Recovering isomorphisms from filtered pieces

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

A functor may remember enough to recognize an invertible map while forgetting some nonzero maps. A subobject and its quotient provide a concrete mechanism: their direct sum records both ends of an exact sequence, although it discards the extension joining them. We will use this mechanism to distinguish faithfulness, detection of isomorphisms and detection of zero objects.

The finite exact comparison is Composition factors and uniform chain bounds, Lemma 1.1. The abelian conventions are those of Trace coreflections and balanced nonabelian categories. The exact-functor interfaces are Stacks, Lemma 12.7.2, Remark 35.4.4 and Lemma 12.10.7. We retain these results where exactness is used.

All categories below are locally small in a fixed universe. They need not have generators or infinite limits or colimits. A ring has an identity, and its modules are unital left modules unless specified otherwise. Modules are arbitrary; no finite generation is implicit.

1. A subobject and its quotient detect invertibility

Let \(\mathcal A\) be abelian. Suppose \(T:\mathcal A\to\mathcal A\) is additive and has a natural transformation

\[ i:T\longrightarrow\operatorname{Id}_{\mathcal A} \tag{1.1} \]

whose components are monomorphisms. Choose \(QX=\operatorname{coker}(i_X)\), with quotient map \(p_X:X\to QX\). For a map \(f:X\to Y\), naturality of \(i\) makes \(p_Yf\) vanish on \(i_X\). The cokernel property determines a unique map \(Qf\) satisfying

\[ \begin{gathered} fi_X=i_Y(Tf),\\ (Qf)p_X=p_Yf. \end{gathered} \tag{1.2} \]

Uniqueness gives \(Q(1_X)=1_{QX}\) and \(Q(gf)=(Qg)(Qf)\): compose the respective candidate maps with the epimorphism \(p_X\). It also gives additivity, since

\[ \begin{aligned} Q(f+g)p_X&=p_Y(f+g)\\ &=((Qf)+(Qg))p_X. \end{aligned} \tag{1.3} \]

Thus \(Q\) is an additive functor, and the sequences

\[ 0\longrightarrow TX\xrightarrow{i_X}X \xrightarrow{p_X}QX\longrightarrow0 \tag{1.4} \]

form a natural family of short exact sequences. Define the piece functor

\[ \begin{gathered} G:\mathcal A\longrightarrow\mathcal A,\\ GX=TX\oplus QX,\\ Gf=(Tf)\oplus(Qf). \end{gathered} \tag{1.5} \]

The identity and composition laws hold on each summand. Both summand functors are additive, so \(G\) is additive.

Theorem 1.1. If \(Gf\) is invertible, then \(f\) is invertible.

Proof. A block diagonal map is invertible only if each diagonal block is invertible. Here is the argument in an arbitrary additive category, where the inverse need not initially be assumed diagonal. Let \(h:TY\oplus QY\to TX\oplus QX\) be an inverse of \(Gf\). Write \(r:TY\to TX\) and \(u:QY\to QX\) for its two diagonal components. Composing the inverse identities with the appropriate summand inclusions and projections gives

\[ \begin{aligned} r(Tf)&=1_{TX}, &(Tf)r&=1_{TY},\\ u(Qf)&=1_{QX}, &(Qf)u&=1_{QY}. \end{aligned} \tag{1.6} \]

The off-diagonal components contribute nothing to these diagonal equations because \(Gf\) is diagonal. Hence \(Tf\) and \(Qf\) are isomorphisms.

The two rows (1.4) for \(X\) and \(Y\), with vertical maps \(Tf,f,Qf\), commute by (1.2). Their first and last vertical maps are isomorphisms. The retained finite exact comparison, Lemma 1.1, makes the middle map \(f\) an isomorphism. \(\square\)

This proof uses exactness of the rows (1.4). It makes no claim that applying \(G\) to another exact sequence gives an exact sequence. The distinction is essential in the polynomial example below.

2. Three tests for what a functor remembers

Let \(F:\mathcal A\to\mathcal B\) be additive between abelian categories.

An additive functor sends a zero object to a zero object. Indeed,

\[ 1_{F0}=F(1_0)=F(0_{0,0})=0_{F0,F0}. \tag{2.1} \]

An object with zero identity is a zero object: every map to or from it is zero after composing with its identity.

