Equivalences and chosen representatives

Written by GPT-6.1 Sol (OpenAI), October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Public domain (CC0).

An equivalence lets us replace objects by convenient models while retaining every map between them. The replacement need not preserve object labels. This distinction explains why matrices model finite-dimensional vector spaces, why subobjects form an ordered set, and why a faithful functor can still fail to identify its source with a subcategory of its target.

We assume the category and size conventions of Universes and small categories, the functor and cancellation results of Relations and cancellation in categories, and the component and lifting results of Natural transformations and composition of functors. For subobjects and connectedness we use the exact proofs in Zero maps, components and subobjects. All object collections and Hom sets are sets in ambient set theory. Ambient choice is available; no fixed universe is required to contain all the object labels.

Basic references are the open [Stacks] treatment of quasi-inverses, Riehl's Category Theory in Context, Section 1.5, and the supplementary discussions in [Gaillard]. The precise open construction used below is linked where it enters. Its existing argument retains its own licence. The calculations and examples written here are independent exposition.

1. Recognizing a reversible change of category

An equivalence is a functor \(F:C\to D\) for which there are a functor \(G:D\to C\) and natural isomorphisms

The functor \(G\) is a quasi-inverse. An isomorphism of categories instead has an inverse with both composites literally equal to the identity functors. The equations in (1.1) allow an object to return with a different label and a specified isomorphism to its original value. Initial choices of \(\alpha,\beta\) need not satisfy any additional coherence equations.

Theorem 1.1. A functor is an equivalence if and only if it is fully faithful and essentially surjective.

Proof of necessity. For \(f:X\to Y\), naturality of \(\alpha\) gives

Thus \(GF\) is faithful, since conjugation by isomorphisms is injective. If \(F(f)=F(f')\), apply \(G\) and use (1.2) to get \(f=f'\). The same argument with \(\beta\) makes \(FG\) faithful and then makes \(G\) faithful.

Given \(u:FX\to FY\), put \(f=\alpha_YG(u)\alpha_X^{-1}\). Equation (1.2) applied to this \(f\) says \(GF(f)=G(u)\). Faithfulness of \(G\) therefore gives \(F(f)=u\). Hence \(F\) is full as well as faithful. Finally \(\beta_Z:FGZ\to Z\) exhibits every \(Z\in D\) as isomorphic to an image object. This proves essential surjectivity.

Construction and checks for sufficiency. We use the chosen-object construction of Stacks, quasi-inverse lemma, as used in its equivalence criterion. In that notation \(j=G\) and \(i_Z=\beta_Z^{-1}\). Choose an object \(GZ\) and an isomorphism \(\beta_Z:FGZ\to Z\) for each \(Z\in D\). Full faithfulness determines the arrow \(G(u):GZ\to GW\) uniquely by

Here are the functor checks left implicit in that open argument. For \(u=1_Z\), the right side is \(1_{FGZ}=F(1_{GZ})\), so faithfulness gives \(G(1_Z)=1_{GZ}\). For \(u:Z\to W\) and \(v:W\to V\),

Faithfulness gives the composition law. Equation (1.3) is exactly naturality of \(\beta:FG\Rightarrow1_D\); its chosen components are invertible.

For the other comparison, postcomposition with \(F\) is fully faithful on functor categories by Theorem 2.2 of Natural transformations and composition of functors. Apply its unique natural-transformation lift to \(\beta F : FGF\Rightarrow F\). This gives \(\alpha:GF\Rightarrow1_C\) with \(F(\alpha_X)=\beta_{FX}\). Each \(\alpha_X\) is invertible because a fully faithful functor reflects isomorphisms, by Proposition 3.2 of Relations and cancellation in categories. The component criterion then makes \(\alpha\) a natural isomorphism. This supplies the second comparison without repeating the already proved natural-lifting argument.

All choices are indexed by the ambient object set of \(D\). If \(D\) is empty, the existence of \(F\) forces \(C\) to be empty and the construction is vacuous. If \(C\) is empty and \(F\) is essentially surjective, \(D\) must be empty. Thus the theorem includes both boundary cases. \(\square\)

The criterion concerns all arrows between each fixed pair of objects. It does not say that a faithful functor is an equivalence onto its literal arrow image. We will examine that distinction in Section 7.

2. Coherent comparisons and detecting natural isomorphisms

The comparisons of an equivalence can be chosen to work together, even when two initially supplied comparisons do not.

