Testing equivalence with arrows and finite characters

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text: CC0.

An equivalence retains the maps that distinguish objects, even when it changes their labels. This gives a way to rule out an equivalence with an opposite category: an incoming-map property of an initial object would have to become an outgoing-map property of a terminal object. Finite abelian groups pass a different test. Their characters recover every point, and precomposition reverses every homomorphism without losing information.

We use categories with ambient sets of objects and locally small Hom sets. Sets, relations, rings and finite groups lie in a fixed universe containing the natural numbers. Rings are associative and unital, their homomorphisms preserve the unit, and the zero ring is allowed. The needed set constructions, including the usual encodings of integers, rationals and their quotients, follow from Universes and small categories. Our categorical prerequisites are the full inverse and cancellation proofs in Relations and cancellation in categories, the universal-object and zero-map proofs in Zero maps, components and subobjects, and the equivalence and skeleton constructions in Equivalences and chosen representatives. For characters we retain the complete group and cyclic calculations specified in Section 5.

1. Extreme objects remember an arrow direction

Proposition 1.1. If a category has an initial object \(I\), a terminal object \(T\), and an arrow \(b:T\to I\), then \(I\) and \(T\) are isomorphic.

Proof. Initiality supplies \(a:I\to T\). The composites \(ba\) and \(ab\) are endomorphisms of \(I\) and \(T\), respectively. Each of those endomorphism sets is a singleton containing its identity, by initiality of \(I\) and terminality of \(T\). Hence

These are both inverse equations. In particular the object is both initial and terminal, with the existing zero-map construction in the prerequisite. \(\square\)

Consider the following property of an initial object:

Lemma 1.2. Equivalences preserve initial objects and property (1.2).

Proof. Let \(F:C\to D\) be an equivalence. The complete equivalence criterion in the prerequisite makes it fully faithful and essentially surjective. For any \(Y\in D\), choose an isomorphism \(c:FX\to Y\). Postcomposition with \(c\) and full faithfulness identify

The final set is a singleton, so \(FI\) is initial. If \(h:Y\to FI\), fullness writes \(hc=F(u)\) for \(u:X\to I\). If (1.2) holds in \(C\), \(u\) is invertible; functoriality preserves its inverse, so \(hc\) is invertible. Since \(c\) is an isomorphism, \(h=F(u)c^{-1}\) is invertible. This proves preservation for every incoming arrow. \(\square\)

Theorem 1.3. The category of sets in the universe is not equivalent to its opposite.

Proof. Its initial object is the empty set. A function \(X\to\varnothing\) can exist only if \(X\) is empty, and then is the identity of the empty set. Thus (1.2) holds.

An initial object in the opposite category is a terminal set, hence a singleton: maps from a singleton into a terminal set identify its elements with a singleton Hom set. For any singleton \(T\), an inclusion \(T\to\{0,1\}\) gives an arrow \(\{0,1\}\to T\) in the opposite category. It is not invertible, since the original function is not a bijection. Thus no initial object there satisfies (1.2). Lemma 1.2 rules out an equivalence. \(\square\)

An equivalence does not preserve the literal name of an initial object. The argument works because every terminal set is a singleton and the obstruction applies to each such object.

2. A category with at most one arrow becomes an order

Theorem 2.1. If every Hom set of \(C\) has at most one element, then \(C\) is equivalent to the category of a partially ordered set. The empty category is included.

Proof. Retain the complete ambient skeleton construction, Corollary 3.2 of Equivalences and chosen representatives. It supplies a full skeletal subcategory \(K\) and an equivalence \(K\hookrightarrow C\), with all needed representative choices, conjugated arrows and functor laws already proved. Fullness means the Hom sets of \(K\) still have at most one element.

On its ambient object set define

Identities prove reflexivity, and composition proves transitivity. If \(X\leq Y\) and \(Y\leq X\), choose the two arrows \(f,g\). Each composite is an endomorphism in a Hom set with at most one element, so it is the corresponding identity. Thus \(X,Y\) are isomorphic. Since \(K\) is skeletal, they are equal, proving antisymmetry.

