Splittings, support and sequential exactness

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

A finite direct sum comes with projections that detect every map into it. An infinite coproduct need not behave this way in an arbitrary abelian category. Exact filtered colimits restore this detection: a map is assembled from its restrictions to the inverse images of finite subcoproducts. That same passage from finite pieces to a colimit explains the telescope sequence of a direct system.

We first study splittings and a scalar-image functor, where a single short exact sequence detects exactness. We then use subobjects to test generators and support. Finally, we distinguish an ordinary sequential colimit from a represented formal colimit. The latter supplies compatible sections after one stage and gives a split telescope sequence, even without exact filtered colimits in the original category.

We use additive categories, abelian categories and modules over unital rings. All index sets and diagrams are small in a fixed universe; categories are locally small in that universe, with larger ambient universes for presheaves. No essential-smallness assumption is imposed. The precise prerequisites are Formal linear combinations and finite sums, for biproduct matrices; Generators and small quotient families, §2; Ind-objects through their elements, Theorem 2.1; and Ind-abelian categories and controlled exact sequences, Theorems 3.2 and 6.1. These give, respectively, matrix composition, the generator convention, compact constants and the abelian structure with exact filtered colimits in the ind-category. Formal colimits and compact presentations, Proposition 1.1 supplies realization when ordinary filtered colimits exist.

1. A split complex is a biproduct decomposition

Let \(\mathcal C\) be additive. Consider a complex

Theorem 1.1. The following conditions are equivalent.

1. There are \(k:Y\to X\) and \(h:Z\to Y\) such that \(kf=1_X\), \(gh=1_Z\), and \(fk+hg=1_Y\).
2. There is an isomorphism \(X\oplus Z\simeq Y\) identifying the inclusion of \(X\) with \(f\) and the projection to \(Z\) with \(g\).
3. For every \(W\), applying \(\operatorname{Hom}(W,-)\) to (1.1) gives a short exact sequence of abelian groups.
4. For every \(W\), applying \(\operatorname{Hom}(-,W)\) gives the short exact sequence with the arrows reversed.

Proof. Under condition 1, compose \(fk+hg=1_Y\) on the left with \(k\). Since \(kf=1\), this gives \(khg=0\). Composing on the right with \(h\), and using \(gh=1\), gives \(kh=0\). Hence

These matrices give condition 2. Conversely, the inverse of the isomorphism in condition 2 has components \(k,g\); its restriction to the other summand gives \(h\). The inverse equations give condition 1. Under this decomposition, either Hom functor identifies the sequence with the inclusion and projection of two abelian groups. Such a sequence is short exact. Thus condition 2 implies both 3 and 4.

Assume condition 3. Surjectivity for \(W=Z\) gives \(h\) with \(gh=1_Z\). The endomorphism \(1_Y-hg\) is killed by \(g\). Exactness for \(W=Y\) therefore gives \(k\) with \(fk=1_Y-hg\). Consequently \(f(kf-1_X)=0\), because \(gf=0\). Injectivity for \(W=X\) gives \(kf=1_X\). These are condition 1.

Assume condition 4 instead. Surjectivity for \(W=X\) gives \(k\) with \(kf=1_X\). The endomorphism \(1_Y-fk\) kills \(f\), so exactness for \(W=Y\) gives \(h\) with \(hg=1_Y-fk\). Then \((gh-1_Z)g=0\). Injectivity of precomposition by \(g\), for \(W=Z\), gives \(gh=1_Z\). Again condition 1 follows. \(\square\)

We call such a complex split. This definition requires no kernels or cokernels. In an abelian category it agrees with the usual split short exact sequence. A biproduct row is short exact, so the theorem's condition implies the usual one. Conversely, suppose (1.1) is short exact and \(g\) has a section \(h\). Since \(f\) is the kernel of \(g\), the map \(1_Y-hg\) factors uniquely as \(fk\). Monicity of \(f\) gives \(kf=1_X\), as in the proof. If instead \(f\) has a retraction \(k\), use the cokernel property of \(g\) to factor \(1_Y-fk\) as \(hg\); epicity gives \(gh=1_Z\). Thus either usual splitting yields the biproduct decomposition.