Proposition 2.1. For quasi-inverse \(F:C\to D\) and \(G:D\to C\), there are natural isomorphisms as in (1.1) satisfying both equations

Proof. Keep any given natural isomorphism \(\beta:FG\Rightarrow1_D\). Theorem 1.1 makes \(F\) fully faithful. Its fully faithful postcomposition functor uniquely lifts \(\beta F\) to a natural transformation \(\alpha:GF\Rightarrow1_C\). Reflection of component isomorphisms makes this lift a natural isomorphism. This proves the first equation.

Naturality of \(\beta\) for the particular arrow \(\beta_Y:FGY\to Y\) gives

Cancel the invertible \(\beta_Y\). By the first equation in (2.1), \(F(\alpha_{GY})=\beta_{FGY}\), while cancellation in (2.2) gives \(FG(\beta_Y)=\beta_{FGY}\). Faithfulness of \(F\) now implies \(\alpha_{GY}=G(\beta_Y)\). This is the second equation. For empty categories every family and equation is empty, so the same proof includes them. \(\square\)

The two equations are the triangle equations for an adjoint equivalence, with unit \(\alpha^{-1}\) and counit \(\beta\). This result applies to two different categories. The tensor-inverse result in Tensor duality and coherent inverses gives the corresponding special case for autoequivalences under composition.

There is another useful detection result that needs essential surjectivity alone.

Proposition 2.2. If \(Q:J\to I\) is essentially surjective, then precomposition

is faithful and conservative. It need not be full.

Proof. For functors \(A,B:I\to C\) and a natural transformation \(\eta:A\Rightarrow B\), choose for any \(i\in I\) an isomorphism \(u:Qj\to i\). Naturality determines the component at \(i\):

If two transformations have the same restriction, their components at every \(Qj\) agree. Equation (2.4) makes their components at every \(i\) agree, proving faithfulness. If \(Q^*\eta\) is invertible, all \(\eta_{Qj}\) are invertible by the component criterion. Equation (2.4) then makes all \(\eta_i\) invertible, so \(\eta\) is invertible by Lemma 2.1 of Natural transformations and composition of functors. This proves conservativity. No lift of any arrow of \(I\) was used. If \(J\) is empty, essential surjectivity forces \(I\) to be empty; their functor categories each have one object and one arrow, and the assertions still hold. Exercise 3 gives a failure of fullness. \(\square\)

3. Choosing convenient object labels

We can replace the values of a functor by isomorphic objects in a full subcategory. The comparison maps, rather than just the chosen objects, determine how arrows must change.

Proposition 3.1. Let \(K\subseteq D\) be full, with inclusion \(j\), and let \(F:C\to D\). Suppose every \(FX\) is isomorphic to some object of \(K\). There are a functor \(T:C\to K\) and a natural isomorphism \(w : F\Rightarrow jT\). Given another pair \((T',w')\), there is a unique natural isomorphism \(a:T'\Rightarrow T\) satisfying \(w=(ja)w'\).

Proof. Choose \(TX\in K\) and \(w_X:FX\to TX\). Define

Fullness puts this arrow in \(K\). An identity goes to \(w_X1_{FX}w_X^{-1}=1_{TX}\), and for composable \(f,g\),

Thus \(T\) is a functor. The equation \(T(f)w_X=w_YF(f)\) is naturality of \(w\), and all its components are isomorphisms.

For the second pair, fullness uniquely regards \(a_X=w_X(w'_X)^{-1}\) as an arrow of \(K\). Using (3.1) for both pairs gives

Hence \(a\) is natural and componentwise invertible. Any comparison compatible with \(w,w'\) must have these components, proving uniqueness. The statement does not assert uniqueness of all isomorphisms between the underlying functors. \(\square\)

Corollary 3.2. Every ambient category has a full subcategory \(K\) containing exactly one object from each isomorphism class, and \(K\hookrightarrow C\) is an equivalence. Isomorphic objects of \(K\) are equal.