The order category has exactly the same objects and existence of arrows as \(K\). Sending its unique arrow to the corresponding unique arrow of \(K\) is a bijection on each Hom set and preserves identities and composition because each relevant Hom set has at most one member. Its inverse has the same properties. This gives a category isomorphism with \(K\), and the retained equivalence with \(C\) proves the assertion. For empty \(C\), its skeleton and this ordered set are empty. \(\square\)

The order is partial, rather than necessarily linear: there can be two objects with no arrow in either direction.

3. Two complete inverse tests

There are two useful stronger results already proved in the prerequisites. We retain their whole arguments and make the substitutions explicit.

Proposition 3.1. If \(X\xrightarrow fY\xrightarrow gZ\), \(gf\) is invertible and \(g\) is monic, then both \(f\) and \(g\) are invertible.

Proof interface. Put \(h=(gf)^{-1}\) and \(s=fh:Z\to Y\). Then

Use the complete one-sided inverse calculation in Relations and cancellation, Section 1, with its monic retraction \(r=g\) and section \(s=fh\). It proves \(sg=1_Y\), hence \(g^{-1}=s\). The same section proves that compositions of invertible arrows are invertible. Since \(f=g^{-1}(gf)\), it proves invertibility of \(f\) too; its inverse is \(hg\). Both statements retain their full existing proof, without an additional balanced-category hypothesis. \(\square\)

Proposition 3.2. In a category with a zero object, \(1_X=0_{XX}\) implies \(X\simeq0\).

Proof interface. Exercise 2 and its full solution in Zero maps, components and subobjects prove the stronger assertion that \(1_X=0_{XX}\) holds exactly when \(X\) is a zero object. That proof uses only the two absorption laws for zero maps and treats both incoming and outgoing Hom sets. Proposition 2.1 of the same lesson gives the unique isomorphism between initial objects, applied to the now initial \(X\) and \(0\). This supplies the required isomorphism in an arbitrary category with a zero object. \(\square\)

Neither test assumes that Hom sets have an addition.

4. A ring epimorphism need not be surjective

For sets, retain the complete proof in Exercise 1 of Relations and cancellation: monomorphisms are precisely injections and epimorphisms are precisely surjections, with the singleton witnesses for a failed injection, two-valued witnesses for a failed surjection and all empty cases.

Theorem 4.1. In the category of associative unital rings, the inclusion \(\mathbb Z\to\mathbb Q\) is both monic and epic and is not invertible. The ring \(\mathbb Z\) is initial, and the zero ring is terminal.

Proof. The inclusion is injective on underlying sets. If two parallel ring maps become equal after this inclusion, equality of their values and injectivity give equality before it. Thus the inclusion is monic.

For epicity, let \(v,w:\mathbb Q\to R\) be ring maps agreeing on \(\mathbb Z\). Every nonzero integer \(b\), regarded in \(\mathbb Q\), is a unit. Its images are therefore units and

The images of \(a\) and \(b\) agree. A two-sided inverse is unique by the elementary multiplication calculation: if \(xy=yx=1\) and \(xz=zx=1\), then \(y=y(xz)=(yx)z=z\). Consequently the two right sides in (4.1) agree. Every rational has this form, so \(v=w\), proving epicity. This proof applies to noncommutative target rings too.

It is not an isomorphism: the underlying function omits \(1/2\), whereas applying an inverse ring map would give an inverse function.

A unital ring map \(\mathbb Z\to R\) is forced to send \(n\) to \(n1_R\). This rule is a ring map: repeated addition and additive inverses give \((a+b)1_R=a1_R+b1_R\); distributivity gives \((a1_R)(b1_R)=ab1_R\), first for nonnegative integers and then for negatives by additive inverses. It sends \(1\) to \(1_R\). Thus exactly one map exists, proving initiality. Every ring has exactly one function to the one-element zero ring. It preserves addition, multiplication and the unit since \(0=1\) there, and is therefore the unique ring map. This proves terminality. \(\square\)

When a target has no map from \(\mathbb Q\), the epimorphism test has no competing pair of maps to that target. It does not require such maps to exist.

5. Finite characters recover every point and map

Let \(E=\mathbb Q/\mathbb Z\), with its additive group structure. For a finite abelian group \(A\), put

Addition is pointwise. For a homomorphism \(f:A\to B\), set \(D(f)(\chi)=\chi f\). Retain the complete stronger precomposition construction and both contravariant functor laws in Character duals and one-sided injectivity, Section 1, with \(R=\mathbb Z\). Thus \(D(gf)=D(f)D(g)\), with identity and addition preserved, already holds for all abelian groups. We will prove that \(D(A)\) is finite, so those same laws restrict to the finite category.