Having a retraction of \(f\) and a section of \(g\) separately does not ensure middle exactness. Exercise 2 identifies the extra summand that can remain.

2. When multiplication images preserve exactness

Let \(A\) be a commutative unital ring and \(a\in A\). On all \(A\)-modules define

for \(u:M\to N\). Linearity gives \(u(am)=au(m)\), so the restriction is defined. Restrictions preserve identities, compositions and sums of maps. Thus \(F_a\) is additive. It always preserves monomorphisms and epimorphisms: restricting an injective map remains injective, while if \(u\) is surjective, every \(an\in aN\) is \(u(am)\) for a lift \(m\) of \(n\).

Theorem 2.1. The following conditions are equivalent: \(F_a\) is left exact; \(F_a\) is right exact; \(a\in Aa^2\); and \(Aa=Ac\) for an idempotent \(c\in A\). Under these conditions \(F_a\) is exact.

Proof. Apply \(F_a\) to the short exact sequence

The three resulting terms are \(a^2A,aA,0\), and the first map is the inclusion. Left exactness forces its image to be the kernel of \(aA\to0\), namely all of \(aA\). Right exactness forces the same inclusion to be surjective. Either condition therefore gives \(aA=a^2A\), equivalently \(a=ba^2\) for some \(b\in A\).

Set \(c=ba\). Commutativity gives \(c^2=b^2a^2=ba=c\). Also \(c\in Aa\) and \(a=ca\in Ac\), so \(Aa=Ac\). Conversely, if these ideals agree, multiplying their elements into an arbitrary module gives \(aM=cM\), naturally in \(M\). The identity

follows from \(m=cm+(1-c)m\) and from \(c(1-c)=0\): applying \(c\) to an element in the intersection makes it both itself and zero. Every module map respects this decomposition.

To check exactness explicitly, consider \(0\to L\xrightarrow{u}M\xrightarrow{v}N\to0\). The restricted first map is injective. If \(x\in cM\) satisfies \(v(x)=0\), write \(x=u(\ell)\). Then \(x=cx=u(c\ell)\), so the restricted kernel is the restricted image. For \(cn\in cN\), choose \(m\) with \(v(m)=n\); then \(v(cm)=cn\). Thus the restricted last map is surjective. This proves exactness, hence both one-sided exactness conditions. \(\square\)

Preserving injective and surjective maps alone is weaker than preserving short exact sequences. For instance, multiplication by \(2\) over \(\mathbb Z\) preserves both types of maps but \(2\notin4\mathbb Z\). Sequence (2.2) locates the failure at the middle term.

3. Subobject tests for generators

Let \((G_i)_{i\in I}\) be a small family in an abelian category. Call it separating when maps \(r,s:X\to Y\) are equal whenever \(ru=su\) for every \(u:G_i\to X\). By the cited generator lesson, separation and detection of isomorphisms by all \(\operatorname{Hom}(G_i,-)\) are equivalent here: abelian categories have equalizers and are balanced. We use either equivalent condition for a generating family.

Theorem 3.1. This family generates if and only if every monomorphism \(m:Z\to X\) for which all maps \(\operatorname{Hom}(G_i,Z)\to\operatorname{Hom}(G_i,X)\) are surjective is an isomorphism. These conditions imply

Proof. Suppose the family separates, and let \(q:X\to Q\) be the cokernel of such an \(m\). Every map \(u:G_i\to X\) lifts through \(m\), so \(qu=0\). Separation gives \(q=0\). Since \(q\) is epic, \(1_Qq=0q\) implies \(1_Q=0\); hence \(Q=0\). Thus \(m\) is epic as well as monic, and is invertible.

Conversely, suppose the asserted subobject test holds. If \(r,s:X\to Y\) agree on all the probes, each \(u:G_i\to X\) factors uniquely through the equalizer \(e:E\to X\). The maps \(\operatorname{Hom}(G_i,e)\) are therefore surjective. The test makes \(e\) an isomorphism, so \(r=s\). Finally, apply the test to \(0\to X\) under the hypothesis of (3.1). \(\square\)

Zero-object detection can hold without generation. Let \(k\) be a field, \(R=k[\varepsilon]/(\varepsilon^2)\), and \(G=R/(\varepsilon)\). Evaluation at \(1\) identifies \(\operatorname{Hom}_R(G,M)\) with \(\ker(\varepsilon:M\to M)\). For every nonzero \(m\in M\), either \(\varepsilon m=0\), or \(\varepsilon m\) is a nonzero element of this kernel. Thus \(G\) detects zero objects among all modules. But multiplication by \(\varepsilon\) on \(R\) is nonzero and vanishes on \(\ker\varepsilon=\varepsilon R\). It cannot be distinguished from zero by maps from \(G\). This is the same weak-detection distinction developed for traces in Trace coreflections and balanced nonabelian categories; no finiteness restriction on modules repairs the distinction in this example.