Proof. Isomorphism is an equivalence relation on the ambient object set: identities, inverses and composition prove its three properties. Ambient choice selects a representative from each class. Let \(r_X\) be the representative of \(X\), and choose \(q_X:X\to r_X\), taking \(q_X=1_X\) on representatives. Proposition 3.1 applied to \(1_C\) defines

On \(K\) this functor is literally the identity, so \(rj=1_K\). The invertible \(q\) supplies the other comparison, proving the equivalence. If two representatives are isomorphic they belong to the same class and hence are the same chosen object. For empty \(C\), \(K\) is empty and the proof is vacuous. \(\square\)

Such a subcategory is called a skeleton. A skeleton does not remove automorphisms or other endomorphisms: it retains every arrow between its chosen objects.

Corollary 3.3. A fully faithful \(F:C\to D\) factors literally through an equivalence \(C\to H\), where \(H\) is the full subcategory of \(D\) on the object set \(F(\operatorname{Ob}C)\).

Proof. This image is an ambient set. A full subcategory on any ambient object subset inherits identities, composition and the category laws. The restriction \(C\to H\) has the same Hom maps as \(F\), hence is fully faithful. Every object of \(H\) is an actual image, so the restriction is essentially surjective. Apply Theorem 1.1. The composite with the inclusion agrees with \(F\) on all objects and arrows. \(\square\)

4. A small model and large labels

A category is locally \(\mathcal U\)-small when each Hom set is bijective to a member of \(\mathcal U\). It is \(\mathcal U\)-small when it is locally \(\mathcal U\)-small and its object set is \(\mathcal U\)-small. These conditions are size bounds up to bijection. An object label or an arrow label need not literally belong to \(\mathcal U\).

A category is essentially \(\mathcal U\)-small if it is equivalent to a \(\mathcal U\)-small category.

Proposition 4.1. An ambient category \(C\) is essentially \(\mathcal U\)-small if and only if it is locally \(\mathcal U\)-small and has a \(\mathcal U\)-small subset of objects meeting every isomorphism class.

Proof. Suppose \(E:K\to C\) is an equivalence with \(K\) \(\mathcal U\)-small. Theorem 1.1 makes its Hom maps bijective. For arbitrary \(X,Y\in C\), choose isomorphisms \(u:EA\to X\), \(v:EB\to Y\). The map

is a bijection, by full faithfulness and invertible conjugation. Thus every target Hom set is \(\mathcal U\)-small. The subset \(S=E(\operatorname{Ob}K)\) meets every isomorphism class by essential surjectivity. It is \(\mathcal U\)-small: it is a quotient of the small object set by equality of images, and the quotient closure is proved in Proposition 3.1 of Universes and small categories.

Conversely, take the full subcategory \(H\) on such a subset \(S\). Its object set is \(\mathcal U\)-small and its Hom sets inherit the local bound. Thus \(H\) is \(\mathcal U\)-small. The inclusion is fully faithful and essentially surjective, so Theorem 1.1 makes it an equivalence. No choice of literally small labels has been added. If such an encoding is wanted, Proposition 4.1 of Universes and small categories supplies the complete construction. \(\square\)

The category with object set \(\mathcal U\) and exactly one arrow between each pair is essentially \(\mathcal U\)-small, despite its large label set: choose one object, and its one-object full subcategory is a small model. The explicit comparison is also given in Exercise 4 of Universes and small categories. By contrast, \(\mathcal U\text{-}\mathsf{Set}\) is locally \(\mathcal U\)-small but is not essentially \(\mathcal U\)-small. Example 4.2 of that lesson gives the complete Cantor argument; a small family of representatives cannot contain all its isomorphism classes. We use that proof directly.

5. Models made from matrices, monoids and orders

Example 5.1. Fix any field \(k\). Let \(\mathsf M_k\) have objects the nonnegative integers and

Composition is matrix multiplication. For an \(m\times n\) matrix \(A\) and an \(\ell\times m\) matrix \(B\), the entry of \(BA\) in row \(i\), column \(j\) is \(\sum_{r=1}^m B_{ir}A_{rj}\). Distributivity and interchange of finite sums show that both bracketings of a triple product have entries \(\sum_{r,t}C_{ir}B_{rt}A_{tj}\). The identity matrix has the required two unit laws. Empty sums are zero; a matrix with zero rows or columns is the unique matrix of that shape. In particular the \(0\times0\) identity is the empty matrix. These observations verify the category laws also when a dimension is zero.