Retained extension and cyclic interfaces. The complete stronger proof of Stacks, Lemma 15.55.1, in the pinned structured source, proves that divisible abelian groups are injective. Here \(E\) is divisible: a rational representative \(q\) of a class has \(q/n+\mathbb Z\) as an \(n\)-th root for every positive integer \(n\). Therefore every character of a subgroup extends to the whole abelian group. We retain that full extension proof; no finite decomposition theorem is needed.

For \(C_n=\mathbb Z/n\mathbb Z\), retain Exercise 1 and its full cyclic calculation in Character duals and one-sided injectivity. It identifies every character as

proves \(|D(C_n)|=n\), computes double evaluation, and includes the trivial group \(n=1\). That exact character count will be the induction base.

Lemma 5.1. For every finite abelian group \(A\), \(D(A)\) is finite with \(|D(A)|=|A|\). Characters separate nonzero elements of \(A\).

Proof. First, if \(a\in A\), the order of its cyclic subgroup divides \(|A|\): its distinct cosets partition \(A\), and translation gives each coset the same cardinality as that subgroup. Thus \(|A|a=0\), so every character value lies in \(E[|A|]\). By the retained cyclic computation this is the finite subgroup consisting of the classes \(j/|A|+\mathbb Z\). The character set is therefore a subset of a finite function set, so is finite and lies in the fixed universe.

We prove the cardinality statement by induction on \(|A|\). The trivial group and all cyclic groups are covered by the retained computation. If \(A\) is not cyclic, choose a nonzero \(a\) and put \(H=\langle a\rangle\). Then \(1<|H|<|A|\), and \(|A/H|<|A|\). Restriction

is surjective by the retained extension theorem. Its kernel is exactly \(D(A/H)\): a character vanishing on \(H\) descends to the quotient by \(x+H\mapsto\chi(x)\), which is well-defined precisely because it vanishes there; conversely precomposition with the quotient gives such a character. These are inverse group maps. Each fibre of a surjective group homomorphism is a coset of its kernel, by translation by any one member of that fibre. Induction and the same coset count now give

For separation, retain the stronger injective evaluation theorem in Stacks, Lemma 15.56.7, whose complete structured proof extends a nonzero cyclic character using the general extension theorem. If every character vanished at \(a\), evaluation would send \(a\) to zero, and that injectivity would give \(a=0\). Therefore each nonzero \(a\) is separated by a character, as required. \(\square\)

Theorem 5.2. Character duality is an equivalence from the category \(\mathsf{Ab}_{\mathrm f}\) of finite abelian groups to its opposite. Its double comparison is point evaluation.

Proof. Define

Retain Proposition 1.1 and its full proof in Character duals and one-sided injectivity, Section 1, specialized to \(R=\mathbb Z\). It proves that \(e_A\) is an additive, injective homomorphism and that these maps are natural for every abelian group. The new finite cardinality result in Lemma 5.1, applied twice, makes its source and target have the same size. It is therefore bijective. The inverse of a bijective group homomorphism is additive: apply the given homomorphism to the desired addition equation and use injectivity. Hence \(e_A\) is a group isomorphism.

For \(f:A\to B\), evaluation at \(\chi\in D(B)\) gives

This is exactly the retained naturality equation under our notation. The functor \(D:\mathsf{Ab}_{\mathrm f}\to\mathsf{Ab}_{\mathrm f}^{\mathrm{op}}\) and its opposite \(D^{\mathrm{op}}\) have both composites given by double duality. The components \(e_A\) give the natural isomorphism from the identity to double duality on the original category. Reversing them in the opposite category gives the natural isomorphism from double duality to its identity there. These are quasi-inverse comparisons, proving the equivalence.

The comparisons satisfy the useful evaluation identity

Indeed its value at \(\chi\) and then at \(a\) is \(e_{D(A)}(\chi)(e_A(a))=e_A(a)(\chi)=\chi(a)\).

