Full subcategories and exact closure

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).

A full subcategory keeps every map between its objects. It need not keep the objects that express the kernel or cokernel of those maps. That distinction controls whether an exact sequence computed in the subcategory remains exact in the ambient category. Extension closure asks a further question: does a middle object belong when its two ends do?

We first identify the ambient operations that a subcategory must retain. A five-term sequence then packages three closure conditions into one test. We prove a useful criterion based on covering quotients from within the subcategory, and construct ordinary exact sequences from generating or cogenerating objects. The exercises show that several plausible closure conditions are genuinely different.

We assume additive and abelian categories, kernels, cokernels and short exact sequences. Use Formal linear combinations and finite sums, Section 1, for biproduct identities and the additivity of a fully faithful functor. The ordinary abelian framework is described in Trace coreflections and balanced nonabelian categories. Basic open references are [Stacks, Abelian categories], [Stacks, Subcategories] and [Mathlib, Abelian subcategories], listed below. All categories are locally small in a fixed universe; their object collections may be larger. No small colimit hypothesis is imposed.

1. Which objects belong to the subcategory?

Fix an abelian category \(\mathcal A\) and a full subcategory \(\mathcal J\). Write \(\mathcal J^{\simeq}\) for the full subcategory of objects isomorphic to an object of \(\mathcal J\). Thus membership in \(\mathcal J^{\simeq}\) does not depend on a chosen representative. Such a subcategory is called replete.

All the following closure conditions use ambient constructions in \(\mathcal A\):

Replacing the ends of a morphism or short exact sequence by isomorphic objects transports its kernel, cokernel and exactness. Hence these conditions hold for \(\mathcal J\) exactly when they hold for \(\mathcal J^{\simeq}\). We may pass to the latter when proving them.

In this lesson, thick means closed under ambient kernels, cokernels and extensions. This definition makes sense for any full subcategory. The structural theorems below impose their stated additive or nonempty hypotheses. A nonempty replete full subcategory closed under subobjects, quotients and extensions is a Serre subcategory. The five-term criterion in Section 3 identifies nonempty thickness with the weak Serre convention of [Stacks, Subcategories].

The difference concerns which maps may have their ends outside the subcategory. A kernel of a map between two objects of \(\mathcal J\) is a special ambient subobject. Subobject closure tests all ambient subobjects, including those whose quotient is outside \(\mathcal J\). Exercise 3 exhibits the distinction.

Proposition 1.1. A full subcategory \(\mathcal J\) is additive if and only if an ambient zero object and every ambient binary biproduct of its objects are isomorphic to objects of \(\mathcal J\). In that case the inclusion is additive and preserves finite biproducts.

Proof. Suppose first that \(\mathcal J\) is additive, even if its addition was initially specified without reference to \(\mathcal A\). Its inclusion is fully faithful between additive categories, so is additive by the criterion in the prerequisite lesson. If \(Z\) is zero in \(\mathcal J\), then \(1_Z=0\). The same equation holds in \(\mathcal A\), and an object whose identity is zero is a zero object: every map into or out of it equals zero.

For a biproduct \(S=X\oplus Y\) in \(\mathcal J\), its inclusions and projections satisfy

\[ \begin{gathered} p_ri_s=\delta_{rs},\\ i_Xp_X+i_Yp_Y=1_S. \end{gathered} \tag{1.1} \]

Here \(\delta_{rs}\) denotes the identity when \(r=s\) and the zero morphism otherwise. The additive inclusion preserves these equations. They give the ambient product and coproduct universal properties: a map with coordinates \(a,b\) is \(i_Xa+i_Yb\), and a map with specified restrictions \(c,d\) is \(cp_X+dp_Y\). Thus \(S\) is an ambient biproduct.

Conversely, restrict the ambient abelian Hom groups and bilinear composition to the full subcategory. Transport the ambient zero and biproduct structure through the specified isomorphisms to objects of \(\mathcal J\). Fullness gives their universal properties inside \(\mathcal J\). This makes \(\mathcal J\) additive, with additive inclusion. Empty and iterated binary biproducts give all finite ones. \(\square\)

This statement requires suitable representatives, rather than literal inclusion of every chosen model of zero or a biproduct. Working with \(\mathcal J^{\simeq}\) makes that qualification automatic.

2. When the inclusion is exact

A full subcategory is fully abelian if it is abelian and its inclusion into \(\mathcal A\) is exact. Here exactness of an additive functor means that it sends every short exact sequence to a short exact sequence.