The functor \(V:\mathsf M_k\to\mathsf{Vect}^{\mathrm{fd}}_k\) sends \(n\) to \(k^n\) and \(A\) to \(x\mapsto Ax\). The entry formula proves the composition law, and identity matrices act as identities. A linear map \(k^n\to k^m\) is uniquely determined by the images of the \(n\) standard basis vectors; their coordinate columns form its matrix. This also holds at \(n=0\) or \(m=0\), where the only map is the zero map. Hence every Hom map is bijective. Every finite-dimensional space has a finite basis and is isomorphic to \(k^n\) for its dimension \(n\). Thus \(V\) is an equivalence by Theorem 1.1, over every field.

Example 5.2. Suppose \(C\) is nonempty and every two of its objects are isomorphic. Choose \(A\in C\). Its endomorphisms form a monoid \(M\), with multiplication given by composition and unit \(1_A\). The one-object monoid category \(BM\) is its full subcategory on \(A\). Its inclusion is fully faithful and essentially surjective, so it is an equivalence by Theorem 1.1. The monoid need not be a group: the one-object category of \(\{1,p\}\), with \(p^2=p\ne1\), already satisfies the object hypothesis and has a noninvertible arrow.

If \(C\) is a connected groupoid, it is nonempty by the convention in Section 4 of Zero maps, components and subobjects. Any two objects are joined by a finite zigzag. Each backward arrow is invertible, so reversing it and composing the resulting forward arrows gives an isomorphism between the two objects. The preceding construction applies, and \(\operatorname{End}(A)\) is now a group because every arrow is invertible. Thus a connected groupoid has a one-object group model.

Example 5.3. Fix \(X\in C\). Let \(\mathsf{Mono}/X\) be the full subcategory of the slice \(C/X\) on the monomorphisms into \(X\). A map \((A,f)\to(B,g)\) is an arrow \(h:A\to B\) satisfying \(f=gh\). Because \(g\) is monic, there is at most one such \(h\).

Let \(\mathsf{Sub}(X)\) be the category of subobject classes, with one arrow \([f]\to[g]\) precisely when \([f]\le[g]\). Section 3 of Zero maps, components and subobjects proves, in full, that the classes form an ambient set, the order is independent of representatives, and the factorization exists precisely for the stated order relation and is unique. It also proves reflexivity, transitivity and antisymmetry. We reuse those results.

The quotient assignment \(P(f)=[f]\) sends \(h\) to the unique order arrow. Identities and composites are preserved since the order category has at most one arrow with prescribed endpoints and the corresponding factorization equations compose. The complete representative-independence proof just cited gives a bijection on every Hom set: either both are empty or both are singletons. The object map is onto the class set. Hence \(P\) is an equivalence by Theorem 1.1. This statement uses the ambient set of monomorphisms and its quotient, without assuming that either is \(\mathcal U\)-small.

6. Opposite diagrams and empty constructions

Taking opposite categories also reverses the direction of a natural transformation. Keeping both reversals visible gives a literal category isomorphism

To verify it, send the object \(F\) to the already constructed functor \(F^{\mathrm{op}}\). An arrow from \(F\) to \(G\) on the left is a transformation \(\eta:G\Rightarrow F\) before taking opposites. Send it to the components \(\eta_X^{\mathrm{op}}:FX\to GX\) in \(D^{\mathrm{op}}\). For \(f:X\to Y\) in \(C\), original naturality is \(F(f)\eta_X=\eta_YG(f)\). Reversing this equality gives

This is the naturality equation for the arrow \(f^{\mathrm{op}}:Y\to X\), with the required reversed components. Identities go to identities. For consecutive arrows on the left represented by \(\eta:G\Rightarrow F\) and \(\zeta:H\Rightarrow G\), their composite is represented by \(\eta\zeta:H\Rightarrow F\). Its image has component

the composite of the two image transformations. Taking opposites again reverses the same component arrows and returns every original functor and transformation. With the convention that opposite categories retain object and arrow labels, these are literal inverse functors. The argument includes empty categories.

For an empty family of categories, the product has exactly one object, the empty object tuple, and exactly one arrow, the empty arrow tuple. Identities and composition are its unique possible operations. Thus it is isomorphic to the one-object one-arrow category \(\mathsf{Pt}\). The disjoint union of an empty family has no objects and no arrows, and is isomorphic to the empty category. The two constructions are different even though their indexing sets are both empty.

7. When a faithful functor has a literal image

Call \(F:C\to D\) half-full when it reflects isomorphism classes of objects: \(FX\cong FY\) implies \(X\cong Y\). A subcategory is half-full when its inclusion has this property. This terminology concerns existence of some source isomorphism. It does not require a given target isomorphism to lift.

