The First Isomorphism Theorem: From Regular Categories to Algebraic Models

The classical First Isomorphism Theorem combines two ingredients. The categorical ingredient identifies the quotient by a kernel pair with the image of a morphism. The algebraic ingredient ensures that images inherit the required operations and that the relevant quotients can be computed on the underlying objects.

Regular categories provide a setting for the first ingredient. Categories of models of finite-product sketches provide a general setting for the second.

Definition (Regular category). A morphism is a regular epimorphism if it is the coequalizer of some pair of parallel morphisms. A category $\mathcal{C}$$\mathcal C$ is regular if:

1. (R1) $\mathcal{C}$$\mathcal C$ has all finite limits.

2. (R2) Every morphism $f:X\to Y$$f:X\to Y$ admits a factorization

   $$X\stackrel{e}{\to }I\stackrel{m}{\to }Y,f=me,$$

   where $e$$e$ is a regular epimorphism and $m$$m$ is a monomorphism.

3. (R3) Regular epimorphisms are stable under pullback.

We first establish three facts that will be used throughout.

Lemma A (Orthogonality). In any category, regular epimorphisms are left orthogonal to monomorphisms: every commuting square

$$\begin{array}{ccc}A& \stackrel{u}{\to }& C\\ e\downarrow & & \downarrow m\\ B& \stackrel{v}{\to }& D,\end{array}$$

with $e$$e$ a regular epimorphism and $m$$m$ a monomorphism, has a unique diagonal filler $d:B\to C$$d:B\to C$ satisfying

$$de=u,md=v.$$

Proof. Write $e$$e$ as the coequalizer of $a,b:S\rightrightarrows A$$a,b:S\rightrightarrows A$. Since $mu=ve$$mu=ve$, we have

$$mua=vea=veb=mub.$$

Monicity of $m$$m$ gives $ua=ub$$ua=ub$. The coequalizer property of $e$$e$ supplies a unique $d:B\to C$$d:B\to C$ with $de=u$$de=u$. Moreover,

$$mde=mu=ve.$$

Every coequalizer is an epimorphism, so cancellation of $e$$e$ gives $md=v$$md=v$. $◻$$\square$

Two consequences will be useful.

First, every regular epimorphism is extremal: if a regular epimorphism $e:X\to Y$$e:X\to Y$ factors as $e=ma$$e=ma$ with $m:I\to Y$$m:I\to Y$ monic, then $m$$m$ is an isomorphism. Indeed, Lemma A applied to

$$\begin{array}{ccc}X& \stackrel{a}{\to }& I\\ e\downarrow & & \downarrow m\\ Y& \stackrel{{\mathrm{id}}_Y}{\to }& Y\end{array}$$

gives $d:Y\to I$$d:Y\to I$ with $md={\mathrm{id}}_Y$$md=\operatorname{id}_Y$. Monicity of $m$$m$ then gives $dm={\mathrm{id}}_I$$dm=\operatorname{id}_I$.

Second, regular epimorphism–monomorphism factorizations are unique up to a unique compatible isomorphism. If

$$f=me=m^{\prime }{e}^{\prime },$$

Lemma A gives a unique comparison $u:I\to I^{\prime }$$u:I\to I'$ satisfying

$$ue=e^{\prime },m^{\prime }u=m.$$

The comparison in the opposite direction is its inverse, by uniqueness.

Accordingly, in a regular category the subobject $m:I\hookrightarrow Y$$m:I\hookrightarrow Y$ in such a factorization is called the image of $f$$f$, written $\mathrm{im}(f)$$\operatorname{im}(f)$.

These facts require no finite limits or pullback-stability assumption.

Lemma B (Regular epimorphisms and kernel pairs). If a regular epimorphism $e:X\to Y$$e:X\to Y$ has a kernel pair

$$r_1,r_2:\mathrm{Eq}(e)\rightrightarrows X,$$

then $e$$e$ is the coequalizer of $r_1$$r_1$ and $r_2$$r_2$.