We use the following precise criterion from [Stacks, Additive functors].

Lemma 2.1. An additive functor between abelian categories is exact if and only if it preserves kernels and cokernels of all morphisms.

Proof. If it preserves both constructions, it sends a pair consisting of a kernel and its cokernel to such a pair. Thus it preserves every short exact sequence.

Conversely, let \(F\) be exact and factor a morphism \(f:A\to B\) through its image \(I\). The two short exact sequences are

\[ \begin{gathered} 0\to K\to A\to I\to0,\\ 0\to I\to B\to Q\to0, \end{gathered} \tag{2.1} \]

where \(K=\ker f\) and \(Q=\operatorname{coker}f\). Their images are short exact. In particular \(F(A)\to F(I)\) is epic and \(F(I)\to F(B)\) is monic. The composite is \(Ff\). Its kernel is the kernel of the first map, namely \(FK\); its cokernel is the cokernel of the second map, namely \(FQ\). The displayed structure maps are the required universal maps. \(\square\)

Theorem 2.2. For a full additive subcategory \(\mathcal J\) of \(\mathcal A\), these are equivalent:

  1. \(\mathcal J\) is fully abelian.
  2. \(\mathcal J\) is closed under ambient kernels and cokernels.

The sufficiency direction is also the full-subcategory construction in [Mathlib, Abelian subcategories].

Proof. Suppose (2). Choose, inside \(\mathcal J\), representatives of the ambient kernels and cokernels of all its maps. Their ambient universal properties restrict to \(\mathcal J\) because it is full. Thus they are kernels and cokernels inside \(\mathcal J\), and inclusion preserves them.

For any \(f:X\to Y\) in \(\mathcal J\), its coimage and image are obtained from these same kernels and cokernels:

\[ \begin{gathered} \operatorname{coim}f =\operatorname{coker}(\ker f\to X),\\ \operatorname{im}f =\ker(Y\to\operatorname{coker}f). \end{gathered} \tag{2.2} \]

They lie in \(\mathcal J\) after choosing representatives. Their canonical comparison is the ambient comparison, which is invertible in the abelian category \(\mathcal A\). Its inverse lies in \(\mathcal J\) by fullness. Therefore \(\mathcal J\) is abelian. Inclusion is additive by Proposition 1.1 and preserves kernels and cokernels, so Lemma 2.1 makes it exact.

If (1) holds, the exact inclusion preserves the internal kernels and cokernels by Lemma 2.1. These are therefore representatives of the ambient constructions, proving (2). \(\square\)

Having internal kernels and cokernels is not enough. They must agree with the ambient ones up to their universal comparisons. Exercise 2 gives a full additive category where both internal constructions exist but the category is not abelian.

3. A test with five terms

The following criterion is the weak Serre characterization of [Stacks, Subcategories]. We keep its distinction from Serre closure and prove the exact interface needed here.

Theorem 3.1. A full additive subcategory \(\mathcal J\) is thick if and only if every ambient sequence

\[ X_0\xrightarrow{f_0}X_1 \xrightarrow{f_1}X_2 \xrightarrow{f_2}X_3 \xrightarrow{f_3}X_4 \tag{3.1} \]

exact at \(X_1,X_2,X_3\), with \(X_0,X_1,X_3,X_4\) in \(\mathcal J\), has \(X_2\in\mathcal J^{\simeq}\).

Proof. Pass to the replete closure. Suppose \(\mathcal J\) is thick. Put

\[ K=\operatorname{coker}f_0, \qquad L=\ker f_3. \tag{3.2} \]

Both objects belong to \(\mathcal J\). Exactness at \(X_1\) identifies the quotient \(K\) with \(\operatorname{coim}f_1\), hence with \(\operatorname{im}f_1\). Its induced map into \(X_2\) is monic. Exactness at \(X_3\) identifies \(L\) with \(\operatorname{im}f_2\), so the induced map \(X_2\to L\) is epic. Exactness at \(X_2\) identifies the first image with the kernel of the second map. We obtain

\[ 0\to K\to X_2\to L\to0. \tag{3.3} \]

Extension closure gives \(X_2\in\mathcal J\).

Conversely, assume the five-term test. For a map \(f:A\to B\) in \(\mathcal J\), apply it to

\[ \begin{gathered} 0\to0\to\ker f\to A\xrightarrow{f}B,\\ A\xrightarrow{f}B\to\operatorname{coker}f\to0\to0. \end{gathered} \tag{3.4} \]

These sequences are exact at their three interior positions. The additive subcategory contains a zero object, so the test gives both kernel and cokernel closure. Applied to \(0\to A\to E\to B\to0\) with \(A,B\in\mathcal J\), it gives extension closure. Thus \(\mathcal J\) is thick. \(\square\)

Corollary 3.2. A nonempty thick full subcategory is additive and fully abelian. After passing to its replete closure, it is a weak Serre subcategory. Every Serre subcategory satisfies these conclusions.

