Internal groups, limits and transport

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

A compatible family of group elements is again a group: multiply its coordinates. This elementary observation has a categorical counterpart. An object can carry group operations on every set of maps into it, and these operations pass to any limit that exists in the underlying category. They also pass through functors with suitable compatibility properties.

There are two useful ways to see the operations. With finite products, multiplication and inverse are ordinary morphisms. Without products, the group laws live in the representable functor. We develop both descriptions, then explain transport through finite products and through cofiltered comma categories. Finally, the Eckmann–Hilton argument shows why a group internal to the category of groups must be abelian.

We assume categories, groups, products and limits. Retain the complete Yoneda proof in the AI-integrated Stacks baseline: evaluation at the identity identifies natural transformations from a representable with its target presheaf value. Section 2 checks that interface for the group operations. For presheaves, use Ind-objects through their elements; §5 below supplies the exact filtered-set interface needed here. Basic open references are [Mathlib, Groups] and [Mathlib, Internal groups], listed below. The proofs here include the constructions we need.

Fix a universe for sets and locally small Hom sets. Categories may have larger object collections. Choose a larger ambient universe containing the categories when forming categories of functors or comma diagrams. A limit indexed by a small category means small in the fixed universe, unless we explicitly say otherwise.

1. Limits of groups are compatible families

Write \(\mathsf{Grp}\) for groups and group homomorphisms, and \(U:\mathsf{Grp}\to\mathsf{Set}\) for the forgetful functor.

Theorem 1.1. Groups admit all small limits. Their forgetful functor creates these limits: a specified limiting cone of underlying sets has exactly one group structure making its projections group homomorphisms, and that cone is limiting among groups.

Proof. Let \(D:\mathsf J\to\mathsf{Grp}\) be a small diagram. Define

\[ \begin{gathered} L=\{(x_j)\in\prod_{j\in\operatorname{Ob}\mathsf J}D_j:\\ D(\alpha)(x_j)=x_k \\ \text{ for every }\alpha:j\to k\}. \end{gathered} \tag{1.1} \]

The identities form a compatible family. If \(x,y\in L\), then

\[ \begin{gathered} D(\alpha)(x_jy_j)\\ =D(\alpha)(x_j)D(\alpha)(y_j)\\ =x_ky_k. \end{gathered} \tag{1.2} \]

The same calculation with inverses proves that \((x_j^{-1})\) is compatible. Thus coordinatewise multiplication, identity and inverse make \(L\) a group. Its projections are homomorphisms and satisfy the diagram relations.

For a group cone \(q_j:B\to D_j\), the unique underlying set map into \(L\) is

\[ q(b)=(q_j(b))_j. \tag{1.3} \]

It preserves multiplication and identity coordinatewise, hence is a group homomorphism. Uniqueness as a set map proves uniqueness as a homomorphism. This is the limit universal property.

Now start with any specified limiting set cone \((S,p_j)\). Its unique comparison with (1.1) is a bijection commuting with all projections. Transport the group structure along that bijection. Any other group structure making all \(p_j\) homomorphisms must have the same products, identity and inverses, because the projections distinguish elements. Thus the transported structure is unique and gives a group limit.

When \(\mathsf J\) is empty, (1.1) is the singleton set of empty families. It carries the trivial group structure and is terminal. The same proof includes this case. \(\square\)

For example, the equalizer of parallel homomorphisms \(f,g:A\to B\) is the subgroup

\[ \{a\in A:f(a)=g(a)\}. \tag{1.4} \]

It need not be normal: the equalizer of the identity of a group and conjugation by a fixed element is that element's centralizer. Normality is unnecessary for a limit. Similarly, a pullback of \(A\to B\leftarrow C\) is the subgroup of \(A\times C\) consisting of pairs with equal images.

2. A group on a representable functor

For \(G\in\mathsf C\), write

\[ \begin{gathered} h_G(X)=\operatorname{Hom}_{\mathsf C}(X,G),\\ h_G(t)(a)=a\circ t. \end{gathered} \tag{2.1} \]

For the retained Yoneda bijection, an element \(x\in H(G)\) gives the natural transformation whose value on \(b\colon T\to G\) is \(H(b)(x)\). Naturality follows from \(H(bt)=H(t)H(b)\). Evaluation at \(1_G\) returns \(x\), while naturality of a transformation \(\alpha\) gives \(\alpha_T(b)=H(b)(\alpha_G(1_G))\). These are the two inverse checks. Taking \(H=h_K\) gives postcomposition by a unique arrow \(G\to K\). We use exactly this complete provider for the operations below; it also detects equality of arrows by evaluating at the identity.