4. Exact filtered colimits detect support

Assume \(\mathcal C\) is abelian, has all small colimits, and its small filtered colimits are exact. This is the AB5 condition, in the convention of Stacks, Definition 19.10.1. No generator is assumed.

For a small family \((Y_i)_{i\in I}\), write \(S=\bigoplus_{i\in I}Y_i\), with inclusions \(j_i\) and coordinate maps \(\pi_i:S\to Y_i\). The latter are defined by identity on the indicated summand and zero on all others. For a finite subset \(J\subseteq I\), let \(D_J=\bigoplus_{i\in J}Y_i\), with its split inclusion \(d_J:D_J\to S\). Compatible maps from all \(D_J\) are exactly a map from each summand; consequently

Lemma 4.1. For any \(f:X\to S\), the subobjects \(X_J=X\times_S D_J\) form a directed union with colimit \(X\).

Proof. The pullback is the kernel of

These kernel diagrams are compatible as \(J\) increases. An exact filtered-colimit functor preserves their kernels: apply exactness to the short exact sequences defining kernels and images, or to the resulting exact complexes. The colimit of the middle terms is \(X\oplus S\), since the constant diagram on \(X\) has colimit \(X\), and (4.1) applies to the other summand. The colimit of the last terms is \(S\). Thus \(\operatorname{colim}_J X_J\) is the kernel of \((f,-1):X\oplus S\to S\). This kernel is the graph \((1,f):X\to X\oplus S\): a pair of maps \((a,b)\) is killed exactly when \(b=fa\). Projection to \(X\) therefore gives the asserted isomorphism, with the original subobject maps. \(\square\)

Theorem 4.2. If \(K\subseteq I\) and \(\pi_i f=0\) for every \(i\notin K\), then \(f\) factors uniquely through the subcoproduct \(S_K=\bigoplus_{i\in K}Y_i\). In particular, if every coordinate of \(f\) is zero, then \(f=0\).

Proof. Let \(a_J:X_J\to X\) and \(b_J:X_J\to D_J\) be the pullback maps. For \(i\in J\setminus K\), the \(i\)-th coordinate of \(b_J\) is \(\pi_i f a_J=0\). Since \(D_J\) is a finite biproduct, its projections determine a map into it. Hence \(b_J\) factors through \(D_{J\cap K}\), and \(fa_J\) factors through \(S_K\). Denote the resulting map by \(g_J:X_J\to S_K\).

The inclusion \(S_K\to S\) has a retraction defined by identity on the summands in \(K\) and zero elsewhere; it is monic. Therefore the \(g_J\) are compatible: their composites into \(S\) are the compatible maps \(fa_J\), and monicity cancels the inclusion. Lemma 4.1 gives a unique induced \(g:X\to S_K\). Its composite into \(S\) equals \(f\), since both agree on every \(X_J\). Monicity also gives uniqueness. Set \(K=\varnothing\) to obtain the last assertion. \(\square\)

Corollary 4.3. If the product \(P=\prod_{i\in I}Y_i\) exists as well, the canonical map \(S\to P\) is a monomorphism.

Proof. Let \(n:N\to S\) be its kernel. Each \(\pi_i n\) is zero, so Theorem 4.2 gives \(n=0\). A zero monomorphism has zero source: \(n1_N=n0\) implies \(1_N=0\). Thus the comparison has zero kernel and is monic. \(\square\)

The theorem does not say that every map into \(S\) has finite support. The identity of an infinite sum of nonzero modules has infinitely many nonzero coordinates. Lemma 4.1 assembles its source from subobjects rather than putting the entire source inside one finite stage. Exercise 3 specifies a hypothesis that does give one finite stage.

