Generators and small quotient families

A small family of probes can detect isomorphisms and encode subobjects by small sets. The probes must retain their separate outputs. An empty Hom set can erase information if all outputs are multiplied together.

We use small Hom sets, a fixed universe of small indexing sets, and the Yoneda identification of maps from representable presheaves with elements. The basic presheaf convention is the one in Ind-objects through their elements. A coproduct indexed by a small set is always meant in the category under discussion.

1. Keep the family of probes

Let $\mathcal C$ be a locally small category and $(G_i)_{i\in I}$ a small family. Its probe functor is $\begin{gathered} H:\mathcal C\longrightarrow\operatorname{Set}^{I},\\ H(X)=(\operatorname{Hom}_{\mathcal C}(G_i,X))_{i\in I}, \end{gathered} \tag{1.1}$ where $I$ is regarded as a discrete category. Isomorphisms in $\operatorname{Set}^{I}$ are componentwise bijections.

Call the family separating if $H$ is faithful, and detecting if $H$ is conservative: whenever every $\operatorname{Hom}(G_i,f)$ is bijective, $f$ is an isomorphism. A single detecting object will be called a detecting generator. A cogenerating family is defined by reversing arrows, using $\operatorname{Hom}(-,G_i)$.

The canonical terminology in Adámek–Rosický–Vitale, Definition 6.1 and Remark 6.2 calls a separating family a generator, and a separating, conservative family a strong generator. The extra condition matters. In §2 we check that, when equalizers exist, a detecting family is automatically separating, so is a strong generator in that canonical sense.

The functor in (1.1) is different from the set-valued functor $P(X)=\prod_{i\in I}\operatorname{Hom}(G_i,X). \tag{1.2}$ In $\operatorname{Set}\times\operatorname{Set}$, take $G_1=(1,\varnothing),\qquad G_2=(\varnothing,1).$ Their separate Hom functors are the two coordinate projections, so jointly detect all isomorphisms. But $P(A,B)=A\times B$. The morphism $(\varnothing,\varnothing)\longrightarrow(\varnothing,1) \tag{1.3}$ is not an isomorphism, although its image under $P$ is the bijection $\varnothing\to\varnothing$. Thus a product of the output sets can hide a failed component.

The coproduct $G_1\amalg G_2=(1,1)$ has Hom functor $P$, so is not a detecting generator. The usual identification $\operatorname{Hom}\!\left(\coprod_iG_i,X\right) \simeq\prod_i\operatorname{Hom}(G_i,X)$ does not justify replacing an arbitrary detecting family by its coproduct.

This replacement does work in a pointed category, which has a zero object. All these Hom sets then have distinguished zero maps. If the product map is bijective, each factor map is injective: place two competing elements in that coordinate and zero maps in all others. Each factor map is surjective: place any desired target element in that coordinate and target zero maps in the others, then lift the resulting tuple. Thus the product is bijective exactly when every factor is bijective. A coproduct of a detecting family is consequently a detecting generator in this pointed setting. This also covers the empty family whenever it is detecting.

Useful examples retain the separate probes. A singleton detects isomorphisms of sets. For left modules over any unital ring $R$, evaluation at $1$ identifies $\operatorname{Hom}_R(R,M)$ with the underlying set of $M$. A bijective module map has a linear inverse: apply the map to each addition and scalar-multiplication equation and use injectivity. Hence $R$ is a detecting generator.

For a small category $\mathcal A$, the representables $h_a$ form a detecting family in $\operatorname{Set}^{\mathcal A^{op}}$. Yoneda gives $\operatorname{Hom}(h_a,F)=F(a)$, and a componentwise bijective natural transformation has a natural inverse by reversing its naturality squares. This is the exact contravariant version of Adámek–Rosický–Vitale, Example 6.6. It asserts a family of generators; it makes no unsupported claim that their coproduct is a single detecting generator. A two-element set is a cogenerator in $\operatorname{Set}$, as Exercise 1 checks, including empty sets.

2. Detection gives faithfulness and tests cancellation

Theorem 2.1. Suppose $\mathcal C$ has equalizers and $(G_i)$ is detecting. Then it is separating. For any morphism $f:X\to Y$, $f$ is a monomorphism exactly when all $\operatorname{Hom}(G_i,f)$ are injective. If all these maps are surjective, $f$ is an epimorphism.

Proof. Let $a,b: X\rightrightarrows Y$ have equal images under every probe, and let $e:E\to X$ be their equalizer. Every $u:G_i\to X$ satisfies $au=bu$, so factors uniquely through $e$. Hence every $\operatorname{Hom}(G_i,e)$ is bijective. Detection makes $e$ an isomorphism. Since $ae=be$, this gives $a=b$, proving faithfulness.