A group object is an object \(G\) together with group structures on all \(h_G(X)\), such that precomposition by every \(t:X\to Y\) is a group homomorphism

\[ h_G(t):h_G(Y)\longrightarrow h_G(X). \tag{2.2} \]

Equivalently, \(h_G\) lifts to a functor \(\mathsf C^{\rm op}\to\mathsf{Grp}\). If a group-valued functor has an underlying presheaf isomorphic to \(h_G\), transport its group laws through that specified isomorphism to obtain this description on the actual Hom sets. No products in \(\mathsf C\) are required.

Let \(m_X\), \(e_X\) and \(a_X\) denote multiplication, identity and inverse in \(h_G(X)\). For \(t:X\to Y\), naturality is

\[ \begin{gathered} m_Y(b,c)\circ t =m_X(b\circ t,c\circ t), \\ b,c:Y\to G. \end{gathered} \tag{2.3} \]

It also gives \(e_Y\circ t=e_X\) and \(a_Y(b)\circ t=a_X(b\circ t)\).

In the presheaf category, these are natural maps

\[ \begin{gathered} m:h_G\times h_G\to h_G,\\ e:\mathbf1\to h_G,\qquad a:h_G\to h_G, \end{gathered} \tag{2.4} \]

where \(\mathbf1\) is the terminal presheaf. The ordinary group axioms hold at every object. Conversely, maps (2.4) satisfying those axioms give groups on every Hom set, and their naturality makes every precomposition map a homomorphism. This proves the equivalence of the two descriptions.

A morphism of group objects \(f:G\to H\) is a morphism in \(\mathsf C\) for which

\[ f\circ(-):h_G(X)\longrightarrow h_H(X) \tag{2.5} \]

is a group homomorphism for every \(X\). Identity and composition preserve this condition. We obtain a category \(\operatorname{Gr}(\mathsf C)\) and a faithful forgetful functor to \(\mathsf C\). Yoneda identifies its morphisms with group-valued natural transformations between the corresponding representable lifts: their underlying set transformations determine a unique \(f\), and the group condition is precisely (2.5).

Suppose now that \(\mathsf C\) admits finite products, with terminal object \(1\). The product universal property gives

\[ h_{G\times G}\cong h_G\times h_G, \qquad h_1\cong\mathbf1. \tag{2.6} \]

Yoneda turns (2.4) into unique morphisms

\[ \begin{gathered} \mu:G\times G\to G,\\ \eta:1\to G,\qquad \iota:G\to G. \end{gathered} \tag{2.7} \]

Their formulas are

\[ \begin{gathered} m_X(b,c)=\mu\langle b,c\rangle,\\ e_X=\eta\,!_X,\qquad a_X(b)=\iota b. \end{gathered} \tag{2.8} \]

Here \(!_X:X\to1\) is the unique map. The group axioms become the following morphism identities, with the canonical identifications of triple products understood:

\[ \begin{aligned} \mu(\mu\times\operatorname{id}) &=\mu(\operatorname{id}\times\mu),\\ \mu\langle\operatorname{id},\eta!_G\rangle &=\operatorname{id}_G,\\ \mu\langle\eta!_G,\operatorname{id}\rangle &=\operatorname{id}_G,\\ \mu\langle\operatorname{id},\iota\rangle &=\eta!_G,\\ \mu\langle\iota,\operatorname{id}\rangle &=\eta!_G. \end{aligned} \tag{2.9} \]

For instance, associativity at every \(X\), applied to the three projections of \(G^3\), proves the first identity. Precomposing that identity with \(\langle b,c,d\rangle\) recovers associativity on \(h_G(X)\). The other identities follow in exactly the same two directions by taking the identity of \(G\) and then arbitrary maps into \(G\). Thus (2.7)–(2.9) are equivalent to the representable definition, with no additional axioms.