5. A sequential colimit and its telescope

Let \(X_0\to X_1\to\cdots\) be a direct system in an abelian category. Write \(u_{n,m}:X_n\to X_m\) for the composite transition when \(n\leq m\), including \(u_{n,n}=1\). Suppose the countable coproduct \(S=\bigoplus_{n\geq0}X_n\) exists. Define its shift by

Proposition 5.1. The cokernel of \(D\), with the maps induced by the \(j_n\), is the ordinary colimit of this system. In particular, countable coproducts suffice for existence of these colimits in an abelian category.

Proof. A map \(a:S\to Y\) satisfies \(aD=0\) exactly when its components satisfy \(aj_n=aj_{n+1}u_{n,n+1}\). These are precisely the compatibility equations for a cocone to \(Y\). The cokernel universal property gives a unique factorization for each such cocone. This identification is natural in \(Y\), and supplies the colimit universal property. \(\square\)

If these sequential colimits are exact, the first map is monic as well. We use the complete open proof of Stacks, Lemma 13.33.6, Tag 093W, whose precise hypotheses are an abelian category with existing exact colimits indexed by \(\mathbb N\). It gives the short exact row

where \(pj_n\) is the colimit cocone. Countable coproducts are obtained from finite prefixes under those hypotheses. Proposition 5.1 shows that the more explicit assumption of countable coproducts together with exact sequential colimits fits the same result.

Here is the finite matrix in that proof, including its signs. Put \(S_m=\bigoplus_{0\leq n\leq m}X_n\) and let \(r_m:S_m\to S_{m+1}\) have column \(j_n-j_{n+1}u_{n,n+1}\) at \(X_n\). Then \([r_m,j_{m+1}]:S_m\oplus X_{m+1}\to S_{m+1}\) is invertible. For a tuple of morphisms \(y_k:W\to X_k\), its inverse has components

Indeed, the first coordinate of the forward matrix is \(z_0\); each intermediate coordinate is \(z_n-u_{n-1,n}z_{n-1}\); the last is \(t-u_{m,m+1}z_m\). Substitution in (5.3) gives each \(y_n\). Conversely, the same recurrence recovers each \(z_n\) and \(t\). Since the identities hold for maps from every \(W\), they are identities of the matrices themselves. The resulting finite row has last map with components \(u_{k,m+1}\); its kernel is \(r_m\), and its section is \(j_{m+1}\).

For any abelian \(\mathcal C\), the corresponding formal row is short exact in \(\operatorname{Ind}(\mathcal C)\):

The ind-abelian prerequisite supplies this category's small colimits and exact filtered colimits, so the exact hypotheses of Tag 093W hold. Constants preserve finite biproducts; consequently \(\widehat S\) is the formal filtered colimit of their finite sums. The shift in (5.4) has the same specified columns as (5.1). A countable coproduct need not exist in \(\mathcal C\) for (5.4) to exist.

6. A represented formal colimit makes the telescope split

Now assume only that \(\mathcal C\) is abelian with countable coproducts. We do not assume exact sequential colimits in \(\mathcal C\).

Theorem 6.1. If \(\widehat L=\operatorname{colim}_n\iota X_n\) is represented by an object \(L\in\mathcal C\), then its representing cocone \(p_n:X_n\to L\) makes (5.2) a split short exact sequence in \(\mathcal C\).

Proof. Choose the representing isomorphism so that the formal cocone has components \(\iota p_n\). Full faithfulness gives the actual compatibility \(p_mu_{n,m}=p_n\). For any \(Y\in\mathcal C\), maps \(\iota L\to\iota Y\) are exactly compatible maps \(\iota X_n\to\iota Y\). Full faithfulness therefore says that this same cocone represents the ordinary colimit \(L\).

