0\to V\), \(r:V\to W_0\), \(W_0\in\mathsf N\), with \(jr=q\), \(rj=1_{W_0}\). Since \(qj=j\) and \(rq=r\), (4.4) gives \(Gj=0\) and \(rF=0\). Consequently

while the opposite composite is \(FG+jr=FG+q=1_V\). We obtain

which is (4.1). \(\square\)

The minus sign in \(G\) removes the actual error \(gf-1_X\). The projector \(q\) then measures the remaining summand. It need not split for arbitrary reasons in \(\mathsf C\); its factorization through \(\mathsf N\) is what supplies a splitting.

5. Which coproducts survive?

Let \(I\) be a small set. Suppose \(\mathsf C\) has coproducts indexed by \(I\), and \(\mathsf N\) is closed under those coproducts in \(\mathsf C\). This section does not require idempotent completeness. We use the usual axiom of choice for small families of representatives and factorizations.

Theorem 5.1. The quotient \(\mathsf C/\mathcal I\) has coproducts indexed by \(I\), and \(Q\) preserves them.

Proof. For a family \((X_i)_{i\in I}\), write \(S=\bigoplus_iX_i\), with injections \(\iota_i\). For any \(Y\), restriction gives a group isomorphism

Under (5.1), the subgroup \(\mathcal I(S,Y)\) is exactly \(\prod_i\mathcal I(X_i,Y)\). One inclusion follows by restricting a single factorization. For the other, choose factorizations \(f_i=v_i u_i\) through \(Z_i\in\mathsf N\). Let \(Z=\bigoplus_iZ_i\), with injections \(\kappa_i\). There are unique maps \(u:S\to Z\) and \(v:Z\to Y\) such that

The map \(vu\) restricts to \(f_i\) for every \(i\); it is the map corresponding to that family in (5.1). Closure puts \(Z\) in \(\mathsf N\).

For groups \(A_i\) and subgroups \(B_i\), the componentwise quotient map has kernel \(\prod_iB_i\) and is surjective by choosing representatives. Thus

Applying this to (5.1) gives, naturally in \(Y\),

The isomorphism is restriction along \(Q\iota_i\). Every object of the quotient is a \(QY\), so (5.4) is the required coproduct universal property. \(\square\)

The subgroup equality is the reason for the closure hypothesis. If each component factors through some discarded object, one needs a single discarded object through which their assembled map factors. Exercise 4 shows precisely what can go wrong.

6. Four exercises with complete solutions

Exercise 1 (warm-up: torsion that survives). Take \(\mathsf C=\mathsf{Ab}\) and let \(\mathsf N\) be the full subcategory of finitely generated projective abelian groups. Show that \(\mathsf N\) is additive and idempotent complete. For positive integers \(m,n\), calculate

Show that \(Q(\mathbb Z\oplus\mathbb Z/n)\simeq Q(\mathbb Z/n)\), whereas \(Q(\mathbb Z/2)\) and \(Q(\mathbb Z/4)\) are not isomorphic.

Solution. A module \(P\) is projective when maps from \(P\) lift through surjections. If it is also finitely generated, choose a surjection \(\mathbb Z^r\to P\); a lift of \(1_P\) splits this surjection. Hence \(P\) is a retract of a finite free group. Conversely such a retract is finitely generated and projective: projectivity of \(\mathbb Z^r\) follows by lifting its basis, and the inclusion and retraction transfer each lifting problem to it. Finite direct sums have the same properties, including the zero group.

For an idempotent \(e:P\to P\), its image is a retract of \(P\), with inclusion and restriction of \(e\) as splitting maps. It is generated by the images of a finite generating set of \(P\), and is projective by the retract argument. The splitting therefore stays inside \(\mathsf N\).