Proof. Choose \(X\) in the subcategory. The kernel of \(1_X\) is zero, so zero belongs up to isomorphism. For two of its objects \(A,B\), the split exact sequence

\[ 0\to A\to A\oplus B\to B\to0 \tag{3.5} \]

puts their biproduct in the subcategory up to isomorphism. Proposition 1.1 proves additivity, Theorem 2.2 proves full abelianness, and Theorem 3.1 gives the canonical weak Serre test.

For a Serre subcategory, an ambient kernel is a subobject of the source and an ambient cokernel is a quotient of the target. Its defining closures therefore imply thickness, and it is nonempty by definition. \(\square\)

The nonempty hypothesis matters in this corollary: the empty full subcategory satisfies the three closure conditions vacuously. Exercise 3 shows that a nonempty thick subcategory need not be Serre.

4. Covering quotients from within

To prove extension closure, one can sometimes construct a presentation whose two defining objects already lie in the subcategory.

Theorem 4.1. Let \(\mathcal J\) be a fully abelian subcategory of \(\mathcal A\). Suppose that whenever \(q:E\to B\) is an ambient epimorphism with \(B\in\mathcal J\), there are \(T\in\mathcal J\) and \(f:T\to E\) such that \(qf\) is epic. Then \(\mathcal J\) is thick.

The map \(f\) need not be epic, and the object \(T\) may depend on \(q\). The premise asks for an epimorphic composite onto \(B\).

Proof. We may replace \(\mathcal J\) by its replete closure; the premise transports through isomorphisms. Fully abelian closure already gives ambient kernels and cokernels by Theorem 2.2. It remains to prove extension closure.

Take a short exact sequence

\[ 0\to A\xrightarrow{j}E\xrightarrow{q}B\to0 \tag{4.1} \]

with \(A,B\in\mathcal J\). Choose \(T,f\) as in the premise and put \(v=qf\). Its ambient kernel \(k:K\to T\) belongs to \(\mathcal J\). Since \(qfk=0\) and \(j\) is the kernel of \(q\), there is a unique \(b:K\to A\) with

\[ jb=fk. \tag{4.2} \]

Form the two morphisms

\[ \begin{gathered} d=\langle k,-b\rangle:K\to T\oplus A,\\ h=(f,j):T\oplus A\to E. \end{gathered} \tag{4.3} \]

Their composite is \(fk-jb=0\). We claim that \(d\) is the kernel of \(h\).

Let \(z:Z\to T\oplus A\) have components \(t:Z\to T\) and \(a:Z\to A\), and suppose \(hz=0\). Applying \(q\) gives \(vt=0\). Thus \(t=k\ell\) for a unique \(\ell:Z\to K\). Substitution and (4.2) give

\[ 0=ft+ja=j(b\ell+a). \tag{4.4} \]

Since \(j\) is monic, \(a=-b\ell\). Hence \(z=d\ell\), uniquely. This is the kernel universal property in the arbitrary abelian category.

Next \(h\) is epic. If \(r:E\to W\) satisfies \(rh=0\), then \(rj=0\). Because \(q\) is the cokernel of \(j\), write \(r=sq\). Also

\[ 0=rf=sqf=sv. \tag{4.5} \]

The epimorphism \(v\) forces \(s=0\), so \(r=0\). Applying this zero-annihilator argument to the difference of any two maps proves right cancellation for \(h\).

An epimorphism in an abelian category is the cokernel of its kernel. Therefore (4.3) presents \(E\) as the ambient cokernel of a morphism \(K\to T\oplus A\) between objects of \(\mathcal J\). Cokernel closure puts \(E\) in \(\mathcal J\). This proves extension closure and the theorem. \(\square\)

The premise is sufficient. Exercise 3 gives a thick fully abelian subcategory for which it fails.

5. Generating and cogenerating objects

A full subcategory \(\mathcal J\) is generating in \(\mathcal A\) if every \(X\in\mathcal A\) receives an epimorphism \(J\to X\) from an object \(J\in\mathcal J\). It is cogenerating if every \(X\) admits a monomorphism \(X\to J\) into an object of \(\mathcal J\).