In particular every homomorphism \(b:D(B)\to D(A)\) comes from exactly one \(f:A\to B\). An explicit inverse to \(f\mapsto D(f)\) is

For a given \(f\), naturality (5.6) makes (5.8) recover \(f\) from \(b=D(f)\). Conversely (5.7), with the comparisons invertible, gives \(D(e_A)=e_{D(A)}^{-1}\) and \(D(e_B^{-1})=e_{D(B)}\). Naturality for \(b:D(B)\to D(A)\) gives \(D(D(b))e_{D(B)}=e_{D(A)}b\). Applying \(D\) to (5.8) and substituting these identities gives \(D(f)=b\). This checks both Hom inverses explicitly. The trivial group has one character and every formula remains valid. \(\square\)

The relation category also reverses arrows without losing information. Retain the exact converse-relation proof in Section 4 of Relations and cancellation: \((SR)^{\mathrm t}=R^{\mathrm t}S^{\mathrm t}\), \((R^{\mathrm t})^{\mathrm t}=R\), and the converse of a diagonal is the same diagonal. These identities provide literal inverse functors between the relation category and its opposite, including empty underlying sets.

For \(C_4\), the complete cyclic pairing is visible as a table. The column headings record the input point. Entries are classes modulo \(\mathbb Z\).

Figure 5.1. Rows are all four characters from (5.2). Column \(a\) is exactly the character \(e_{C_4}(a)\) of the row group. All four columns are distinct, showing injectivity of evaluation; they exhaust the four characters of that group. The editable array gives the full exact pairing, with no sampled or approximate values.

6. Four exercises with full solutions

Exercise 1 — quotient four labels to two positions

Foundation. Let \(C\) have objects \(a,b,c,d\). There is one arrow \(x\to y\) exactly when either \(x,y\in\{a,b\}\), or \(x,y\in\{c,d\}\), or \(x\in\{a,b\}\) and \(y\in\{c,d\}\). Composition is the unique arrow with the resulting endpoints. Identify its initial and terminal objects, construct its ordered model, and determine whether an arrow from \(a\) to \(c\) is monic, epic or invertible.

Solution. The displayed relation is reflexive and transitive, so identities exist and every composable pair has the asserted composite. Both associativity expressions have the same endpoints and hence are the same unique arrow. The identity laws follow the same way.

There is exactly one arrow from each of \(a,b\) to every object, making both initial. Neither \(c\) nor \(d\) is initial, since neither maps to \(a\). Dually \(c,d\) are exactly the terminal objects. The arrows each way between \(a,b\) compose to the unique endomorphisms, which are identities, so \(a\simeq b\). Likewise \(c\simeq d\); the two classes are not isomorphic, since no arrow goes from the upper class to the lower one.

Send \(a,b\) to \(0\) and \(c,d\) to \(1\) in the order \(0<1\). This assignment preserves arrow existence, identities and composites. For every pair of source objects, both Hom sets are either empty or singletons in exactly the same cases; its Hom maps are thus bijective. Both target objects occur. The complete equivalence criterion proves it is an equivalence.

Every arrow of \(C\) is monic and epic because every parallel test pair already consists of the same arrow. In particular \(a\to c\) has both cancellation properties. It has no inverse because \(c\to a\) does not exist. Also there is no arrow from any terminal object to any initial one, consistent with Proposition 1.1.

Exercise 2 — targets which accept rational scalars

Intermediate. Determine the number of unital ring maps from \(\mathbb Q\) to each of \(\mathbb Q\), \(M_2(\mathbb Q)\), \(\mathbb F_p\) for prime \(p\), and the zero ring. Explain how these examples fit the epimorphism test for \(\mathbb Z\to\mathbb Q\).

Solution. Theorem 4.1 implies that two such maps to any fixed target must agree: both restrict to the unique unital map from \(\mathbb Z\). So every count is at most one.

For \(\mathbb Q\), the identity is one map. For \(M_2(\mathbb Q)\), \(q\mapsto qI_2\) preserves addition, multiplication and the unit, so is one map, even though the target ring is noncommutative. The unique map to the zero ring preserves the unit because its sole element is also its unit.

There is no map to \(\mathbb F_p\). If \(v\) were one, applying it to \(p(1/p)=1\) would give

which is impossible in this nonzero ring. The four counts are therefore \(1,1,0,1\). Epicity asserts uniqueness when a competing pair exists; the zero count is permitted. The inclusion itself still omits \(1/2\), so these uniqueness assertions do not make it an isomorphism.