Compactness of the constant \(\iota L\) gives an index \(n_0\) and a map \(j:L\to X_{n_0}\) whose composite into the formal colimit is the identity. Thus \(p_{n_0}j=1_L\). For \(n\geq n_0\), set

The \(s_n\) are compatible sections of the \(p_n\). The split decomposition gives \(X_n\simeq L\oplus N_n\), with projection \(p_n\); under it every transition is \(1_L\oplus v_{n,m}\). The tail is cofinal in the sequence. Taking its formal colimit gives \(\iota L\oplus\operatorname{colim}_n\iota N_n\), and the map to the represented \(\iota L\) is projection onto the first factor. Because that map is an isomorphism,

The compact-constant Hom formula applied to \(\iota N_n\) says that its structural map to this colimit is represented by \(1_{N_n}\). Equality with zero in a filtered colimit of Hom groups means equality at a later stage. Hence, for each \(n\geq n_0\), there is \(m_n>n\) with \(v_{n,m_n}=0\). All subsequent transitions from \(N_n\) are then zero too. If \(N_n=0\), any later index works.

Let \(T=\bigoplus_{n\geq n_0}N_n\), with inclusions \(b_n\), and let \(D_N=1-\operatorname{sh}_N\). Define a map \(Q:T\to T\) by its columns

Each sum is finite. Increasing \(m_n\) changes nothing, since the added transitions vanish. To compute \(QD_Nb_n\), increase the bounds for columns \(n\) and \(n+1\) to a common bound; all terms except \(b_n\) cancel. To compute \(D_NQb_n\), consecutive terms cancel, leaving \(b_n-b_{m_n}v_{n,m_n}=b_n\). Thus

This uses finite sums separately on each summand and the coproduct universal property, not an infinite sum in a Hom group.

For the constant \(L\)-part put \(B=\bigoplus_{n\geq n_0}L\), with inclusions \(a_n\). Let \(D_La_n=a_n-a_{n+1}\), let \(p_Ba_n=1_L\), and put \(s=a_{n_0}\). Define

where the sum is empty at \(n=n_0\). Direct finite cancellation gives

Theorem 1.1 therefore splits \(0\to B\xrightarrow{D_L}B\xrightarrow{p_B}L\to0\). Countable coproducts commute with the finite direct-sum decompositions by their universal property. On the tail \(\bigoplus_{n\geq n_0}X_n\simeq B\oplus T\), the telescope map is \(D_L\oplus D_N\) and its cocone is \([p_B,0]\). Equations (6.4) and (6.6) split this row: its retraction on the first arrow is \(H\oplus Q\), and the section of the last arrow is \((s,0)\).

It remains to restore the finitely many omitted terms. If \(n_0=0\), there are none. Otherwise write the full coproduct as \(A\oplus S'\), where \(A=\bigoplus_{n<n_0}X_n\) and \(S'\) is the tail. The full telescope map has matrix

where \(V\) is the tail telescope map and \(t\) is zero on all prefix summands except the last, whose transition enters the first tail summand. The finite shift is nilpotent of exponent \(n_0\), so \(U^{-1}=\sum_{r=0}^{n_0-1}\operatorname{sh}_{\mathrm{prefix}}^r\). The codomain automorphism

Write the original cocone on this sum as \([p_A,p']\). Compatibility gives \([p_A,p']D=0\), whence \(p_AU=p't\). Therefore \([p_A,p']E^{-1}=[0,p']\). With the codomain changed by \(E\), the full row is the direct sum of the isomorphism row \(0\to A\xrightarrow{U}A\to0\) and the split tail row. It is split exact. Transporting the splitting back by \(E\) proves the assertion for the original \(D\) and the original cocone. \(\square\)

The representability hypothesis controls transition maps, not just the value of an ordinary colimit. Exercise 4 gives an ordinary colimit for which the telescope row is exact but does not split; the formal colimit is consequently not represented.

7. Four graded exercises with full solutions

Exercise 1 (easy: two scalar images). Let \(A=\mathbb Z/12\mathbb Z\). Decide whether \(F_4\) and \(F_6\) are left exact, right exact and exact on all \(A\)-modules. For each failure, exhibit the failed image of a short exact sequence.

Solution. In \(A\), \(4^2=4\), so \(4\) is idempotent. Theorem 2.1 gives all three exactness properties for \(F_4\), with the natural decomposition \(M=4M\oplus9M\). Indeed \(1-4=9\), \(4+9=1\), \(4\cdot9=0\), and both scalars are idempotent modulo \(12\).