These two definitions exchange under opposite categories: a monomorphism in \(\mathcal A^{\rm op}\) is an epimorphism in \(\mathcal A\). They use a possibly different object of \(\mathcal J\) for each \(X\). No common generating object or supply of coproducts is assumed.

Proposition 5.1. If \(\mathcal J\) is cogenerating, every \(X\) has an ambient exact sequence

\[ 0\to X\to J^0\to J^1\to J^2\to\cdots \tag{5.1} \]

with all \(J^n\in\mathcal J\). If \(\mathcal J\) is generating, every \(X\) has an exact sequence

\[ \cdots\to J^{-1}\to J^0\to X\to0 \tag{5.2} \]

with all \(J^{-n}\in\mathcal J\).

Proof. In the cogenerating case, put \(C_0=X\). Choose a monomorphism \(a_n:C_n\to J^n\) and let \(c_n:J^n\to C_{n+1}\) be its cokernel. Induction supplies all these choices. Each pair is short exact:

\[ 0\to C_n\xrightarrow{a_n}J^n \xrightarrow{c_n}C_{n+1}\to0. \tag{5.3} \]

Define \(d_n=a_{n+1}c_n\). Then \(d_{n+1}d_n=0\), since \(c_{n+1}a_{n+1}=0\). Because \(a_{n+1}\) is monic, \(\ker d_n=\ker c_n=\operatorname{im}a_n\). For \(n\ge1\), the epimorphism \(c_{n-1}\) gives \(\operatorname{im}d_{n-1}=\operatorname{im}a_n\). At \(J^0\), the incoming map is \(a_0\). These equalities prove exactness of (5.1), including its first terms.

For completeness, the generating construction starts with \(K_0=X\). Choose an epimorphism \(p_n:J^{-n}\to K_n\), with kernel \(b_n:K_{n+1}\to J^{-n}\). The map from \(J^{-(n+1)}\) to \(J^{-n}\) is \(b_np_{n+1}\). Consecutive composites vanish because \(p_nb_n=0\). Its image is \(\operatorname{im}b_n=\ker p_n\); at \(J^{-n}\) for \(n\ge1\), the outgoing map is \(b_{n-1}p_n\), whose kernel is \(\ker p_n\) since \(b_{n-1}\) is monic. The last map \(p_0\) is epic. Thus (5.2) is exact. \(\square\)

This construction uses successive objectwise choices. It asserts no functorial choice and no injectivity, projectivity or acyclicity of the terms. Its content is an ordinary exact sequence under the stated generating or cogenerating hypothesis.

6. Four exercises with complete solutions

Exercise 1 (warm-up: a set of allowed primes). Fix a set \(P\) of prime numbers. Let \(\mathcal T_P\) consist of abelian groups in which every element has finite order whose prime divisors belong to \(P\). Prove that it is a Serre subcategory of \(\mathsf{Ab}\). Is it generating or cogenerating? Include the case \(P=\varnothing\).

Solution. Zero belongs, and the condition is preserved by isomorphisms. An element of a subgroup has its same order in the larger group. An element of a quotient has order dividing the order of any chosen representative. Thus subobjects and quotients remain in \(\mathcal T_P\).

Consider \(0\to A\to E\to B\to0\) with \(A,B\in\mathcal T_P\). For \(x\in E\), its image in \(B\) is killed by a positive integer \(m\) whose prime divisors lie in \(P\). Then \(mx\) belongs to \(A\), where it is killed by another such integer \(n\). Consequently \(mnx=0\). The order of \(x\) divides \(mn\), so has only primes in \(P\). This proves extension closure, without a uniform exponent bound on the groups. Hence \(\mathcal T_P\) is Serre.

A quotient of a torsion group is torsion, so no object of \(\mathcal T_P\) can surject onto \(\mathbb Z\). Also any map \(\mathbb Z\to T\) into a torsion group sends \(1\) to an element of finite order \(m\), and kills \(m\ne0\). Such a map cannot be monic. Thus the subcategory is neither generating nor cogenerating in \(\mathsf{Ab}\).

If \(P\) is empty, a permitted element order has no prime divisor and must be \(1\). Every element is zero, so \(\mathcal T_P\) consists exactly of zero groups and their isomorphic copies. The same conclusions hold.

Exercise 2 (intermediate: internal cokernels change). Let \(\mathcal F\) be the full subcategory of torsion-free abelian groups. Prove that it is additive, closed under ambient kernels and extensions, and has internal cokernels of every morphism. Compute these cokernels. Show that \(\mathcal F\) is not abelian, and that extension closure together with kernel closure does not imply thickness.