Proof. Write $e$$e$ as the coequalizer of $a,b:S\rightrightarrows X$$a,b:S\rightrightarrows X$. Since $ea=eb$$ea=eb$, the kernel-pair universal property gives a unique $u:S\to \mathrm{Eq}(e)$$u:S\to\operatorname{Eq}(e)$ with

$$r_{1}u=a,r_{2}u=b.$$

Thus the following triangles commute:

$$\begin{array}{ccccc}& & S& & \\ & a\swarrow & \downarrow u& \searrow b& \\ X& \stackrel{r_1}{\leftarrow }& \mathrm{Eq}(e)& \stackrel{r_2}{\to }& X.\end{array}$$

If $h:X\to Z$$h:X\to Z$ satisfies $h{r}_1=h{r}_2$$hr_1=hr_2$, then

$$ha=h{r}_{1}u=h{r}_{2}u=hb.$$

Hence $h$$h$ factors uniquely through $e$$e$. Since $e{r}_1=e{r}_2$$er_1=er_2$ by definition of the kernel pair, this is precisely the required coequalizer universal property. $◻$$\square$

Only the existence of this kernel pair is needed.

Lemma C (Composition and finite products). In a category satisfying (R2), regular epimorphisms are closed under composition. In a regular category, they are also closed under finite products.

Proof. Let $e:X\to Y$$e:X\to Y$ and $d:Y\to Z$$d:Y\to Z$ be regular epimorphisms. By (R2), factor their composite as

$$de=mr,X\stackrel{r}{\to }I\stackrel{m}{\to }Z,$$

where $r$$r$ is a regular epimorphism and $m$$m$ is monic. Lemma A gives a filler $h:Y\to I$$h:Y\to I$ in

$$\begin{array}{ccc}X& \stackrel{r}{\to }& I\\ e\downarrow & & \downarrow m\\ Y& \stackrel{d}{\to }& Z.\end{array}$$

In particular, $d=mh$$d=mh$. Since $d$$d$ is a regular epimorphism and therefore extremal, $m$$m$ is an isomorphism. Thus $de=mr$$de=mr$ is a regular epimorphism.

Now suppose $\mathcal{C}$$\mathcal C$ is regular, and let $e:A\to B$$e:A\to B$ and $e^{\prime }:A^{\prime }\to B^{\prime }$$e':A'\to B'$ be regular epimorphisms. Factor their product as

$$A\times A^{\prime }\stackrel{e\times {\mathrm{id}}_{A^{\prime }}}{\to }B\times A^{\prime }\stackrel{{\mathrm{id}}_B\times e^{\prime }}{\to }B\times B^{\prime }.$$

The first map is a pullback of $e$$e$:

$$\begin{array}{ccc}A\times A^{\prime }& \stackrel{e\times {\mathrm{id}}_{A^{\prime }}}{\to }& B\times A^{\prime }\\ {\pi }_A\downarrow & & \downarrow {\pi }_B\\ A& \stackrel{e}{\to }& B.\end{array}$$

Similarly, the second map is a pullback of $e^{\prime }$$e'$. Both are regular epimorphisms by (R3), and their composite is regular epic by the first part.

Induction proves the result for every nonempty finite product. The empty product is the identity of the terminal object, which is a regular epimorphism. $◻$$\square$

The composition argument uses (R2). The finite-product argument additionally uses finite products from (R1) and pullback stability from (R3).

Examples.

• Sets and algebraic structures. The category $\mathbf{Set}$$\mathbf{Set}$ is regular, with surjections as its regular epimorphisms. Every finitary variety of algebras is regular as well. Examples include groups, monoids, rings, and Lie algebras over a fixed field. We will obtain this algebraic statement from the proposition below.

• Abelian categories. Every abelian category is regular. In particular, $\mathbf{Ab}$$\mathbf{Ab}$, $R\text{-}\mathbf{Mod}$$R\text{-}\mathbf{Mod}$, and ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$ are regular. Every epimorphism is regular, and the usual coimage–image isomorphism supplies (R2).

• Toposes. Every elementary topos is regular. Examples include presheaf categories and categories of sheaves of sets on a site.

• Compact Hausdorff spaces. The category $\mathbf{CompHaus}$$\mathbf{CompHaus}$ is regular. Continuous surjections are quotient maps and coequalizers of their kernel pairs; they are also stable under pullback. The image of a continuous map carries the subspace topology. This agrees with the quotient topology obtained from its fibers: the resulting continuous bijection from a compact space to a Hausdorff space is a homeomorphism. By contrast, $\mathbf{Top}$$\mathbf{Top}$ is not regular, because quotient maps need not remain quotient maps after pullback.