A morphism \(f:G\to H\) of these objects satisfies

\[ \begin{gathered} f\mu_G=\mu_H(f\times f),\\ f\eta_G=\eta_H,\qquad f\iota_G=\iota_Hf. \end{gathered} \tag{2.10} \]

Preserving multiplication and identity already forces the last identity: a homomorphism sends the inverse of an element to the inverse of its image. Apply that fact on every Hom set and use Yoneda.

The object is commutative if all its Hom groups are abelian, equivalently if the representable lift takes values in \(\mathsf{Ab}\). In presheaves this means \(m\tau=m\), where \(\tau\) swaps the two factors. With products in \(\mathsf C\), it means

\[ \mu\tau_G=\mu, \qquad \tau_G\langle b,c\rangle=\langle c,b\rangle. \tag{2.11} \]

The equivalence follows by evaluating on all pairs, or on the two projections of \(G\times G\).

3. Every existing limit carries its group structure

Theorem 3.1. Let \(D:\mathsf J\to\operatorname{Gr}(\mathsf C)\) be a small diagram. If its underlying diagram has a specified limit \((L,p_j)\) in \(\mathsf C\), then \(L\) has a unique group-object structure making all \(p_j\) group-object morphisms. This lifted cone is limiting in \(\operatorname{Gr}(\mathsf C)\).

Proof. For every \(X\), the underlying limit property gives a natural bijection

\[ \begin{gathered} \operatorname{Hom}_{\mathsf C}(X,L)\\ \cong\lim_{j\in\mathsf J} \operatorname{Hom}_{\mathsf C}(X,D_j). \end{gathered} \tag{3.1} \]

The right side is the compatible-family group from Theorem 1.1. Transport it to the left. A map \(t:X'\to X\) acts by precomposition in every coordinate; those coordinate maps are homomorphisms because each \(D_j\) is a group object. Therefore precomposition on the left is also a homomorphism. This defines a group object on \(L\), and (3.1) makes every projection a morphism of group objects.

For a group-object cone \(q_j:B\to D_j\), let \(q:B\to L\) be the underlying unique cone map. On every Hom set, the map induced by \(q\) has homomorphisms in all coordinates. Theorem 1.1 makes it a homomorphism into the compatible-family group. Thus \(q\) is a group-object morphism, proving the lifted universal property.

Finally, (3.1) and its projections force every group operation on \(\operatorname{Hom}(X,L)\). No other group-object structure can make those projections homomorphisms. For an empty diagram, these Hom sets are singletons and have their unique trivial group law. This also proves uniqueness in the empty case. \(\square\)

In particular, if \(\mathsf C\) admits finite products, then so does \(\operatorname{Gr}(\mathsf C)\), and its forgetful functor preserves them. The terminal object has singleton Hom groups. For two group objects \(G,H\), the product law on maps into \(G\times H\) is

\[ (g,h)(g',h')=(gg',hh'). \tag{3.2} \]

This proof works for any existing limit, even if \(\mathsf C\) has no finite products in general. If all the diagram objects are commutative, their compatible-family groups are abelian, so the same theorem applies to commutative group objects.

4. Transport through finite products

Theorem 4.1. Suppose \(\mathsf C\) admits finite products and \(F:\mathsf C\to\mathsf D\) preserves them. Then \(F\) sends group objects and their morphisms to group objects and their morphisms. It also sends commutative group objects to commutative group objects.

The hypothesis includes the empty product: \(F1\) is terminal in \(\mathsf D\). For every \(A,B\), the canonical cone \((F(A\times B),Fp_1,Fp_2)\) is a product cone of \(FA,FB\). We need only these product objects in \(\mathsf D\); arbitrary products of objects outside the image of \(F\) are unnecessary.

Proof. Use \(F1\) as terminal object and \(F(G\times G)\) as the product of \(FG\) with itself. Define multiplication, identity and inverse by \(F\mu\), \(F\eta\) and \(F\iota\). Product uniqueness gives

\[ F\langle b,c\rangle=\langle Fb,Fc\rangle \tag{4.1} \]

relative to these product cones, and \(F(!_X)\) is the terminal map from \(FX\). The corresponding statement for three factors follows by comparing their three projections. Applying \(F\) to (2.9) therefore gives all the group-object identities in \(\mathsf D\). Applying it to (2.10) gives the morphism identities, and applying it to (2.11) preserves commutativity.