First consider a condition that really does give a literal image.

Proposition 7.1. If \(F\) is faithful and injective on objects, it induces an isomorphism of \(C\) with a subcategory \(H\subseteq D\) containing exactly its object and arrow images.

Proof. Each image object has a unique source preimage. Therefore the subsets

are well-defined. They contain identities because \(F(1_X)=1_{FX}\). For composable image arrows \(F(f):FX\to FY\) and \(F(g):FY\to FZ\), injectivity on objects ensures that their middle source object is the same \(Y\). Their composite is \(F(gf)\), so the subsets are closed under composition. Inherited operations give a subcategory. The restriction \(C\to H\) is bijective on objects and, by faithfulness and (7.1), on every Hom set. The inverse bijections preserve identities and composition: the unique lift of \(F(g)F(f)=F(gf)\) is \(gf\). Thus they form a functor and give a literal inverse. No half-fullness hypothesis was needed. \(\square\)

Without injectivity on objects, even faithfulness together with half-fullness does not give (7.1).

Example 7.2. Let \(E\) have objects \(0,1\) and one arrow \(h_{ij}:i\to j\) for every pair. Define \(h_{jk}h_{ij}=h_{ik}\). The identity at \(i\) is \(h_{ii}\); each arrow has inverse \(h_{ji}\). All category laws hold because there is exactly one arrow between given endpoints.

Let \(BC_2\) be the one-object group category of \(\{1,s\}\), where \(s^2=1\ne s\). Define

with exponents modulo two. Identity exponents are zero, and \(s^{k-j}s^{j-i}=s^{k-i}\) proves the composition law. Each source Hom set is a singleton, so \(F\) is faithful. Every pair of source objects is isomorphic, so \(F\) is half-full.

The four Hom images are nevertheless

All four pairs have the same target endpoint pair \((*,*)\). Formula (7.1) would therefore assign different subsets to the same target Hom set. The table displays the complete obstruction; the issue is comparison between different source pairs, not injectivity within a fixed Hom set.

Any subcategory of \(BC_2\) containing every arrow image contains both \(1\) and \(s\), so it is the whole target. The restricted \(F\) is not full: its map from \(\operatorname{End}_E(0)\) misses \(s\). Moreover no equivalence between these categories exists, since full faithfulness would biject a singleton endomorphism set with a two-element one.

Remark. For a faithful half-full functor, the literal image need not be equivalent to the source: Example 7.2 is a counterexample, and Theorem 8.1 gives a replacement with a specified natural comparison.

8. Transported images preserve all object values

The obstruction can be removed by changing the arrow assignment through specified comparison isomorphisms. Half-fullness then supplies exactly the representative consistency that is needed.

Theorem 8.1. If \(F:C\to D\) is faithful and half-full, there are a subcategory \(H\subseteq D\), an equivalence \(T:C\to H\), and a natural isomorphism \(\theta:F\Rightarrow jT\), where \(j\) is the inclusion, such that

The inclusion \(j\) is faithful and half-full. The theorem does not require \(H\) to contain all the original arrows \(F(f)\).

Proof. Choose the skeleton and comparisons of Corollary 3.2. Write \(r_X\) for the representative, \(q_X:X\to r_X\) for the chosen isomorphism, and \(r(f)=q_Yfq_X^{-1}\). This last map is a bijection from \(\operatorname{Hom}_C(X,Y)\) to \(\operatorname{Hom}_K(r_X,r_Y)\), with inverse \(a\mapsto q_Y^{-1}aq_X\).

Put \(S=F(\operatorname{Ob}C)\). For each actual object \(Z\in S\), choose \(X_Z\) with \(FX_Z=Z\), and set

If \(FX=Z\), half-fullness makes \(X\) isomorphic to \(X_Z\), so their chosen representatives are equal: \(r_X=r_Z\). Thus the representative is independent of the preimage. The transport \(t_Z\) is a single fixed choice for the actual target object \(Z\).

Define subsets of target Hom sets by

The identity is obtained from \(a=1_{r_Z}\). For \(a:r_Z\to r_W\), \(b:r_W\to r_V\),

Hence the subsets are composition-closed; inherited operations make \(H\) a subcategory on \(S\).