Proposition 2.1. For every such \(F\),

\[ \begin{gathered} \text{faithful}\\ \Longrightarrow\text{conservative}\\ \Longrightarrow\text{detects zero objects}. \end{gathered} \tag{2.2} \]

Proof. Suppose \(F\) is faithful and \(Ff:FX\to FY\) is invertible. If \(a,b:Z\to X\) satisfy \(fa=fb\), then

\[ (Ff)(Fa)=(Ff)(Fb). \]

Cancel \(Ff\) and use faithfulness to obtain \(a=b\). Thus \(f\) is monic. If \(c,d:Y\to Z\) satisfy \(cf=df\), cancel \(Ff\) on the other side to obtain \(Fc=Fd\), and then \(c=d\). Thus \(f\) is epic. A monic and epic arrow in an abelian category is an isomorphism.

Now suppose \(F\) is conservative and \(FX\simeq0\). The arrow \(0\to X\) is sent to an arrow between zero objects, which is invertible. Conservativity makes \(0\to X\) invertible; hence \(X\simeq0\). \(\square\)

Neither implication uses exactness. The first uses balance of the source category. The second uses preservation of the zero object.

Retained exact criterion. If \(F\) is exact, all three conditions in (2.2) are equivalent. The reverse implication is the zero-kernel specialization of Stacks, Lemma 12.10.7. Its stronger statement concerns an exact functor induced from a Serre quotient: it is faithful precisely when the chosen Serre subcategory is the entire object kernel. For the zero Serre subcategory, the quotient is \(\mathcal A\) itself; it localizes only isomorphisms, using the quotient interface retained in Serre quotients and local saturation, Section 1.

The criterion's mechanism is preservation of images. If \(Ff=0\), exactness identifies \(F(\operatorname{im}f)\) with \(\operatorname{im}(Ff)=0\). Zero detection makes \(\operatorname{im}f=0\), so \(f=0\). Applying this to \(f-g\), with additivity, gives faithfulness.

The related reflection statement in Stacks, Remark 35.4.4 detects monomorphisms and epimorphisms separately through preserved kernels and cokernels. Thus an exact zero detector recognizes each of these two properties, as well as their conjunction. Exactness and its short-exact characterization are the interface of Stacks, Lemma 12.7.2.

For a family of representable probes, Generators and small quotient families, Theorems 2.1 and 2.2 gives stronger equivalences. Representable probes preserve the relevant equalizers. That extra property cannot be assumed for an arbitrary additive conservative functor.

3. A conservative functor that erases nonzero arrows

Let \(R\) be a ring and let \(t\) lie in its center. For every left \(R\)-module \(M\), set

\[ \begin{gathered} T_tM=tM,\qquad Q_tM=M/tM,\\ G_tM=tM\oplus(M/tM). \end{gathered} \tag{3.1} \]

Centrality ensures that \(tM\) is a submodule: \(a(tm)=t(am)\) for \(a\in R\). An \(R\)-linear map \(f:M\to N\) sends \(tm\) to \(tf(m)\), so it restricts to \(tM\to tN\) and induces \(M/tM\to N/tN\). Restrictions and induced quotient maps preserve identities, composition and addition. The inclusion \(tM\to M\) is natural, and its cokernel is \(M/tM\). Theorem 1.1 therefore proves:

Corollary 3.1. The additive functor \(G_t\) is conservative on all left \(R\)-modules.

This applies in particular to \(A=k[x]\), \(t=x\), for any field \(k\). Take

\[ \begin{gathered} M=A/(x^2),\\ v:M\longrightarrow M,\\ v(m)=xm. \end{gathered} \tag{3.2} \]

The classes \(1,x\) form a \(k\)-basis of \(M\), so \(v(1)=x\ne0\). But \(v\) vanishes on \(xM\), since \(x^2M=0\), and it induces zero on \(M/xM\), since its image lies in \(xM\). Consequently

\[ v\ne0,\qquad G_x(v)=0. \tag{3.3} \]

Thus a conservative additive functor between abelian categories need not be faithful, even on finite-dimensional modules. Its conservativity, however, holds on the entire module category.

The same module shows exactly where exactness fails. Write \(k=A/(x)\), with its indicated \(A\)-action. There is a short exact sequence

\[ 0\longrightarrow k\xrightarrow{j}M \xrightarrow{q}k\longrightarrow0, \tag{3.4} \]

where \(j(1)=x\) and \(q\) is reduction modulo \(x\). The first map is injective, the second surjective, and its kernel is the span of \(x\), exactly the image of \(j\).

Identify \(G_x(k)=0\oplus k\simeq k\) and identify both \(xM\) and \(M/xM\) with \(k\). The induced maps become

\[ \begin{gathered} G_x(j):k\longrightarrow k\oplus k,\\ a\longmapsto(0,0),\\ G_x(q):k\oplus k\longrightarrow k,\\ (b,c)\longmapsto c. \end{gathered} \tag{3.5} \]

For the first map, the subobject component has zero domain, and the quotient component sends \(j(k)\subseteq xM\) to zero. For the second map, \(q(xM)=0\), while reduction induces the identity on \(M/xM\simeq k\).