If other product objects are chosen in \(\mathsf D\), use the unique comparison

\[ c: F(G\times G)\xrightarrow{\sim}FG\times FG. \tag{4.2} \]

Multiplication then reads \(F\mu\,c^{-1}\). The identity uses the unique terminal comparison, and inverse remains \(F\iota\). These formulas describe the same Hom-set group structure. \(\square\)

For example, forgetting a topological group's topology preserves finite products, so it gives its underlying abstract group. An arbitrary functor need not transport a group object; Exercise 4 gives an obstruction that holds for every attempted choice of group laws.

5. Transport without finite products

There is a stronger condition on \(F\) that makes sense without assuming finite products in \(\mathsf C\). For \(Y\in\mathsf D\), define the comma category \(\mathsf K_Y=(Y\downarrow F)\). Its objects are

\[ (X,u),\qquad u:Y\to FX, \tag{5.1} \]

and an arrow \((X,u)\to(X',u')\) is a map \(r:X\to X'\) with \(Fr\,u=u'\). Assume each \(\mathsf K_Y\) is cofiltered: it is nonempty, every pair of objects has a common source, and every pair of parallel arrows can be equalized by a map into their source. We call this the comma left-exact condition. We spell it out to distinguish it from simply preserving finite limits that happen to exist.

First retain the disjoint-union and finite-zigzag description of a set colimit from Ind-objects through their elements, Proposition 1.2. Its universal map sends a class \([i,a]\) to the value of any compatible family at \(a\): the generating arrow relations make this well defined, and every class has a representative, so the map is unique. This is the full set-colimit universal property.

For a filtered index, retain the complete finite-graph cocone proof in Hom, tensor and module limits, §3. Apply its categorical statement to the finite graph underlying a zigzag of set representatives. At the cocone vertex, each adjacent pair has equal images by the corresponding diagram-arrow relation; hence its two endpoints have equal images there. Conversely, equal images at one common vertex are already two generating colimit identifications. Thus two set representatives are equal exactly when they agree at a common later stage. This uses the entire filtered-category condition, including equalization of parallel arrows, and imposes no module structure on the diagram.

Here is the group-colimit construction over that set colimit.

Lemma 5.1. For a filtered diagram \(A:\mathsf I\to\mathsf{Grp}\), its colimit of underlying sets has a unique group structure making the stage maps homomorphisms. It is also the group colimit.

Proof. Write an element as \([i,a]\). Two representatives agree precisely when they have equal images at a common later stage. For two elements, choose arrows \(r:i\to k\) and \(s:j\to k\) and set

\[ [i,a]\,[j,b] =[k,A(r)(a)A(s)(b)]. \tag{5.2} \]

To compare two choices \(k,k'\), move them to a common stage. The resulting two maps from \(i\), and then the two from \(j\), can be equalized after one further stage. Their products then agree. If a representative is replaced by an equal one, first move both representatives to a stage witnessing that equality; apply the same common-stage and equalization argument to the multiplication. Thus (5.2) is well defined.

The identities at any two stages become the same identity at a common stage. Define the identity by that class and the inverse of \([i,a]\) by \([i,a^{-1}]\). These definitions respect all representative relations. Three elements can be represented at one stage, where associativity holds; their two products have the same class. The same-stage identity and inverse equations prove the remaining group laws. Every stage map is a homomorphism.

A compatible family of homomorphisms \(A_i\to B\) induces its unique set-colimit map. Formula (5.2) shows that it preserves multiplication and identity, so it is a homomorphism. Conversely, any group structure making the stage maps homomorphisms is forced by (5.2), since finitely many representatives can be brought to one stage. This proves both uniqueness and the group universal property. \(\square\)

Theorem 5.2. Under the comma left-exact condition, every group object \(G\) of \(\mathsf C\) gives a group object on \(FG\) in \(\mathsf D\), without a finite-product hypothesis. Group-object morphisms and commutativity are preserved.

Proof. Fix \(Y\). Over \(\mathsf I_Y=\mathsf K_Y^{\rm op}\), take the group diagram

\[ (X,u)\longmapsto\operatorname{Hom}_{\mathsf C}(X,G). \tag{5.3} \]

An arrow in \(\mathsf I_Y\) arising from \(r:X\to X'\) in \(\mathsf K_Y\) acts by precomposition \(a'\mapsto a'r\). This is a group homomorphism by the group-object definition. The index \(\mathsf I_Y\) is filtered. Lemma 5.1 gives a group structure on its set colimit.