• Torsion-free abelian groups. The full subcategory ${\mathbf{Ab}}_{\mathrm{tf}}\subseteq \mathbf{Ab}$$\mathbf{Ab}_{\mathrm{tf}}\subseteq\mathbf{Ab}$ is regular. Finite limits and ordinary images remain torsion-free. For $f:A\to B$$f:A\to B$, the surjection $A\to \mathrm{im}(f)$$A\to\operatorname{im}(f)$ is the coequalizer of its kernel pair in $\mathbf{Ab}$$\mathbf{Ab}$; all objects in this diagram are torsion-free, so the same universal property holds in the full subcategory. Conversely, Lemma A and this factorization show that every regular epimorphism in ${\mathbf{Ab}}_{\mathrm{tf}}$$\mathbf{Ab}_{\mathrm{tf}}$ is surjective. Pullbacks preserve these surjections, proving (R3). General coequalizers, however, need not be computed in $\mathbf{Ab}$$\mathbf{Ab}$: one forms the ordinary quotient and then divides out its torsion subgroup.

Regular logic. The structure of a regular category also supports a language for describing images and relations.

Regular formulas are built from equations and applications of relation symbols using

$$\mathrm{\top },\wedge ,\mathrm{\exists }.$$

Here $\mathrm{\top }$$\top$ is the empty conjunction. A regular theory consists of a signature of sorts, function symbols, and relation symbols, together with axioms expressed as sequents

$$\varphi {⊢}_{\mathrm{\Gamma }}\psi ,$$

where $\varphi ,\psi$$\phi,\psi$ are regular formulas and $\mathrm{\Gamma }$$\Gamma$ contains their free variables. In set-theoretic semantics, such a sequent means

$$\mathrm{\forall }\overrightarrow{x}(\varphi (\overrightarrow{x})\Rightarrow \psi (\overrightarrow{x})).$$

Universal quantification and implication belong to this external reading of a sequent; they are not constructors of regular formulas.

In a regular category, sorts are interpreted as objects, function symbols as morphisms, and relation symbols as subobjects of products. A formula in context $\mathrm{\Gamma }=(x_1:X_1,\dots ,x_n:X_n)$$\Gamma=(x_1:X_1,\ldots,x_n:X_n)$ is interpreted as a subobject

$$[[\varphi ]]\hookrightarrow [[\mathrm{\Gamma }]]:=X_1\times \cdots \times X_n.$$

Logical construction	Categorical interpretation
Atomic relation $R(t_1,\dots ,t_n)$$R(t_1,\ldots,t_n)$	Pullback of the interpreting subobject for $R$$R$ along the tuple of interpreting term maps
Equality $s=t$$s=t$, with values in $A$$A$	Pullback of the diagonal ${\mathrm{\Delta }}_A:A\to A\times A$$\Delta_A:A\to A\times A$ along $⟨s,t⟩$$\langle s,t\rangle$
Truth $\mathrm{\top }$$\top$	The whole context object
Conjunction $\varphi \wedge \psi$$\phi\wedge\psi$	Intersection of subobjects
Substitution along a tuple of terms $\tau :\mathrm{\Delta }\to \mathrm{\Gamma }$$\tau:\Delta\to\Gamma$	Pullback ${\tau }^{\ast }[[\varphi ]]$$\tau^*[\![\phi]\!]$
Existential quantification $\mathrm{\exists }y\varphi$$\exists y\,\phi$	Image under the projection forgetting $y$$y$
Validity of $\varphi {⊢}_{\mathrm{\Gamma }}\psi$$\phi\vdash_\Gamma\psi$	Inclusion $[[\varphi ]]\le [[\psi ]]$$[\![\phi]\!]\leq[\![\psi]\!]$

For a morphism $f:A\to B$$f:A\to B$ and a subobject $p:P\hookrightarrow A$$p:P\hookrightarrow A$, define

$${\mathrm{\exists }}_f(P):=\mathrm{im}(fp).$$

This operation is left adjoint to pullback:

$${\mathrm{\exists }}_f(P)\le Q⟺P\le f^{\ast }(Q).$$

Indeed, both conditions say that $fp$$fp$ factors through the subobject $Q\hookrightarrow B$$Q\hookrightarrow B$.