For \(6\), the ideal \(6A=\{0,6\}\) is nonzero whereas \(6^2A=0\). Applying \(F_6\) to \(0\to6A\to A\to A/6A\to0\) gives

Here multiplication by \(6\) kills both \(6A\) and \(A/6A\). The kernel of the map from \(6A\) to zero is \(6A\), but the preceding image is zero. Thus the required middle exactness fails for both left and right exactness. In particular \(F_6\) is not exact. It still preserves all monomorphisms and epimorphisms, by (2.1).

Exercise 2 (moderate: the missing middle summand). For modules over any unital ring, suppose \(gf=0\), \(kf=1_X\) and \(gh=1_Z\). Put \(h'=h-fkh\) and \(E=\ker k\cap\ker g\subseteq Y\). Prove \(Y\simeq X\oplus Z\oplus E\) compatibly with \(f\) and \(g\), and determine when the original complex splits. Give a concrete example with \(E\ne0\).

Solution. The hypotheses give \(kh'=kh-kfkh=0\) and \(gh'=gh-gfkh=1_Z\). For \(y\in Y\), set

Applying \(k\) gives zero, since \(kf=1\) and \(kh'=0\); applying \(g\) gives zero, since \(gf=0\) and \(gh'=1\). Thus \(e(y)\in E\). All maps are linear. Define

Their composite on \(Y\) is the identity by (7.2). On \(X\oplus Z\oplus E\), the first two coordinates recover \(x,z\); also \(e(fx)=0\), \(e(h'z)=0\), and \(e(e)=e\). Hence the other composite is the identity as well. The maps \(f,g\) become inclusion of \(X\) and projection to \(Z\). Their middle homology is precisely \(E\). If \(E=0\), this is the canonical biproduct row and splits. If the complex splits, it is exact by Theorem 1.1, so \(E=0\) by the displayed decomposition.

For example, over a field take \(X=Z=k\), \(Y=k^3\), \(f(x)=(x,0,0)\), \(g(x,z,e)=z\), \(k(x,z,e)=x\), and \(h(z)=(0,z,0)\). All three given identities hold, but \(E=\{(0,0,e):e\in k\}\ne0\). The complex is not exact at \(Y\), so cannot split.

Exercise 3 (hard: finite generation and coproducts). In an AB5 abelian category with small colimits, call \(X\) finitely generated if whenever it is the union of a small directed family of subobjects, one member is all of \(X\). Prove that every map from such an \(X\) to a coproduct factors through a finite subcoproduct. Then prove that \(\bigoplus_{i\in I}Y_i\) is finitely generated exactly when only finitely many \(Y_i\) are nonzero and each of these is finitely generated.

Solution. Lemma 4.1 writes the source of any \(f:X\to\bigoplus_iY_i\) as the directed union of \(X_J\). Finite generation gives \(X_J=X\) for one finite \(J\), and its pullback map supplies the factorization through \(D_J\).

We need two consequences of AB5 about directed unions. If \(M=\operatorname{colim}_\lambda M_\lambda\) is a directed union and \(a:T\to M\) is any map, then

Use the kernels of \((a,-\mathrm{incl}):T\oplus M_\lambda\to M\), exactly as in Lemma 4.1. Also a quotient \(q: X\twoheadrightarrow Q\) of a finitely generated object is finitely generated. Indeed, for a directed union \(Q=\bigcup Q_\lambda\), (7.4) says its inverse images cover \(X\). One inverse image must be \(X\). Then \(q\) factors through \(Q_\lambda\to Q\); since \(q\) is epic, this inclusion is epic as well as monic, and hence is an isomorphism.

Suppose \(S=\bigoplus_iY_i\) is finitely generated. Its finite subcoproducts cover it by (4.1), so \(D_J\to S\) is an isomorphism for some finite \(J\). For \(i\notin J\), \(\pi_i\) is zero on \(D_J\), hence zero on \(S\). But \(\pi_i j_i=1_{Y_i}\), so \(Y_i=0\). Each \(Y_i\) is a quotient of \(S\) by its split projection, so each is finitely generated by the quotient argument.