Every \(P\in\mathsf N\) is torsion-free, since its split inclusion into \(\mathbb Z^r\) is injective. Thus every map \(\mathbb Z/n\to P\) is zero. The ideal in (6.1) is zero and its group is the original group of maps. A map is determined by the image \(x\) of \(1\), with \(nx=0\) in \(\mathbb Z/m\). Put \(d=\gcd(m,n)\), \(m=dm'\), \(n=dn'\). The condition \(m\mid nx\), together with coprimality of \(m',n'\), is equivalent to \(m'\mid x\). The \(d\) multiples of \(m'\) form a cyclic subgroup of order \(d\). Therefore (6.1) is isomorphic to \(\mathbb Z/d\).

The inclusion and projection of the torsion summand in \(\mathbb Z\oplus\mathbb Z/n\) compose one way to its identity; the error in the other composite is the projection onto the \(\mathbb Z\) summand, which factors through \(\mathbb Z\in\mathsf N\). They become inverse under \(Q\).

Finally every group of maps between finite torsion groups is unchanged by this quotient, by the same torsion-free argument. Inverses between the two cyclic groups would thus be actual inverse group maps. The groups have different cardinalities, so there are none. \(\square\)

Exercise 2 (intermediate: a missing splitting hypothesis). Let \(k\) be a field, \(\mathsf C\) the category of finite-dimensional \(k\)-vector spaces, and \(\mathsf N\) its full subcategory of even-dimensional spaces. Prove that all quotient groups of maps are zero. Nevertheless show that \(k\) and the zero space are not stably isomorphic relative to \(\mathsf N\). Identify an idempotent of an object of \(\mathsf N\) that cannot split in \(\mathsf N\).

Solution. The zero space and finite direct sums of even-dimensional spaces belong to \(\mathsf N\), so it is a full additive subcategory. Every space \(V\) is a retract of an even-dimensional space: use \(V\) itself when its dimension is even, and \(V\oplus k\) otherwise. Hence \(1_V\) factors through an object of \(\mathsf N\). Any map out of \(V\) is its composite with \(1_V\), so it too factors through \(\mathsf N\). Every quotient Hom group is zero. Its sole map between any two objects is an isomorphism, since each identity is also that sole map.

A stable isomorphism \(k\oplus Z_1\simeq 0\oplus Z_2\), \(Z_1,Z_2\in\mathsf N\), would equate an odd dimension with an even dimension. This is impossible. The rank-one projection on \(k^2\) exhibits the missing hypothesis. In any splitting \(ab=e\), \(ba=1_W\), the injection \(a\) identifies \(W\) with the image of \(e\), so \(\dim W=1\). Such \(W\) cannot belong to \(\mathsf N\). \(\square\)

Exercise 3 (hard: square-zero operators and a quotient equivalence). Let \(R=k[\varepsilon]/(\varepsilon^2)\). Take \(\mathsf C\) to be the category of finite-dimensional left \(R\)-modules, and \(\mathsf N\) its full subcategory of finite free \(R\)-modules. Prove that \(\mathsf N\) is idempotent complete and that

induces an equivalence \(\mathsf C/\mathcal I\simeq\mathsf{Vect}^{\mathrm{fd}}_k\). Calculate what \(H\) does to \(0\to k\to R\to k\to0\), where the first map sends \(1\) to \(\varepsilon\) and the second takes residue modulo \(\varepsilon\).

Solution. An \(R\)-module here is a finite-dimensional space with an operator \(T\), multiplication by \(\varepsilon\), satisfying \(T^2=0\). Choose a basis \(u_1,\ldots,u_a\) of \(\operatorname{im}T\), choose \(v_i\) with \(Tv_i=u_i\), and extend the \(u_i\) to a basis of \(\ker T\) by \(w_1,\ldots,w_b\). The vectors \(v_i,u_i,w_j\) form a basis of \(M\). Applying \(T\) proves independence of the \(v_i\) modulo the kernel; independence inside the kernel proves the rest. For spanning, subtract the appropriate combination of the \(v_i\) from any vector to put it in the kernel. Each pair \(v_i,u_i\) forms a free rank-one module, and each \(w_j\) forms a copy of the residue module \(k\). Thus

The residue module \(k\) is not projective over \(R\). A section of the surjection \(R\to k\) would send \(1\) to an element annihilated by \(\varepsilon\), hence an element of \(\varepsilon R\); its residue is zero, contradicting the section identity. Finite free modules are projective by lifting basis vectors, and their retracts are projective. If such a retract had a \(k\) summand in (6.3), that summand would be projective too, a contradiction. Therefore every retract of a finite free module is finite free. Taking the image of an idempotent gives such a retract and splits it inside \(\mathsf N\).