In particular, if $\varphi (\overrightarrow{x},y)$$\phi(\vec x,y)$ defines $P\hookrightarrow X\times Y$$P\hookrightarrow X\times Y$, then

$$[[\mathrm{\exists }y\varphi (\overrightarrow{x},y)]]=\mathrm{im}(P\hookrightarrow X\times Y\stackrel{{\pi }_X}{\to }X).$$

Axiom (R3) makes this interpretation compatible with substitution. For a pullback square

$$\begin{array}{ccc}{A}^{\prime }& \stackrel{u}{\to }& A\\ f^{\prime }\downarrow & & \downarrow f\\ B^{\prime }& \stackrel{v}{\to }& B,\end{array}$$

pulling back an image factorization gives another image factorization. Consequently,

$$v^{\ast }{\mathrm{\exists }}_f(P)={\mathrm{\exists }}_{f^{\prime }}{u}^{\ast }(P).$$

This is the Beck–Chevalley condition.

Pullback stability also gives Frobenius reciprocity:

$${\mathrm{\exists }}_f(P\wedge f^{\ast }Q)={\mathrm{\exists }}_f(P)\wedge Q.$$

To see this, factor $P\to B$$P\to B$ through its image $J\hookrightarrow B$$J\hookrightarrow B$ and pull the factorization back along $Q\hookrightarrow B$$Q\hookrightarrow B$. The map

$$P{\times }_{B}Q⟶J{\times }_{B}Q$$

is a regular epimorphism by (R3), so the image of $P{\times }_{B}Q\to B$$P\times_B Q\to B$ is precisely $J\wedge Q$$J\wedge Q$.

For projection maps, this expresses the logical equivalence

$$\mathrm{\exists }y(\varphi (\overrightarrow{x})\wedge \psi (\overrightarrow{x},y))⊣⊢\varphi (\overrightarrow{x})\wedge \mathrm{\exists }y\psi (\overrightarrow{x},y),$$

where $y$$y$ does not occur freely in $\varphi$$\phi$.

Thus (R1) interprets equality, conjunction, and substitution; (R2) supplies existential images; and (R3) supplies the compatibility laws needed for the regular logical rules.

Theorem (First Isomorphism Theorem). Let $\mathcal{C}$$\mathcal C$ be a regular category and let $f:X\to Y$$f:X\to Y$. Write

$$\mathrm{Eq}(f):=X{\times }_{Y}X$$

for its kernel pair, with projections $r_1,r_2$$r_1,r_2$. Then their coequalizer exists, and

$$\boxed{{\displaystyle X/\mathrm{Eq}(f)\cong \mathrm{im}(f).}}$$

Proof. By (R2), choose a factorization

$$X\stackrel{e}{\to }I\stackrel{m}{\to }Y,f=me,$$

with $e$$e$ regular epic and $m$$m$ monic.

The square defining the kernel pair of $f$$f$,

$$\begin{array}{ccc}\mathrm{Eq}(f)& \stackrel{r_2}{\to }& X\\ r_1\downarrow & & \downarrow f\\ X& \stackrel{f}{\to }& Y,\end{array}$$

remains a pullback when $Y,f$$Y,f$ are replaced by $I,e$$I,e$. Indeed, for every pair $s,t:T\to X$$s,t:T\to X$,

$$fs=ft⟺es=et$$

by monicity of $m$$m$. Therefore

$$\mathrm{Eq}(f)\cong \mathrm{Eq}(e)$$

compatibly with their projections.

By Lemma B, $e$$e$ is the coequalizer of this kernel pair. Hence the quotient exists. Any chosen coequalizer $q:X\to X/\mathrm{Eq}(f)$$q:X\to X/\operatorname{Eq}(f)$ is uniquely isomorphic to $e$$e$, giving a commuting diagram

$$\begin{array}{ccccc}X& \stackrel{q}{\to }& X/\mathrm{Eq}(f)& \stackrel{\overline{f}}{\to }& Y\\ \Vert & & \downarrow \theta \cong & & \Vert \\ X& \stackrel{e}{\to }& I& \stackrel{m}{\to }& Y.\end{array}$$

In particular, $\overline{f}=m\theta$$\bar f=m\theta$ is monic. $◻$$\square$

Remark on the hypotheses. This proof does not use (R3). It works in any category with kernel pairs and regular epimorphism–monomorphism factorizations. Pullback stability strengthens the conclusion by making these image factorizations stable under base change.

Why kernel pairs matter. In $\mathbf{Set}$$\mathbf{Set}$, the kernel pair is the relation

$$x{\sim }_f{x}^{\prime }⟺f(x)=f(x^{\prime }).$$

For groups and modules, this relation is determined by the ordinary kernel, and its quotient is the familiar quotient by $\ker (f)$$\ker(f)$.