Define \(TX=FX\) and

Its endpoints are valid since \(r_{FX}=r_X\) and \(r_{FY}=r_Y\). The identity and composition calculations for \(r\) and (8.4) give the functor laws. On a fixed Hom set, (8.5) is a composite of three bijections: conjugation to the representative Hom set; the faithful map \(F\) onto its image; and conjugation by the fixed \(t\)'s onto (8.3). Thus \(T\) is fully faithful. Its object map is onto \(S\), so Theorem 1.1 makes it an equivalence, including the full quasi-inverse and natural comparisons constructed there.

Set \(\theta_X=t_{FX}^{-1}F(q_X):FX\to FX\). This is invertible, and

So \(\theta\) is the specified natural isomorphism. The inclusion is faithful because its Hom maps are inclusions. If \(Z,W\in H\) are isomorphic in \(D\), then (8.2) makes \(F(r_Z),F(r_W)\) isomorphic. Half-fullness and the skeleton property imply \(r_Z=r_W\). Now \(t_W^{-1}t_Z\) and \(t_Z^{-1}t_W\) are inverse arrows in \(H\), obtained from the identity of this common representative. Thus \(j\) is half-full.

If \(C\) is empty, \(K,S,H\) are empty and all the constructions are vacuous. A functor into empty \(D\) already forces this case. No extra size or nonemptiness hypothesis has been imposed. \(\square\)

In Example 7.2, take \(H\) to have only its object \(*\) and identity arrow. The constant arrow assignment \(T(h_{ij})=1\) is an equivalence. The components \(\theta_i=s^i\) satisfy \(\theta_jF(h_{ij})=s^i=T(h_{ij})\theta_i\). Thus the comparison preserves object values while changing arrow values. The missing arrow \(s\) is precisely why this does not rescue the failed literal-image assertion.

9. Tags separate objects for every faithful functor

Even half-fullness is unnecessary if we may enlarge the target by harmless object tags.

For a nonempty ambient set \(S\), let \(E(S)\) have objects \(S\) and one arrow \(e_{st}:s\to t\) per pair, with \(e_{tu}e_{st}=e_{su}\). Unique arrows prove associativity, identities and inverses. The product \(D\times E(S)\) has object pairs \((Y,s)\) and arrow pairs \((f,e_{st})\). Its coordinatewise identities and composition obey the category laws because the two factors do.

Proposition 9.1. The projection \(\pi:D\times E(S)\to D\) is an equivalence.

Proof. Choose \(s_0\in S\), and define \(L(Y)=(Y,s_0)\), \(L(f)=(f,e_{s_0s_0})\). Its identity and composition laws hold coordinatewise, and \(\pi L=1_D\). The arrows \((1_Y,e_{s_0s}):L\pi(Y,s)\to(Y,s)\) are invertible. For any \((f,e_{st})\), both sides of their naturality equation are \((f,e_{s_0t})\). They therefore give \(L\pi\cong1\), proving the equivalence. Empty \(D\) is allowed. Nonempty \(S\) is essential when \(D\) is nonempty. \(\square\)

Proposition 9.2. Every faithful functor \(F:C\to D\) factors literally through an isomorphism onto a subcategory of \(D\times E(S)\), followed by projection, for some nonempty ambient set \(S\).

Proof. Take the disjoint union \(S=\operatorname{Ob}C\sqcup\{\star\}\), so source labels have distinct tags and \(S\) is nonempty even for empty \(C\). Define

Here \(X,Y\) denote their tags in the disjoint union. The functor laws follow from those of \(F\) and \(E(S)\). The tagged object map is injective; faithfulness follows from the first arrow coordinate. Proposition 7.1 therefore identifies \(C\) isomorphically with the literal image subcategory \(H\) of \(\widehat F\). Projecting its objects and arrows gives exactly \(F\), and the ambient projection is an equivalence by Proposition 9.1. \(\square\)

A change of target is sometimes necessary. For a concrete example, let \(\mathsf A\) have two objects \(0,1\), their identities, and one arrow \(a:0\to1\). Let \(\mathsf P\) be the one-object monoid category with arrows \(1,p\), where \(p^2=p\ne1\). Send both objects to its one object, their identities to \(1\), and \(a\) to \(p\). Every source Hom set is empty or a singleton, so the functor is faithful; its only nonidentity arrow composes only with identities, which verifies the functor law.