Solution. The zero group and finite direct sums are torsion-free. Proposition 1.1 gives additivity. A subgroup of a torsion-free group is torsion-free, so ambient kernels belong to \(\mathcal F\).

For an extension \(0\to A\to E\to B\to0\) with torsion-free ends, let \(nx=0\) for \(x\in E\) and \(n>0\). Its image in \(B\) is zero, so \(x\) comes from \(A\). Injectivity identifies the equation with \(na=0\) in \(A\), giving \(a=0\) and \(x=0\). Thus \(E\) is torsion-free.

For \(f:A\to B\), put \(C=B/\operatorname{im}f\) and let

\[ \begin{gathered} \operatorname{Tor}(C)\\ =\{c\in C:\exists n>0,\ nc=0\}. \end{gathered} \tag{6.1} \]

This is a subgroup: if positive integers \(m,n\) kill \(c,d\), then \(mn\) kills \(c+d\), and \(-c\) has the same annihilating integers as \(c\). The quotient \(C/\operatorname{Tor}(C)\) is torsion-free. Indeed, if \(n\bar c=0\), then \(nc\) is torsion and some \(m>0\) kills it. Hence \(mn c=0\), so \(c\) is torsion and \(\bar c=0\).

Every map \(C\to H\) into a torsion-free group kills \(\operatorname{Tor}(C)\). Therefore the composite

\[ B\to C\to C/\operatorname{Tor}(C) \tag{6.2} \]

has exactly the universal property of a cokernel of \(f\) among torsion-free groups. Thus it is the internal cokernel.

For multiplication by \(2\) on \(\mathbb Z\), the ambient cokernel is \(\mathbb Z/2\), while the internal cokernel is zero. The map is monic in \(\mathcal F\) because it is injective. It is epic there as well: if two maps \(r,s:\mathbb Z\to H\) agree after multiplication by \(2\), then \(2(r(1)-s(1))=0\), so \(r=s\) in the torsion-free target. It is not invertible, since no integer maps to \(1\) under multiplication by \(2\).

In an abelian category a map which is both mono and epi is invertible: its zero kernel and cokernel make its coimage and image its source and target, and the canonical comparison is the map itself. Hence \(\mathcal F\) is not abelian. It has internal kernels and cokernels, but the latter differ from ambient cokernels. In particular it is not fully abelian and not thick, despite its kernel and extension closure.

Exercise 3 (advanced: a sufficient premise need not be necessary). Let \(\mathcal V\) be the full subcategory of \(\mathsf{Ab}\) consisting of underlying groups of rational vector spaces. Prove that it is fully abelian and thick but not Serre. Show that the epi-lifting premise of Theorem 4.1 fails, using the epimorphism

\[ \begin{gathered} q:\bigoplus_{r\in\mathbb Q}\mathbb Z\to\mathbb Q,\\ q(e_r)=r. \end{gathered} \tag{6.3} \]

Solution. An abelian group is the underlying group of a rational vector space exactly when multiplication by every positive integer is bijective. In one direction this follows by multiplying by its reciprocal. Conversely, unique division defines

\[ \frac{a}{b}x=a\,y, \qquad by=x,\quad b>0. \tag{6.4} \]

Multiplication by \(bb'\) verifies independence of equivalent fractions; multiplication by common denominators verifies the vector-space addition and scalar laws. Bijectivity lets us cancel those denominators, so this gives a unique rational structure. Any group homomorphism between such groups respects unique division and hence is rational linear.

Thus \(\mathcal V\) is full in \(\mathsf{Ab}\), contains zero and finite sums, and is additive. Kernels of rational-linear maps are rational subspaces, and their group cokernels are their ordinary rational-vector-space quotients. Theorem 2.2 gives full abelianness.

For extension closure, take \(0\to A\xrightarrow{j}E\xrightarrow{p}B\to0\) with rational-vector-space ends, and fix \(n>0\). If \(nx=0\), then \(np(x)=0\), so \(p(x)=0\). Write \(x=j(a)\); injectivity gives \(na=0\), hence \(x=0\). To prove surjectivity of multiplication by \(n\), take \(x\in E\). Choose \(b\in B\) with \(nb=p(x)\), then \(y\in E\) with \(p(y)=b\). The difference \(x-ny\) is \(j(a)\). Choose \(a'\) with \(na'=a\); then

\[ n\bigl(y+j(a')\bigr)=x. \tag{6.5} \]

Every positive integer acts bijectively on \(E\), so (6.4) makes it a rational vector space. This proves extension closure and thickness. But \(\mathbb Z\subset\mathbb Q\) is an ambient subobject whose underlying group is not uniquely divisible. Therefore \(\mathcal V\) is not Serre.