This need not hold for other algebraic structures. Consider the monoid homomorphism

$$\begin{array}{c}f:(\mathbb{N},+,0)⟶(\{0,1\},max,0),f(n)=\{\begin{array}{ll}0,& n=0,\\ 1,& n\ge 1.\end{array}\end{array}$$

Its kernel object, the inverse image of the unit, is the trivial monoid $\{0\}$$\{0\}$, but $f$$f$ is not injective. Its kernel pair records the additional information that all positive integers are identified.

Kernel-pair quotients and exactness. The logical interpretation distinguishes the two objects occurring in the theorem:

$$\mathrm{Eq}(f)(x,x^{\prime }):⟺f(x)=f(x^{\prime }),$$

whereas

$$\mathrm{im}(f)(y):⟺\mathrm{\exists }x(f(x)=y).$$

The theorem says that quotienting by the first relation produces the object defined by the second formula.

It does not say that every internal equivalence relation arises this way. An internal equivalence relation is effective if it is the kernel pair of some morphism. A regular category is Barr-exact if every internal equivalence relation is effective. By the theorem, this is equivalent to requiring every internal equivalence relation to be the kernel pair of its coequalizer.

The category ${\mathbf{Ab}}_{\mathrm{tf}}$$\mathbf{Ab}_{\mathrm{tf}}$ illustrates the distinction. On the object $\mathbb{Z}$$\mathbb Z$, consider the regular formula

$$R(x,y):⟺\mathrm{\exists }z:\mathbb{Z}(x-y=z+z).$$

It defines the internal equivalence relation of congruence modulo $2$$2$:

$$R=\{(x,y)\in {\mathbb{Z}}^2:x-y\in 2\mathbb{Z}\}.$$

As a subgroup of ${\mathbb{Z}}^2$$\mathbb Z^2$, $R$$R$ is torsion-free.

Suppose $h:\mathbb{Z}\to A$$h:\mathbb Z\to A$, with $A$$A$ torsion-free, coequalizes the projections of $R$$R$. Since $(2,0)\in R$$(2,0)\in R$,

$$h(2)=0,\text{so}2h(1)=0.$$

Torsion-freeness forces $h(1)=0$$h(1)=0$, and hence $h=0$$h=0$. Therefore the coequalizer of $R\rightrightarrows \mathbb{Z}$$R\rightrightarrows\mathbb Z$ in ${\mathbf{Ab}}_{\mathrm{tf}}$$\mathbf{Ab}_{\mathrm{tf}}$ is

$$\mathbb{Z}⟶0.$$

Its kernel pair is all of ${\mathbb{Z}}^2$$\mathbb Z^2$, strictly larger than $R$$R$. Thus $R$$R$ has a coequalizer but is not effective, and ${\mathbf{Ab}}_{\mathrm{tf}}$$\mathbf{Ab}_{\mathrm{tf}}$ is regular without being Barr-exact.

Regularity guarantees effective quotients of kernel pairs. Barr-exactness extends this guarantee to all internal equivalence relations.

Proposition (Regularity of categories of algebraic models). Let $\mathcal{C}$$\mathcal C$ be a regular category and let $\mathbb{T}$$\mathbb T$ be a small finite-product sketch. Then

$$\mathrm{Mod}(\mathbb{T},\mathcal{C})$$

is regular. Moreover:

1. Finite limits are computed pointwise.

2. Regular epimorphism–monomorphism factorizations are computed pointwise.

3. A homomorphism of models is a regular epimorphism if and only if each component is a regular epimorphism in $\mathcal{C}$$\mathcal C$.

Proof. Present $\mathbb{T}$$\mathbb T$ by a small category $\mathcal{D}$$\mathcal D$ with specified finite-product cones. Its models form a full subcategory of $[\mathcal{D},\mathcal{C}]$$[\mathcal D,\mathcal C]$.

For a finite diagram of models $(F_j)$$(F_j)$, set

$$L(s):=\underset{j}{lim}{F}_j(s).$$

If $(s\to s_i{)}_{i=1}^n$$(s\to s_i)_{i=1}^n$ is a specified product cone, then

$$L(s)\cong \underset{j}{lim}\prod _i{F}_j(s_i)\cong \prod _i\underset{j}{lim}{F}_j(s_i)=\prod _{i}L(s_i).$$

This includes $n=0$$n=0$. Hence $L$$L$ is a model and gives the required pointwise limit. This part uses only (R1).