Exercise 3 — extend a character across a cyclic subgroup

Intermediate. Let \(A=C_4\), \(H=\{0,2\}\), and let \(\theta:H\to E\) send \(2\) to \(1/2+\mathbb Z\). List every extension to \(A\), compute restriction \(D(A)\to D(H)\), and identify its kernel with \(D(A/H)\).

Solution. The complete cyclic coordinates (5.2) write every character of \(A\) as \(\chi_j(a)=aj/4+\mathbb Z\), \(j\in C_4\). Restriction at the generator \(2\) of \(H\) has value \(j/2+\mathbb Z\). It equals \(\theta(2)\) exactly when \(j\) is odd. Hence the two extensions are \(\chi_1,\chi_3\).

Under \(H\simeq C_2\), the full restriction map is reduction

Its kernel consists of \(\chi_0,\chi_2\). The quotient \(A/H\) is \(C_2\); pulling its character with value \(t/2+\mathbb Z\) at the quotient generator back to \(A\) gives \(\chi_{2t}\). This is an injective map whose image is exactly the kernel just listed. Each of the two restriction fibres has two elements, checking the product count in (5.4). Evaluation still distinguishes \(1\) from \(3\), since \(\chi_1\) takes the different values \(1/4,3/4\) there.

Exercise 4 — reverse a noninvertible finite-group map

Advanced. For a homomorphism of finite abelian groups \(f:A\to B\), prove that \(D(f)\) is surjective exactly when \(f\) is injective, and injective exactly when \(f\) is surjective. For \(f:C_4\to C_6\) with \(f(1)=3\), compute \(D(f)\), its kernel and image in cyclic coordinates, and compare them with the original map.

Solution. If \(f\) is injective, identify \(A\) with the subgroup \(f(A)\). A character on \(A\) defines a character there by \(f(a)\mapsto\chi(a)\). The retained extension theorem extends it to \(B\), proving surjectivity of \(D(f)\). If \(f\) is not injective, choose nonzero \(a\in\ker f\). A character on \(A\) separating \(a\) from zero exists by Lemma 5.1. Every character pulled back from \(B\) vanishes at \(a\), so this character cannot be in the image. Thus \(D(f)\) is not surjective.

If \(f\) is surjective, two characters on \(B\) agreeing after \(f\) agree at every element of \(B\), proving injectivity of \(D(f)\). If \(f\) is not surjective, the finite quotient \(B/f(A)\) is nontrivial. Choose a nonzero class and a character separating it from zero. Pulling that character back to \(B\) gives a nonzero character vanishing on \(f(A)\). It and the zero character have the same pullback along \(f\), disproving injectivity. These arguments include maps involving the trivial group.

For the given \(f\), the relation \(4\cdot3=0\) in \(C_6\) makes it well-defined. Its kernel is \(\{0,2\}\), and its image is \(\{0,3\}\). Precomposing \(\chi_j:C_6\to E\) gives value \(3j/6=j/2\) at \(1\in C_4\). In the \(C_4\) coordinates this is \(2j/4\), so

Its kernel is \(\{0,2,4\}\), and its image is \(\{0,2\}\). The original map is neither injective nor surjective; its dual has neither property, consistently with both proved equivalences. The sizes also agree with the annihilator calculations: characters vanishing on the two-element image form a group of size \(6/2=3\), whereas the characters factoring through the two-element quotient by \(\ker f\) form its two-element dual.

7. References

• Categorical proofs retained in full: Relations and cancellation in categories, Zero maps, components and subobjects, and Equivalences and chosen representatives.
• Cyclic characters retained in full: Character duals and one-sided injectivity, Exercise 1, including cyclic and finite direct-sum evaluation, restriction and reduction.
• Stronger general extension proof: The Stacks Project authors, Lemma 15.55.1, divisible abelian groups are injective. The consulted pinned structured source retains the complete maximal-extension and one-generator argument. Its licence is GFDL 1.2 or later; no source expression is incorporated here.
• Stronger evaluation proof: The Stacks Project authors, Lemma 15.56.7, with its complete cyclic-character extension argument. It has the same GFDL licence; no source expression is incorporated here.