For every functor \(F\), there is a bijection of underlying sets

\[ \begin{gathered} \Phi_Y:\operatorname*{colim}_{(X,u)\in\mathsf I_Y} \operatorname{Hom}_{\mathsf C}(X,G) \\ \longrightarrow\operatorname{Hom}_{\mathsf D}(Y,FG),\\ [X,u,a]\longmapsto Fa\,u. \end{gathered} \tag{5.4} \]

It respects the representative relation: \(Fa'r\,u=Fa'\,u'\) whenever \(Fr\,u=u'\). A map \(v:Y\to FG\) is the image of \([G,v,\operatorname{id}_G]\), proving surjectivity. Further, \(a:X\to G\) is an arrow

\[ (X,u)\longrightarrow(G,Fa\,u) \quad\text{in }\mathsf K_Y. \tag{5.5} \]

In the opposite index this identifies \([X,u,a]\) with \([G,Fa\,u,\operatorname{id}_G]\). Hence representatives with the same image in (5.4) have the same class. This proves injectivity. This bijection itself does not require filteredness; filteredness supplies the group law on its left side.

Transport that law to \(\operatorname{Hom}_{\mathsf D}(Y,FG)\). For \(t:Y'\to Y\), the functor

\[ \begin{gathered} \mathsf K_Y\longrightarrow\mathsf K_{Y'},\\ (X,u)\longmapsto(X,ut) \end{gathered} \tag{5.6} \]

and its opposite give a map between the group colimits, induced by the identity on each stage Hom group. It is a homomorphism by Lemma 5.1. Under (5.4) it sends \(Fa\,u\) to \(Fa\,ut\), exactly precomposition by \(t\). Thus \(FG\) is a group object.

For a group-object morphism \(g:G\to H\), postcomposition with \(g\) gives a natural transformation of stage group diagrams. Its colimit map corresponds under (5.4) to postcomposition with \(Fg\). Thus \(Fg\) is a group-object morphism. Identity and composition follow on underlying maps. If the original Hom groups are abelian, the multiplication of any two classes can be performed at one abelian stage, where they commute; the resulting object is commutative.

All comma diagrams can be formed in the chosen ambient universe. Their colimit sets in (5.4) are isomorphic to the given locally small Hom sets in \(\mathsf D\). Thus this argument requires no claim that the comma categories are small in the original universe. \(\square\)

The two transport constructions agree whenever both hypotheses hold. Indeed, for \(v,w:Y\to FG\), use the product comparison to form

\[ u=c^{-1}\langle v,w\rangle: Y\to F(G\times G). \tag{5.7} \]

The maps \(p_1,p_2:G\times G\to G\) represent \(v,w\) at this common stage of (5.3). Their product in that stage is \(\mu\), so (5.4) gives \(F\mu\,u\), the multiplication in Theorem 4.1. The identity is represented by the unique map \(Y\to F1\) and \(\eta:1\to G\); inverse is represented at \((G,v)\) by \(\iota:G\to G\). Thus all three operations agree.

6. A group inside the category of groups

An internal group in \(\mathsf{Grp}\) has two apparent multiplications: the original group multiplication on its underlying group, and the internal multiplication. Their compatibility forces them to agree and to commute.

Theorem 6.1 (Eckmann–Hilton). A group object in \(\mathsf{Grp}\) is precisely an abelian group with its ordinary multiplication, identity and inverse as the internal structure. The resulting category \(\operatorname{Gr}(\mathsf{Grp})\) is equivalent to \(\mathsf{Ab}\).

Proof. Let \(A\) be the original group, with multiplication written by juxtaposition and identity \(e\). The internal multiplication

\[ m:A\times A\to A \tag{6.1} \]

is a homomorphism for the original group laws. The internal identity is a homomorphism from the trivial group, so it picks the original \(e\). Its two identity axioms give \(m(a,e)=a\) and \(m(e,b)=b\).