The map (6.3) is onto, since any rational number is the image of its indexed basis element. Any map from a rational vector space \(T\) to a free abelian group is zero. For \(t\in T\), its image is divisible by every positive integer in that free group. A finite-support integer vector with this property is zero: each coordinate integer is divisible by every positive integer, forcing it to vanish. Thus every \(T\to\bigoplus_{r\in\mathbb Q}\mathbb Z\) is zero. Its composite with \(q\) cannot be epic onto the nonzero group \(\mathbb Q\). This disproves the epi-lifting premise for this thick subcategory.

Exercise 4 (challenge: an idempotent kernel and a trace). Let \(R\) be a unital ring and \(e^2=e\). Prove that \(M\mapsto eM\), regarded as a functor from left \(R\)-modules to abelian groups, is exact, and that its zero class is Serre. For the upper-triangular matrix ring over a field \(k\), with \(e=\operatorname{diag}(1,0)\), describe its zero class and compare it with the class of modules satisfying \(M=ReM\). Give the actual module maps and actions.

Solution. The map \(m\mapsto em\) is an additive idempotent on the underlying group. Its image \(eM\) is an abelian subgroup, even when it is not an \(R\)-submodule. An \(R\)-linear map \(f\) satisfies \(f(em)=ef(m)\), so restricts to \(eM\to eN\). This makes an additive functor to \(\mathsf{Ab}\).

Apply it to \(0\to A\xrightarrow{i}B\xrightarrow{p}C\to0\). The restricted \(i\) is injective. If \(eb\) is killed by \(p\), write \(eb=i(a)\); applying \(e\) gives \(eb=i(ea)\), so the kernel is the image of \(eA\). For \(ec\in eC\), choose \(b\) with \(p(b)=c\); then \(p(eb)=ec\). Thus

\[ 0\to eA\to eB\to eC\to0 \tag{6.6} \]

is exact. Its zero class contains zero and is replete and full. If \(eB=0\), exactness forces \(eA=eC=0\). If \(eA=eC=0\), it forces \(eB=0\). These observations give subobject, quotient and extension closure, hence the zero class is Serre. This is the exact-functor kernel principle in [Stacks, Subcategories], proved directly for this functor.

Now write

\[ \begin{gathered} R=\left\{\begin{pmatrix}a&b\\0&c\end{pmatrix}:a,b,c\in k\right\},\\ e_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \quad e_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}. \end{gathered} \tag{6.7} \]

A left module is a \(k\)-vector space. The two complementary idempotents give \(M=V\oplus W\), where \(V=e_1M\) and \(W=e_2M\). The off-diagonal matrix \(u=E_{12}\) satisfies \(e_1u=u=ue_2\), so defines a linear map \(t:W\to V\). Conversely, any such map gives the action

\[ \begin{gathered} \begin{pmatrix}a&b\\0&c\end{pmatrix}(v,w)\\ =\bigl(av+b\,t(w),cw\bigr). \end{gathered} \tag{6.8} \]

To check associativity, the product of two matrices has diagonal entries \(aa',cc'\) and off-diagonal entry \(ab'+bc'\). Applying (6.8) twice gives

\[ \bigl(aa'v+(ab'+bc')t(w),cc'w\bigr), \tag{6.9} \]

the same result as their matrix product. The identity acts as the identity. This constructs the module and recovers every original module action.

A module map is exactly a pair of linear maps \(\alpha:V\to V'\), \(\beta:W\to W'\) with \(\alpha t=t'\beta\): commuting with the diagonal idempotents gives the pair, and commuting with \(u\) gives that equation. Conversely the equation makes the pair commute with (6.8).

For \(e=e_1\), the kernel condition \(eM=0\) means \(V=0\), so \(t=0\) and the objects are \((0,W,0)\). On the other hand \(ReM=V\oplus0\): applying any matrix to \((v,0)\) gives \((av,0)\), and the identity already supplies every such vector. Thus \(M=ReM\) exactly when \(W=0\), giving objects \((V,0,0)\). For nonzero \(V\) or \(W\), these are two different classes; their intersection contains only zero. The second condition expresses trace generation by \(Re\), as in Trace coreflections and balanced nonabelian categories, while the first expresses the kernel of an exact functor.

References