A monomorphism induces injections under every Hom functor. Conversely, if the probe maps of $f$ are injective and $fr=fs$ for $r,s: Z\rightrightarrows X$, then $\operatorname{Hom}(G_i,r)=\operatorname{Hom}(G_i,s)$ for every $i$. Faithfulness gives $r=s$.

Finally suppose all probe maps of $f$ are surjective. If $af=bf$ for $a,b: Y\rightrightarrows Z$, every $u:G_i\to Y$ has a lift $v:G_i\to X$. Thus $au=afv=bfv=bu$; faithfulness gives $a=b$. This is the epimorphism cancellation condition. No converse about surjectivity is asserted. $\square$

The equalizer argument checks the exact interface from our detecting convention to the canonical strong-generator convention. It requires no coproduct and preserves all empty Hom sets.

Theorem 2.2. In a balanced category, where every morphism that is both monic and epic is invertible, any separating family is detecting. Consequently, if equalizers also exist, separation and detection are equivalent.

Proof. Suppose all probe maps of $f$ are bijective. The cancellation arguments in Theorem 2.1 use only faithfulness: their injectivity makes $f$ monic, and their surjectivity makes it epic. Balancedness makes $f$ invertible. The converse uses Theorem 2.1. $\square$

Balancedness cannot be dropped. A point separates continuous maps of topological spaces, while a continuous bijection need not have a continuous inverse. Exercise 3 gives the full witness.

3. Canonical coproduct presentations

Assume $\mathcal C$ has small coproducts and the family is separating. For $X$, form $\begin{gathered} E_X=\coprod_{\substack{i\in I\\u:G_i\to X}}G_i,\\ e_X:E_X\longrightarrow X, \end{gathered} \tag{3.1}$ whose component indexed by $(i,u)$ is $u$. The index is small because $I$ and all relevant Hom sets are small.

We retain Adámek–Rosický–Vitale, Proposition 6.3: a separating family supplies these epimorphic presentations; a strong generating family supplies extremal epimorphic presentations. An epimorphism $e$ is extremal if every factorization $e=mt$ with $m$ monic forces $m$ to be an isomorphism.

Here are the full interface checks. If $ae_X=be_X$, restricting to every coproduct component gives $au=bu$ for every probe map. Separation gives $a=b$, so $e_X$ is epic. If the family is also detecting and $e_X=mt$ with $m:B\to X$ monic, every $u:G_i\to X$ factors through $m$. Monicity makes that factorization unique. Therefore every $\operatorname{Hom}(G_i,m)$ is bijective; detection makes $m$ invertible. Thus $e_X$ is extremal.

Conversely, suppose every object is the target of an epimorphism from a small coproduct of family objects. Two arrows equal on all probe maps agree after such an epimorphism, hence agree. The family is separating. If every such presentation can be chosen extremal, a monomorphism $m:B\to X$ through which all probe maps factor receives a factorization of the presentation map; extremality makes $m$ invertible. For a morphism whose probe maps are all bijective, separation makes it monic by cancellation, and then this last property makes it invertible. This proves the complete canonical equivalence without presuming any Hom factor nonempty.

In a balanced category with equalizers and small coproducts, these arguments give three equivalent conditions for a single $G$: it is detecting; $\operatorname{Hom}(G,-)$ is faithful; every $X$ has an epimorphic presentation by a small coproduct of copies of $G$. In the forward direction (3.1) is the specified canonical presentation.

The components in (3.1) vary with $X$. This construction is valid for an arbitrary detecting family; it does not replace the family by the fixed coproduct $\coprod_iG_i$.

4. Small subobjects and strict quotients

A subobject of $X$ is a monomorphism $m:A\to X$, with two such arrows identified when there is an isomorphism of their domains commuting with their maps to $X$.

Theorem 4.1. If $\mathcal C$ has pullbacks and a small detecting family, then the subobjects of each $X$, up to this identification, form a small collection.

Proof. Associate to a subobject $m$ the tuple of subsets $\begin{gathered} S_i(m)=\operatorname{im}\bigl(\\ \operatorname{Hom}(G_i,A)\to\operatorname{Hom}(G_i,X)\bigr). \end{gathered} \tag{4.1}$ Each probe map is injective because $m$ is monic. The possible tuples belong to the small set $\prod_{i\in I}\mathcal P(\operatorname{Hom}(G_i,X)). \tag{4.2}$ This product stores subsets separately; it does not multiply the Hom sets themselves.