The original product group has the two decompositions

\[ (a,b)=(a,e)(e,b)=(e,b)(a,e). \tag{6.2} \]

Applying the homomorphism \(m\) gives

\[ m(a,b)=ab=ba. \tag{6.3} \]

So the internal product is the original product and \(A\) is abelian. The internal inverse satisfies the inverse equations for that product, hence is the original inverse by uniqueness.

Conversely, if \(A\) is abelian, its multiplication is a homomorphism \(A\times A\to A\), since

\[ (aa')(bb')=(ab)(a'b'). \tag{6.4} \]

Its identity and inverse are also homomorphisms, and the ordinary group axioms give the internal identities. Every homomorphism between abelian groups preserves those operations, so gives an internal-group morphism. Conversely, an internal-group morphism is already an underlying group homomorphism. The two constructions agree on objects and morphisms after identifying additive and multiplicative notation. This gives the asserted equivalence. \(\square\)

Theorem 1.1 did not assume commutativity. Commutativity here comes from requiring multiplication itself to be a homomorphism for another group law.

7. Four exercises with complete solutions

Exercise 1 (warm-up: a marked identity). Let \(\mathsf{Set}_*\) be pointed sets and maps preserving the distinguished point. Classify its group objects and their morphisms. Does the distinguished point admit any choice once the group-object structure is fixed?

Solution. Finite products of pointed sets have product distinguished point; the terminal object is the singleton pointed set. Therefore an internal identity map \(1\to A\) must select the distinguished point \(*\). After forgetting the pointing, Theorem 4.1 gives an ordinary group on the set \(A\), whose identity is consequently \(*\).

Conversely, any ordinary group becomes a pointed set by marking its identity. Its multiplication sends \((e,e)\) to \(e\), its inverse sends \(e\) to \(e\), and its identity map is pointed. These maps satisfy all internal group identities because they are the ordinary group operations. A group homomorphism preserves the identity and is automatically pointed. A morphism of the resulting internal groups is an underlying pointed map preserving multiplication and identity, hence exactly a group homomorphism. Thus \(\operatorname{Gr}(\mathsf{Set}_*)\) is equivalent to \(\mathsf{Grp}\), and the distinguished point is forced to be the group identity. No commutativity condition is imposed.

Exercise 2 (intermediate: monotone inverses). Let \(\mathsf{Pos}\) be partially ordered sets and monotone maps, with its ordinary finite products. Show that every group object has a discrete order. Deduce an equivalence \(\operatorname{Gr}(\mathsf{Pos})\simeq\mathsf{Grp}\).

Solution. Forgetting order preserves finite products. Hence an internal group gives an ordinary group \(A\) for which multiplication and inverse are monotone. For every \(c\), left translation \(x\mapsto cx\) is monotone: pair a fixed \(c\) with the varying \(x\), then apply monotone multiplication. Its inverse is translation by \(c^{-1}\), also monotone.

Suppose \(a\le b\). Left translation by \(a^{-1}\) gives \(e\le g\), where \(g=a^{-1}b\). Monotone inversion gives \(e\le g^{-1}\). Left translation by \(g\) then gives \(g\le e\). Antisymmetry forces \(g=e\), hence \(a=b\). Thus two elements are comparable only when equal.

Conversely, put the discrete order on any group. Every map from a discrete poset to a discrete poset is monotone, so its multiplication, inverse and identity are morphisms in \(\mathsf{Pos}\) and give a group object. Morphisms between these objects are exactly group homomorphisms. Forgetting the discrete order and equipping a group with that order are inverse constructions up to their evident identity comparisons. This gives the equivalence. The conclusion used monotone inversion as well as monotone multiplication; an ordered group whose inverse reverses order is a different structure.

Exercise 3 (advanced: a noncommutative pullback). Let \(\varepsilon:S_3\to\{\pm1\}\) be the sign homomorphism. Describe the pullback \(P=S_3\times_{\{\pm1\}}S_3\), its order, the kernel and a section of its first projection. Give an explicit isomorphism \(A_3\rtimes S_3\to P\), using conjugation for the action on \(A_3\), and determine the center of \(P\).