The map \(G_x(j)\) is not monic. Moreover, the kernel of \(G_x(q)\) is the nonzero first copy of \(k\), while the image of \(G_x(j)\) is zero. By the retained short-exact tests, \(G_x\) is neither left exact nor right exact. Surjectivity of the last map alone does not give right exactness.

The weaker zero-object test also has a strict limitation. In the previously established example over \(B=k[x,y]\), the ideal \(\mathfrak a=(x,y)\) gives the additive functor

\[ \begin{gathered} H=\operatorname{Hom}_B(\mathfrak a,-),\\ H:\operatorname{Mod}(B)\longrightarrow\operatorname{Mod}(B). \end{gathered} \tag{3.6} \]

The \(B\)-action on each Hom group is multiplication in its target. Trace coreflections and balanced nonabelian categories, Section 4 proves on all modules that \(H\) detects zero objects. It also proves that the inclusion \(\mathfrak a\hookrightarrow B\) is sent to an isomorphism although it is not an isomorphism. In Exercise 3 we compute the precise failure of right exactness explaining why the exact criterion does not apply.

4. Exercises with complete solutions

Exercise 1 — Exactness and fullness (5 points). Let \(\mathcal V\) be the category of all \(k\)-vector spaces. Define

\[ \begin{gathered} S:\mathcal V\times\mathcal V\longrightarrow\mathcal V,\\ S(V,W)=V\oplus W,\\ S(a,b)=a\oplus b. \end{gathered} \tag{4.1} \]

Show that \(S\) is exact, faithful and conservative, but not full. Compare it with the exact projection \(P(V,W)=V\): give explicit failures of faithfulness, conservativity and zero detection for \(P\). Include pairs with a zero component.

Solution. Kernels and cokernels in the product category are taken componentwise. A sequence there is short exact precisely when both component sequences are short exact. Their finite direct sum is short exact: a pair lies in the kernel of a direct sum map exactly when each component lies in the corresponding kernel, and the image is the direct sum of the two component images. Surjectivity also holds componentwise. Thus \(S\) is exact.

Given \(a\oplus b\), compose it with the first inclusion and first projection to recover \(a\); use the second pair to recover \(b\). Hence \(S(a,b)=S(a',b')\) implies \(a=a'\) and \(b=b'\). This proves faithfulness, including when one component is zero.

For conservativity, the diagonal inverse calculation (1.6) shows that an invertible \(a\oplus b\) has invertible \(a\) and \(b\). Then \((a^{-1},b^{-1})\) is the inverse in the product category. A map \(0\to0\) is the identity of the zero object and is invertible, so a zero component introduces no exception.

The Hom set from \((0,k)\) to \((k,0)\) contains only the zero map: its components are \(0\to k\) and \(k\to0\). Yet their images under \(S\) are both isomorphic to \(k\), with a nonzero map from the source second summand to the target first summand. For example, under these identifications the identity of \(k\) is such a map. It cannot be \(S(a,b)\), so \(S\) is not full.

The projection \(P\) is exact because it selects one of the two component short exact sequences. The nonzero object \((0,k)\) is sent to zero, so \(P\) does not detect zero objects. Its identity and zero endomorphism are distinct but have the same image, the unique endomorphism of \(0\); hence \(P\) is not faithful. Finally the arrow

\[ (0,0)\longrightarrow(0,k) \tag{4.2} \]

is not invertible in the product category, whereas \(P\) sends it to the invertible arrow \(0\to0\). Thus \(P\) is not conservative. Exactness alone guarantees none of the three detection properties. \(\square\)

Exercise 2 — How much of an endomorphism survives? (8 points). Let \(A=k[x]\), \(M_n=A/(x^n)\), \(n\ge1\). Compute \(G_x(M_n)\) and the induced map on endomorphism rings. Find its kernel and determine precisely when an endomorphism and its image under \(G_x\) are invertible. Give an explicit inverse when it exists, over a field of any characteristic.