If $m:A\to X$ and $n:B\to X$ give the same tuple, form $D=A\times_XB$. Maps $G_i\to D$ are pairs with the same image in $\operatorname{Hom}(G_i,X)$. Equality of the two subsets and injectivity of both probe maps show that each projection $D\to A$ and $D\to B$ induces a bijection for every $i$. Detection makes both projections isomorphisms. Their composite gives the required isomorphism $A\simeq B$ over $X$. Thus the code (4.1) is injective on subobject classes, proving smallness. No small coproduct hypothesis is needed. $\square$

For a morphism $q:X\to Y$, call $q$ a strict epimorphism if, for every $Z$, precomposition with $q$ identifies $\operatorname{Hom}(Y,Z)$ with the maps $r:X\to Z$ such that $\begin{gathered} qa=qb\ \Longrightarrow\ ra=rb\\ \text{for every pair }a,b: W\rightrightarrows X. \end{gathered} \tag{4.3}$ In particular the factorization through $q$ is unique, so $q$ is epic. The definition makes sense without assuming limits or colimits.

The same condition has an exact presheaf formulation. Put $Q_q(W)=\operatorname{im}\!\left( \operatorname{Hom}(W,X)\to\operatorname{Hom}(W,Y) \right).$ A map $r:X\to Z$ satisfying (4.3) defines a natural transformation $Q_q\to h_Z$ by $qa\mapsto ra$. Equal representatives give equal results by (4.3), and precomposition proves naturality. Conversely, evaluate such a transformation at $q\in Q_q(X)$ to obtain $r$. Naturality along $a:W\to X$ gives its value $ra$ at $qa$, so these constructions are inverse. Restriction of $h_Y\to h_Z$ corresponds to precomposition with $q$. Thus strictness is exactly the assertion that, for every $Z$, $\operatorname{Hom}(Y,Z)\simeq \operatorname{Nat}(Q_q,h_Z). \tag{4.4}$

If the kernel pair $X\times_YX\rightrightarrows X$ exists, (4.3) says exactly that $q$ is its coequalizer. Indeed each competing pair $a,b$ factors through the kernel pair, while its own projections are one of the competing pairs. Thus the two equalizing conditions coincide, and the required bijection of Hom sets is precisely the coequalizer universal property.

A coequalizer of any pair is also strict: a map satisfying (4.3) equalizes that particular pair, so descends uniquely. Hence, when kernel pairs exist, strict epimorphisms are exactly regular epimorphisms, meaning coequalizers of some parallel pair. No assertion that all epimorphisms are regular is needed.

A strict quotient of $X$ is a strict epimorphism $q:X\to Y$, up to an isomorphism of targets commuting with $q$. A strict monomorphism is a strict epimorphism in the opposite category; its class, with a fixed target and isomorphisms of domains over that target, is a strict subobject.

Corollary 4.2. If $\mathcal C$ has finite limits and a small detecting family, its strict quotients of each $X$ form a small collection.

Proof. Send $q$ to its kernel pair regarded as a subobject $R_q=X\times_YX\longrightarrow X\times X. \tag{4.5}$ This arrow is monic: its two coordinates uniquely determine any arrow into the pullback. Theorem 4.1 makes the possible subobject classes small. If two quotients give the same subobject, they coequalize the same two coordinate maps $R_q\rightrightarrows X$. Since both are their coequalizer, the universal property gives unique compatible arrows in both directions. Uniqueness makes their composites identities. The quotients are therefore equivalent. The kernel-pair code is injective, proving the corollary. $\square$

This proves the strict-quotient bound without a small coproduct assumption. It does not bound arbitrary epimorphic quotients. Reversing the argument gives the corresponding strict-subobject statement under finite colimits and a small detecting cogenerating family.

5. Graded exercises with complete solutions

Exercise 1 (easy). Classify the detecting generators and detecting cogenerators of $\operatorname{Set}$. Include the empty set.

Solution. A detecting generator is exactly a nonempty set $G$. If $\operatorname{Hom}(G,f)$ is bijective, injectivity separates constant maps to two elements of the source, so $f$ is injective. Every constant map from $G$ to an element of the target has a lift, so $f$ is surjective. Its inverse is a set map. This argument also handles an empty source: a nonempty target would supply a constant map with no lift. In contrast $\operatorname{Hom}(\varnothing,X)$ is always a singleton, so $\varnothing$ does not detect a noninvertible map.

A detecting cogenerator is exactly a set $Q$ with at least two elements. Suppose precomposition $Q^Y\to Q^X$ along $f:X\to Y$ is bijective. If $f(x)=f(x')$ with $x\ne x'$, choose a function $X\to Q$ taking different values there. It cannot be pulled back from $Y$, contradicting surjectivity. If $y\notin f(X)$, two functions $Y\to Q$ differing only at $y$ pull back equally, contradicting injectivity. Hence $f$ is bijective. For $Q$ a singleton, every probe map is a bijection. For $Q=\varnothing$, the noninvertible inclusion $1\to2$ induces the bijection $\varnothing\to\varnothing$. Thus neither detects all isomorphisms.