Nevertheless no subcategory of \(\mathsf P\) is equivalent to \(\mathsf A\). An empty subcategory cannot receive an equivalence from nonempty \(\mathsf A\). A nonempty subcategory has one object. Any equivalence from \(\mathsf A\) to it would send \(0,1\) to that same object, and its full faithfulness would lift the identity in the missing reverse Hom set \(\operatorname{Hom}_{\mathsf A}(1,0)=\varnothing\), a contradiction.

10. Exercises with solutions

Exercise 1 — Introductory: four labels and two positions

Let \(C\) have objects \(a,b,c,d\). Assign height zero to \(a,b\) and height one to \(c,d\). There is one arrow \(X\to Y\) if the height of \(X\) is at most the height of \(Y\), and none otherwise; composition is the unique possible arrow. Prove that the height functor \(q:C\to[1]\) is an equivalence but not an isomorphism. Give an explicit quasi-inverse and both comparisons. Determine the endomorphism monoids and the automorphism groups of the objects of \(C\).

Solution. Reflexivity of the height order supplies identities, transitivity supplies composites, and uniqueness proves the unit and associativity laws. The height assignment preserves the endpoints of every arrow and sends it to the unique corresponding order arrow in \([1]\), so it is a functor. For each fixed \(X,Y\), its Hom map is either the bijection between singletons or the unique bijection between two empty sets. Hence it is fully faithful. Both heights occur, making it essentially surjective; Theorem 1.1 gives an equivalence. It is not an isomorphism because a functor with a literal inverse must be bijective on objects, while this object map identifies two labels at each height.

Choose \(r(0)=a\), \(r(1)=c\), and let \(r(0\to1)\) be the unique arrow \(a\to c\). Its identity and composition laws follow from uniqueness, and \(qr=1_{[1]}\). At any \(X\), the unique arrow \(rqX\to X\) connects two objects of the same height; the unique reverse arrow is its inverse. For \(f:X\to Y\), both composites in the comparison square are the unique arrow \(rqX\to Y\), so these components give \(rq\cong1_C\). This is the second comparison, while the first is the identity of \(qr\). Every endomorphism Hom set of \(C\) is a singleton. Its monoid consists only of the identity, and its automorphism group is therefore the trivial group. The two labels at a height are isomorphic, but they are different objects.

Exercise 2 — Intermediate: changing a monoid model

Let \(C\) be nonempty with every two objects isomorphic. Fix \(A\) and isomorphisms \(q_X:X\to A\). Construct the functor \(T:C\to B\operatorname{End}(A)\) determined by these choices. For a second family \(q'_X\), find the specified natural isomorphism between the two functors. Finally, for an isomorphism \(u:A\to B\), describe the monoid isomorphism \(\operatorname{End}(A)\to\operatorname{End}(B)\) and compute how it changes when \(u\) changes.

Solution. Send every object to the unique object and set \(T(f)=q_Yfq_X^{-1}\). An identity goes to \(1_A\), and cancellation of the middle \(q_Y^{-1}q_Y\) proves the composition law. Each Hom map is a bijection, with inverse \(m\mapsto q_Y^{-1}mq_X\), and the target object has a source preimage because \(C\) is nonempty. Thus \(T\) is an equivalence by Theorem 1.1, with no assertion that \(\operatorname{End}(A)\) is a group.

Put \(a_X=q'_Xq_X^{-1}\in\operatorname{Aut}(A)\). Then \(T'(f)=a_YT(f)a_X^{-1}\), so \(T'(f)a_X=a_YT(f)\). The \(a_X\)'s are therefore the components of a natural isomorphism \(T\Rightarrow T'\). Their expression is forced if the isomorphism must be compatible with the two chosen families; it need not be the only natural isomorphism between the underlying functors.

Conjugation \(m\mapsto umu^{-1}\) preserves the monoid unit and products, and conjugation by \(u^{-1}\) is its inverse. If another isomorphism is \(u'=vu\), where \(v=u'u^{-1}\in\operatorname{Aut}(B)\), its conjugation map is

Thus the two monoid identifications differ by inner conjugation on the target. When \(C\) is a groupoid these are group isomorphisms; without that hypothesis the calculation is still a monoid calculation.

Exercise 3 — Advanced: an invertible restricted map need not extend

Let \(I=[1]\), let \(J\) be the discrete category on its two object labels, and let \(Q:J\to I\) be the identity on objects. Let \(K=\{0,1\}\) and let \(A:I\to\mathsf{Set}\) be constant at \(K\), with its nonidentity arrow acting as \(1_K\). Compute the endomorphism sets and automorphism groups of \(A\) and \(Q^*A\), and identify the restriction map. Explain why this does not contradict faithfulness or conservativity of \(Q^*\).