An \(R\)-linear map commutes with \(T\), so it sends its kernel and image to the corresponding kernel and image, and gives a well-defined map on (6.2). This makes \(H\) an additive \(k\)-linear functor. On \(R\), kernel and image both equal \(\varepsilon R\), so \(H(R)=0\). On \(k\), \(T=0\), so \(H(k)=k\). Theorem 2.1 therefore gives \(\overline H\) on the quotient.

For maps \(k^b\to k^c\), every factorization through \(R^n\) is zero. The first map lands in the annihilator of \(\varepsilon\), namely \(\varepsilon R^n\); the second map kills that subspace because \(\varepsilon\) acts trivially on \(k^c\). Thus their quotient Hom group is exactly the space of \(k\)-linear maps \(k^b\to k^c\). For arbitrary decompositions (6.3) of source and target, every matrix block incident with a free summand factors through a finite free module. Only the block \(k^b\to k^c\) survives, and \(H\) induces precisely that block map. It follows that \(\overline H\) is fully faithful. It is essentially surjective because every finite-dimensional \(k\)-space, with \(\varepsilon\) acting by zero, has \(H\) equal to that space. This proves the equivalence.

The displayed sequence is short exact as a sequence of \(R\)-modules: the residue map has kernel \(\varepsilon R\), and multiplication of \(1\) by \(\varepsilon\) identifies \(k\) with that kernel. Applying \(H\) gives

Both middle maps are zero. This sequence fails exactness at both nonzero terms. Since \(\overline H\) is an additive equivalence with an abelian category, it transports the abelian structure to the quotient; equivalences preserve kernel and cokernel universal properties. Therefore \(Q\) itself is not exact in this example. \(\square\)

Exercise 4 (advanced: the converse and an infinite failure). Assume \(\mathsf N\) is idempotent complete, and that \(\mathsf C\) has coproducts indexed by a small set \(I\). Prove that \(Q\) preserves all those coproducts if and only if \(\mathsf N\) is closed under them up to isomorphism. Then take \(\mathsf C=\mathsf{Ab}\), with \(\mathsf N\) as in Exercise 1. Show that \(Q\) fails to preserve the countable coproduct of copies of \(\mathbb Z\).

Solution. Closure implies preservation by Theorem 5.1. Conversely, suppose \(Q\) preserves these coproducts, and choose \(Z_i\in\mathsf N\). Each \(QZ_i\) is zero. Their coproduct is zero too: for any target there is exactly one family of maps out of the zero objects, hence exactly one map out of their coproduct. A preadditive initial object is also zero. Preservation implies that \(Q(\bigoplus_iZ_i)\) is zero. By the conclusion after Lemma 3.1, \(\bigoplus_iZ_i\) is isomorphic to an object of \(\mathsf N\). This is the required closure.

For the concrete case, let \(S=\bigoplus_{n\geq0}\mathbb Z\). If \(QS\) were zero, its identity would factor as \(S\to P\to S\) with \(P\) finitely generated projective. The image of a finite generating set of \(P\) would then generate \(S\), since the composite is its identity. But \(S\) is not finitely generated: any finite collection of its vectors is supported on a finite set of coordinates, and cannot generate a basis vector outside that set. Thus \(QS\ne0\).

Every \(Q\mathbb Z\) is zero, so their countable coproduct in the quotient exists and is zero. It cannot be \(QS\). The original countable coproduct is therefore not preserved. This argument does not claim that all countable coproducts fail to exist in the quotient. \(\square\)

References

• The Stacks Project. Preadditive and additive categories, especially the zero-object criterion, biproduct identities, and preservation by additive functors.
• The Mathlib Community. Category quotients by congruences and preadditive quotient categories. These provide the general quotient and additive-congruence constructions.