Conversely, let \(J\) be finite and suppose its summands are finitely generated. For a directed union \(S=\bigcup M_\lambda\), pull back along each inclusion \(Y_i\to S\). By (7.4) the inverse images cover \(Y_i\), so some \(M_{\lambda_i}\) contains that summand. A common upper index for the finitely many \(\lambda_i\) gives a member containing all summands. Every inclusion \(Y_i\to S\) factors through it; the finite coproduct property then factors \(1_S\) through it. Its inclusion is consequently a split epimorphism and a monomorphism, so it equals \(S\). The empty sum is zero and is finitely generated by the same definition. This proves the equivalence.

Exercise 4 (advanced: an exact telescope that cannot split). Consider \(\mathbb Z\xrightarrow{2}\mathbb Z\xrightarrow{2}\cdots\). Identify its ordinary colimit. Prove its ordinary telescope sequence is short exact but does not split. Prove that its formal colimit in \(\operatorname{Ind}(\mathsf{Ab})\) is not represented by an abelian group.

Solution. The compatible cocone \(p_n(z)=z/2^n\) takes values in \(L=\mathbb Z[1/2]\). Each element belongs to some \(2^{-n}\mathbb Z\), and a compatible family of maps from those groups gives a well-defined homomorphism on their union. This proves the colimit universal property directly.

On \(S=\bigoplus_{n\geq0}\mathbb Z e_n\), the telescope map has columns \(De_n=e_n-2e_{n+1}\), and \(p(e_n)=2^{-n}\). If \(Dx=0\), its zeroth coordinate gives \(x_0=0\); successive coordinates give \(x_n-2x_{n-1}=0\), hence all \(x_n=0\). Thus \(D\) is injective. The map \(p\) is surjective because every element of \(L\) is \(p(ze_n)\).

For middle exactness, let \(y\in S\) have support in \(0,\ldots,m\) and satisfy \(p(y)=0\). If \(m=0\), then \(y=0\). Otherwise set

The relation \(\sum_{k=0}^{m}2^{m-k}y_k=0\) gives \(-2z_{m-1}=y_m\). The remaining coordinate recurrences give \(Dz=y\). Conversely \(pD=0\) by the compatibility of the cocone. This proves short exactness without invoking a general module-colimit theorem.

Every homomorphism \(b:L\to S\) is zero. For \(x\in L\), and every \(r\geq0\), there is \(x/2^r\in L\), so \(b(x)\) belongs to \(2^rS\). An integer divisible by every \(2^r\) is zero. Applying this fact to each coordinate shows \(\bigcap_r2^rS=0\), hence \(b(x)=0\). Since \(L\ne0\), no section of \(p\) exists and the sequence does not split.

If the formal colimit were represented by \(Q\), the proof of Theorem 6.1 identifies its representing cocone with an ordinary colimit, so \(Q\simeq L\). Compactness of \(\iota Q\) would give a finite-stage map \(Q\to\mathbb Z\) whose composite with \(\mathbb Z\to Q\) is \(1_Q\). But the same divisibility argument gives \(\operatorname{Hom}(L,\mathbb Z)=0\), so such a section is impossible. Thus the formal colimit is not represented. Ordinary colimit existence and formal representability are different conditions, even for this elementary sequence.

8. References and scope

• The Stacks Project, Lemma 13.33.6, Tag 093W, supplies the complete proof of (5.2) under exact sequential-colimit hypotheses, and hence of (5.4) through the exact ind-abelian interface. Its proof remains at this exact open source; its text is distributed under GNU Free Documentation License 1.2. That license remains attached to the referenced proof. Formula (5.3) records the finite matrix and its inverse needed to identify the signs and the cocone.
• The Stacks Project, §19.10, Tag 079A, gives the AB conditions and their exactness conventions. §12.5, Tag 00ZX, Definition 12.5.9 and Lemma 12.5.10, gives the usual abelian split-sequence convention. The general additive Hom criterion is proved in §1 above.
• Pierre Schapira, An Introduction to Categories and Homological Algebra, lecture notes, version of 1 March 2026, Sections 4.1, 5.1 and 5.4, for additive and abelian categories, split exact sequences and generators.
• The exact programme prerequisites linked in the introduction retain the proofs of formal filtered Hom, compact constants, realization and the ind-abelian structure. Formal colimits and test objects, Section 2 supplies the presheaf Hom and composition calculation. The sequential representative proof above concerns its actual transition maps; no derived inverse-limit assertion is needed.