Solution. The functor \(Q\) is essentially surjective, since every target object is an actual image. A transformation \(A\Rightarrow A\) consists of functions \(f_0,f_1:K\to K\) satisfying naturality for \(0\to1\), namely \(f_1=f_0\). There are four functions \(K\to K\), so the endomorphism set has four elements; composition is ordinary function composition on their common component. Its invertible elements are the identity and the transposition, so the automorphism group is \(S_2\).

In the discrete restriction there is no equation relating the two components. Its endomorphism set is the product of two four-element function monoids, hence has sixteen elements. Its automorphism group is \(S_2\times S_2\). Restriction is the diagonal map \(f\mapsto(f,f)\) on both the endomorphism monoid and the automorphism group. It is injective but not surjective. In particular the restricted automorphism \((1_K,(0\ 1))\) has no extension to a transformation of \(A\).

Faithfulness asserts injectivity for transformations that already exist on \(I\); this diagonal map has that property. Conservativity concerns a transformation on \(I\) whose restriction is invertible. Such a transformation is \((f,f)\); if its restricted components are bijective, then \(f\) is bijective and its inverse also gives a transformation on \(I\). Conservativity does not assert that every invertible arrow in the restricted functor category lifts. Hence there is no contradiction.

Exercise 4 — Expert: all comparisons for weighted arrows

Let \(S\) be any nonempty ambient set, let \(G\) be any group, and choose elements \(g_s\in G\). Define \(F_g:E(S)\to BG\) by sending every object to the unique object and

Prove that this is faithful and half-full. Determine exactly when it induces an equivalence to some subcategory of \(BG\) containing every literal arrow image. Let \(H\) be the identity-only subcategory of \(BG\) and \(T:E(S)\to H\) its unique functor. Classify all natural isomorphisms \(F_g\Rightarrow jT\). Your answer must allow noncommutative groups and infinite \(S\).

Solution. Identities map to \(g_sg_s^{-1}=1\). For a composable pair,

so the assignment is a functor. Every source Hom set is a singleton, proving faithfulness. Every pair of source objects is isomorphic, proving half-fullness.

A subcategory containing the image object has an endomorphism submonoid \(M\subseteq G\). If the restricted \(F_g:E(S)\to BM\) is an equivalence, full faithfulness on \(\operatorname{End}_{E(S)}(s)\) makes \(M=\{1\}\), since that source monoid contains only its identity. The required containment of all arrow images then gives \(g_tg_s^{-1}=1\) for every \(s,t\), hence the family \(g_s\) is constant. Conversely, for a constant family every arrow image is \(1\); the functor to the identity-only subcategory is fully faithful and essentially surjective, so it is an equivalence. This proves the exact criterion. It does not assume that a general subcategory of \(BG\) is itself a group category.

The unique functor \(T\) is an equivalence: choose one \(s_0\in S\), and the unique arrows supply its quasi-inverse comparison, as in Proposition 9.1 with a one-object first factor. A natural isomorphism \(\theta:F_g\Rightarrow jT\) consists of elements \(r_s\in G\) satisfying

Thus \(r_sg_s\) must be a single constant \(c\in G\), and all solutions are

Conversely, any \(c\in G\) gives these invertible components and satisfies (10.4) by cancellation in the displayed order. Since \(S\) is nonempty, one component recovers \(c\), so the comparisons are in bijection with \(G\). No factors were commuted, and no finiteness was used. Even though the transported functor is fixed, its comparison with \(F_g\) need not be unique without a further compatibility condition.

References

• [Stacks] The Stacks Project Authors, The Stacks Project, Categories, “Constructing a quasi-inverse” and the equivalence criterion, in the AI-integrated pinned source edition. The exact native locators are lemma-construct-quasi-inverse and lemma-equivalence-categories in Categories. The linked open argument retains the GNU Free Documentation License.
• [Riehl] Emily Riehl, Category Theory in Context, Section 1.5, for equivalences of categories and the image of a functor.
• [Gaillard] Pierre-Yves Gaillard, About Categories and Sheaves, 2019, supplementary notes. Discussions of labelled arrows and smallness.