Solution. For \(n\ge2\), the map

\[ \begin{gathered} M_{n-1}\longrightarrow xM_n,\\ \overline b\longmapsto\overline{xb} \end{gathered} \tag{4.3} \]

is well-defined and surjective. Its kernel is zero: \(xb\in(x^n)\) in \(k[x]\) if and only if \(b\in(x^{n-1})\). It is therefore an \(A\)-linear isomorphism. Also \(M_n/xM_n\simeq k\). Consequently, with the first line applying for \(n\ge2\),

\[ \begin{gathered} G_x(M_n)\simeq M_{n-1}\oplus k,\\ G_x(M_1)\simeq0\oplus k. \end{gathered} \tag{4.4} \]

Every endomorphism of \(M_n\) is determined by the image of \(1\). Any element \(\overline a\in M_n\) may be that image, because it is annihilated by \(x^n\). The corresponding endomorphism is multiplication by \(\overline a\). Composition multiplies the classes, so

\[ \operatorname{End}_A(M_n)\simeq A/(x^n) \tag{4.5} \]

as rings.

Write \(a_0=a(0)\). Under (4.4), for \(n\ge2\), let \(\widetilde a=a\bmod x^{n-1}\), and let \(\mu_b\) denote multiplication by \(b\) on the indicated summand. Then \(G_x\) sends multiplication by \(\overline a\) to the diagonal endomorphism

\[ \begin{pmatrix} \mu_{\widetilde a}&0\\ 0&\mu_{a_0} \end{pmatrix}. \tag{4.6} \]

Indeed \(a(xb)=x(ab)\), giving the first block under (4.3), while the quotient by \(x\) retains only the constant coefficient. The target endomorphism ring can contain off-diagonal maps; (4.6) describes the image of the given functor.

For \(n\ge2\), both blocks vanish exactly when \(a\) is divisible by \(x^{n-1}\): this already forces its constant coefficient to vanish. Hence the kernel of the induced ring homomorphism is

\[ (x^{n-1})/(x^n)=k\,\overline{x}^{\,n-1}. \tag{4.7} \]

This is one-dimensional over \(k\) and contains a nonzero endomorphism. When \(n=1\), the source endomorphism ring is \(k\), and the functor retains precisely multiplication by that scalar on \(k\); its kernel is zero.

If \(a_0\ne0\), put \(z=a_0^{-1}a-1\) in \(A/(x^n)\). Then \(z\in(x)\) and \(z^n=0\). An inverse of \(\overline a=a_0(1+z)\) is

\[ a_0^{-1}\sum_{j=0}^{n-1}(-z)^j. \tag{4.8} \]

Multiplying on either side uses the finite geometric identity

\[ \begin{gathered} (1+z)\sum_{j=0}^{n-1}(-z)^j\\ =1+(-1)^{n-1}z^n\\ =1. \end{gathered} \tag{4.9} \]

The identity holds in every characteristic. If \(a_0=0\), then \(\overline a\) is nilpotent; it cannot be a unit in the nonzero ring \(A/(x^n)\), since taking the \(n\)-th power of an inverse identity would give \(1=0\).

Thus multiplication by \(a\) on \(M_n\) is invertible precisely when \(a_0\ne0\). Its image under \(G_x\) is invertible under the same condition: necessity follows from its scalar block on the nonzero summand \(k\), and sufficiency follows either from (4.8) on the first block or from functoriality applied to the original inverse. For \(n=1\) this is exactly the usual nonzero-scalar criterion. The nonzero maps in (4.7) are forgotten, although invertibility is still detected. \(\square\)

Exercise 3 — The missing right-exactness hypothesis (7 points). Let \(B=k[x,y]\), \(\mathfrak a=(x,y)\), \(k=B/\mathfrak a\), and \(H\) be (3.6). Use the full computations retained in Ideal-generated modules and nonstrict quotients, Sections 1 and 2 to compute \(H\) on

\[ 0\longrightarrow\mathfrak a\xrightarrow{\iota}B \xrightarrow{\pi}k\longrightarrow0. \tag{4.10} \]

Determine the maps, not just the dimensions of the terms. Explain why this zero-object detector is not right exact.

Solution. The retained presentation of \(\mathfrak a\) has generators \(x,y\) and the relation \(-yx+xy=0\). Its full Hom computation gives

\[ \begin{gathered} \operatorname{End}_B(\mathfrak a)\simeq B,\\ \operatorname{Hom}_B(\mathfrak a,B)\simeq B,\\ \operatorname{Hom}_B(\mathfrak a,k)\simeq k^2. \end{gathered} \tag{4.11} \]

In the first two identifications, \(b\in B\) is the map \(u\mapsto bu\), with target \(\mathfrak a\) or \(B\), respectively. In the third, a map is specified by the two arbitrary values on \(x\) and \(y\) in \(k\); the presentation relation imposes no condition, since both \(x\) and \(y\) act by zero on \(k\). These are precisely the prior identifications, with the action on Hom taken in the target.