Solution. By Theorem 1.1,

\[ \begin{gathered} P=\{(g,h)\in S_3\times S_3:\\ \varepsilon(g)=\varepsilon(h)\}, \end{gathered} \tag{7.1} \]

with coordinatewise multiplication. There are three even and three odd permutations, so \(|P|=3^2+3^2=18\). The first projection is onto, with diagonal section \(g\mapsto(g,g)\). Its kernel is \(\{1\}\times A_3\), a cyclic group of order three. This is an actual group pullback: two homomorphisms into \(S_3\) with equal signs give a unique homomorphism into (7.1) by pairing.

Conjugation by \(g\in S_3\) preserves the normal subgroup \(A_3\). On \(A_3\rtimes S_3\), use the multiplication

\[ (a,g)(b,k) =\bigl(a(gbg^{-1}),gk\bigr). \tag{7.2} \]

Define

\[ \begin{gathered} \theta(a,g)=(g,ag),\\ \theta^{-1}(g,h)=(hg^{-1},g). \end{gathered} \tag{7.3} \]

Since \(a\) is even, \(g\) and \(ag\) have the same sign. Conversely, equal signs imply \(hg^{-1}\in A_3\). Thus both formulas are well defined and inverse. Moreover,

\[ \begin{aligned} \theta(a,g)\theta(b,k)&=(gk,agbk),\\ \theta\bigl(a(gbg^{-1}),gk\bigr)&=(gk,agbk), \end{aligned} \tag{7.4} \]

so \(\theta\) is a homomorphism and hence an isomorphism. The identity is \((1,1)\); the inverse in (7.2) is \((g^{-1}a^{-1}g,g^{-1})\).

If \((g,h)\) is central in \(P\), it commutes with every diagonal pair \((k,k)\). Thus \(g,h\) are central in \(S_3\). The center of \(S_3\) is trivial: a central permutation must commute with every transposition, so conjugation must preserve each unordered two-element subset of \(\{1,2,3\}\); their intersections force it to fix every element. Hence \(g=h=1\), and \(Z(P)\) is trivial. This limit remains noncommutative; Theorem 6.1 concerns a second internal group law, which this calculation does not supply.

Exercise 4 (challenge: a functor that cannot lift). Let \(\mathcal P:\mathsf{Set}\to\mathsf{Set}\) be covariant powerset, with \(\mathcal P(f)(S)=f(S)\). Show that the two sets \(\mathcal P(C_2)\) and \(\mathcal P(1)\) cannot be given group laws making \(\mathcal P(q)\) a homomorphism for the unique map \(q:C_2\to1\). Deduce that covariant powerset cannot transport all ordinary groups through any choice of group laws. Also show directly that it fails both the finite-product hypothesis and the comma left-exact condition above.

Solution. The map \(\mathcal P(q)\) sends the empty subset to the empty subset and each of the three nonempty subsets of \(C_2\) to the singleton. It is a surjective function from a four-element set to a two-element set, with fibers of sizes one and three.

For a surjective homomorphism \(p:A\to B\) of finite groups, every nonempty fiber is a coset of \(\ker p\). Indeed, choose \(a\) in the fiber of \(b\); then \(p(x)=b\) exactly when \(a^{-1}x\in\ker p\). Multiplication by \(a\) gives a bijection from the kernel to that fiber. All fibers therefore have the same cardinality, which contradicts the sizes one and three. This excludes every possible pair of group laws, without guessing their identities. If powerset transported all groups and group homomorphisms, it would in particular make this map a homomorphism, so no such transport exists.

For \(A=B=\{0,1\}\), the canonical product comparison is

\[ \begin{gathered} \mathcal P(A\times B)\longrightarrow \mathcal P(A)\times\mathcal P(B),\\ S\longmapsto(p_A(S),p_B(S)). \end{gathered} \tag{7.5} \]

The diagonal subset \(\{(0,0),(1,1)\}\) and the full square have the same two projections. The comparison is not injective. Also \(\mathcal P(1)\) has two elements, so powerset does not preserve the empty product either.

Finally take \(Y=1\) in \((Y\downarrow\mathcal P)\). Consider the object with \(X=\varnothing\) and selected subset \(\varnothing\), and the object with \(X=1\) and selected subset \(1\). A common source \((W,u)\) would require a map \(W\to\varnothing\), forcing \(W=\varnothing\). Its selected subset must then be empty, whose direct image in \(1\) is empty, not the selected singleton. This is impossible. The comma category is not cofiltered.

References