Exercise 2 (moderate). Prove that $\operatorname{Set}\times\operatorname{Set}$ has the two-object detecting family in §1 and has no single detecting generator.

Solution. Maps from $(1,\varnothing)$ to $(A,B)$ are precisely elements of $A$; maps from $(\varnothing,1)$ are elements of $B$. Both probe maps are bijective exactly when both coordinate maps are bijective.

For an arbitrary proposed generator $G=(G_1,G_2)$, if both coordinates are nonempty, use the noninvertible map (1.3). Both its source and target Hom sets from $G$ are empty because the first coordinate is empty. If $G_1=\varnothing$, use $(\varnothing,1)\longrightarrow(1,1).$ The first-coordinate Hom sets are both singletons, and the second-coordinate Hom sets into $1$ are also singletons, whatever $G_2$ is. The induced map is bijective but the original map is not invertible. If $G_2=\varnothing$, the symmetric map $(1,\varnothing)\to(1,1)$ gives the same failure. These cases exhaust all $G$.

This category is also the presheaf category on a discrete two-object category. Its representables are a detecting family, whose fixed coproduct fails to detect isomorphisms. Under a literal product-valued family test even the representable example fails; under the componentwise test it is valid and the coproduct replacement fails. The two readings cannot be interchanged.

Exercise 3 (hard). Give a continuous map that is monic and epic and induces a bijection on point probes, but is not invertible or a strict epimorphism. Explain which hypothesis of Theorem 2.2 fails.

Solution. Let $X=\{0,1\}$ have the discrete topology and $Y=\{0,1\}$ the indiscrete topology. The identity set map $f:X\to Y$ is continuous and bijective. It is monic by injectivity of its underlying function, and epic by surjectivity: two continuous maps agreeing after $f$ agree at every point. The inverse is not continuous, because the singleton $\{0\}$, open in $X$, has a nonopen preimage in $Y$. Thus the category of all topological spaces is not balanced.

Maps from a point are elements, so its probe map is bijective. Point probes are faithful, because distinct continuous maps differ at a point; they are not conservative. Finally the kernel pair of $f$ is the diagonal pair of $X$, since $f$ is injective. If $f$ were strict, it would coequalize that identical pair universally; the map $1_X$ would then factor continuously through $f$, giving its forbidden inverse. Hence it is not strict. This also shows why Corollary 4.2 distinguishes strict quotients from arbitrary epimorphic quotients.

Exercise 4 (expert). Let $H$ be any group. In the category of left $H$-sets, classify the strict quotients of the regular left $H$-set $H$, retaining the specified map from $H$. Describe when one quotient factors through another.

Solution. For any $H$-set $X$, evaluation at the group identity identifies $\operatorname{Hom}_H(H,X)$ with its underlying set. The inverse sends $x$ to $h\mapsto hx$. A bijective equivariant map has an equivariant inverse, so $H$ is a detecting generator, consistently with the Cayley probe in Riehl, Example 2.3.7.

A strict epimorphism is surjective here. Indeed its kernel-pair coequalizer is the set quotient by the kernel equivalence relation, with the induced $H$-action, hence is surjective. Conversely a surjective equivariant map is the coequalizer of its kernel pair: a map constant on every fibre descends uniquely as a set map, and the descended map is equivariant by applying it to $hx$. Thus it is strict.

For a strict quotient $q:H\to X$, put $x=q(1)$. Surjectivity and equivariance give $X=Hx$. The stabilizer $K=\{h\in H:hx=x\}$ is a subgroup. The map $H/K\to X$, $hK\mapsto hx$, is well-defined, equivariant and bijective: $hx=h'x$ exactly when $h'^{-1}h\in K$. It carries the specified quotient map $H\to H/K$ to $q$. Conversely every subgroup $K$ gives such a strict quotient.

A compatible isomorphism must carry $q(1)$ to the other distinguished point and therefore preserve its stabilizer. Hence quotient classes correspond to the actual subgroups $K$, with no conjugacy identification and no normality requirement. A compatible factorization $H/K\to H/L$ exists exactly when $K\subseteq L$; then it is uniquely $hK\mapsto hL$. Necessity follows because every element fixing the distinguished source point must fix its image. This concretely realizes the small kernel-pair code: the relation is $h\sim h'$ exactly when $h'^{-1}h\in K$.

Original exposition and solutions are dedicated to CC0 1.0. Canonical results are retained through the cited interfaces without reproducing their text. Self-checked; no independent review has occurred.