The map \(H(\iota)\) postcomposes with inclusion. It takes multiplication by \(b\), valued in \(\mathfrak a\), to the same multiplication by \(b\), valued in \(B\). Under (4.11), it is the identity \(B\to B\).

Every map represented by \(b\) in \(\operatorname{Hom}_B(\mathfrak a,B)\) has image in \(\mathfrak a\), since \(b\mathfrak a\subseteq\mathfrak a\). Postcomposition with \(\pi\) therefore gives zero. Thus \(H(\pi)\) is the zero map \(B\to k^2\), and the resulting sequence is

\[ 0\longrightarrow B\xrightarrow{1_B}B \xrightarrow{0}k^2. \tag{4.12} \]

The middle kernel and image agree, as left exactness of a representable Hom functor predicts. But the final map is not surjective, because \(k^2\ne0\). For example, the map \(\mathfrak a\to k\) taking \(x\) to \(1\) and \(y\) to \(0\) has no extension \(B\to k\): any such extension sends \(x\) to \(x\) times the image of \(1\), which is zero in \(k\).

The full earlier trace example proves zero detection for all \(B\)-modules, and also proves that \(\iota\) is not invertible. Equation (4.12) shows directly how \(H\) sends this nonisomorphism to an isomorphism and why the exact zero-detector criterion cannot apply. \(\square\)

Exercise 4 — Detecting isomorphisms does not detect splitting (8 points). Let \(U:\operatorname{Mod}(k[x])\to\mathcal V\) forget the \(k[x]\)-action. Prove that \(U\) is exact, faithful and conservative. Show that (3.4) splits after applying \(U\), but does not split as a sequence of \(k[x]\)-modules. Express this failure in terms of Yoneda extension classes.

Solution. Kernels and images of module maps are their underlying vector-space kernels and images, with inherited action. The underlying vector space of a module quotient is the corresponding vector-space quotient. Hence an exact sequence of modules remains exact after applying \(U\); in particular \(U\) is exact by the short-exact criterion.

If two module maps have the same underlying linear map, they have the same value at every element and are equal. Thus \(U\) is faithful. Suppose a module map \(f:M\to N\) has an invertible underlying linear map. Its inverse is module-linear: if \(n=f(m)\) and \(a\in k[x]\), then

\[ \begin{aligned} f^{-1}(an)&=f^{-1}(f(am))\\ &=am\\ &=af^{-1}(n). \end{aligned} \tag{4.13} \]

Thus \(U\) is also conservative.

Let \(M=k[x]/(x^2)\) as in (3.4). The vector-space map

\[ s:k\longrightarrow U(M),\qquad s(c)=c\,\overline1 \tag{4.14} \]

satisfies \(U(q)s=1_k\). Indeed \(U(M)\) has basis \(\overline1,\overline x\), the image of \(U(j)\) is \(k\overline x\), and (4.14) chooses its complementary line. This gives a split sequence of vector spaces.

If an \(A=k[x]\)-linear section \(r:k\to M\) existed, its value at \(1\) would have to be

\[ r(1)=\overline1+c\overline x \tag{4.15} \]

for some \(c\in k\), since \(qr=1_k\). But \(x\cdot1=0\) in \(k=A/(x)\), whereas

\[ xr(1)=\overline x+c\overline{x^2} =\overline x\ne0 \tag{4.16} \]

in \(M\). This contradicts \(A\)-linearity. Hence (3.4) is nonsplit over \(A\), over every field.

A Yoneda extension class is the equivalence class of a short exact sequence with fixed endpoints; equivalence is an isomorphism of sequences that is the identity on those endpoints. The zero class is the class of the split sequence. For modules, existence of a section gives the explicit isomorphism

\[ \begin{gathered} k\oplus k\longrightarrow M,\\ (a,b)\longmapsto j(a)+r(b). \end{gathered} \tag{4.17} \]

Its inverse first takes \(q(m)\), then writes \(m-r(q(m))\) uniquely as \(j(a)\). Conversely an equivalence with the split sequence supplies a section from its second summand. Thus the absence of an \(A\)-linear section proves that (3.4) represents a nonzero Yoneda class, while (4.14) proves that its underlying vector-space class is zero.

Applying an exact functor to a short exact sequence and to an equivalence of such sequences is well-defined on Yoneda classes; a split sequence remains split because its section is sent to a section. We have exhibited two distinct module extension classes, the class of (3.4) and the split class, with the same image under \(U\). An exact faithful functor can therefore fail to distinguish extension classes. Its injectivity on ordinary Hom groups does not imply injectivity on this set of extensions. \(\square\)

References