Now let $\alpha :F\to G$$\alpha:F\to G$ be a homomorphism of models. By (R2), factor each component:

$$F(s)\stackrel{e_s}{\to }I(s)\stackrel{m_s}{\to }G(s),{\alpha }_s=m_s{e}_s.$$

For $u:s\to t$$u:s\to t$, Lemma A applied to

$$\begin{array}{ccc}F(s)& \stackrel{e_{t}F(u)}{\to }& I(t)\\ e_s\downarrow & & \downarrow m_t\\ I(s)& \stackrel{G(u)m_s}{\to }& G(t)\end{array}$$

gives a unique $I(u):I(s)\to I(t)$$I(u):I(s)\to I(t)$ satisfying

$$I(u)e_s=e_{t}F(u),m_{t}I(u)=G(u)m_s.$$

Uniqueness gives the functor laws. Thus $I$$I$ is a functor and $e:F\to I$$e:F\to I$, $m:I\to G$$m:I\to G$ are natural transformations.

For a specified product cone, let

$$c:I(s)⟶\prod _{i}I(s_i)$$

be the comparison induced by $I$$I$. We have a commuting diagram

$$\begin{array}{ccccc}F(s)& \stackrel{e_s}{\to }& I(s)& \stackrel{m_s}{\to }& G(s)\\ \downarrow \cong & & \downarrow c& & \downarrow \cong \\ \prod _{i}F(s_i)& \stackrel{\prod _i{e}_{s_i}}{\to }& \prod _{i}I(s_i)& \stackrel{\prod _i{m}_{s_i}}{\to }& \prod _{i}G(s_i).\end{array}$$

By Lemma C, $\prod _i{e}_{s_i}$$\prod_i e_{s_i}$ is a regular epimorphism. Also, $\prod _i{m}_{s_i}$$\prod_i m_{s_i}$ is monic. The two rows therefore give regular epimorphism–monomorphism factorizations of the same morphism, after identifying their endpoints. Lemma A implies that $c$$c$ is an isomorphism.

This includes nullary products, so $I$$I$ is a model. This is the step that uses finite-product preservation from Lemma C, and hence (R3).

We have obtained a factorization

$$F\stackrel{e}{\to }I\stackrel{m}{\to }G$$

in the category of models. Componentwise monicity makes $m$$m$ monic there.

To prove that $e$$e$ is regular epic, form its pointwise kernel pair

$$\mathrm{Eq}(e)=F{\times }_{I}F.$$

It is a model by the finite-limit argument. By Lemma B, each $e_s$$e_s$ is the coequalizer of its kernel pair, so $e$$e$ is a pointwise coequalizer in the functor category. Fullness of the category of models gives the same universal property there.

The same argument applies to every componentwise regular epimorphism between models. Conversely, if $\alpha$$\alpha$ is a regular epimorphism of models, its factorization $\alpha =me$$\alpha=me$ forces $m$$m$ to be an isomorphism by Lemma A. Thus each component ${\alpha }_s=m_s{e}_s$$\alpha_s=m_se_s$ is a regular epimorphism in $\mathcal{C}$$\mathcal C$.

Finally, pullbacks of models are computed pointwise. Pulling back a regular epimorphism of models therefore pulls back regular epimorphisms in $\mathcal{C}$$\mathcal C$. By (R3), all resulting components are regular epimorphisms, so the resulting homomorphism of models is regular epic.

This proves all three regularity axioms and the pointwise assertions. $◻$$\square$

Corollary (First Isomorphism Theorem for algebraic models). Under the hypotheses above, every homomorphism $\alpha :F\to G$$\alpha:F\to G$ induces a canonical isomorphism of models

$$\boxed{{\displaystyle F/\mathrm{Eq}(\alpha )\cong \mathrm{im}(\alpha ).}}$$

The kernel pair, image, and quotient by the kernel pair are computed pointwise. In particular,

$$\mathrm{Eq}(\alpha )(s)\cong F(s){\times }_{G(s)}F(s),$$

and

$$(F/\mathrm{Eq}(\alpha ))(s)\cong F(s)/\mathrm{Eq}({\alpha }_s)\cong \mathrm{im}({\alpha }_s).$$

Proof. Apply the First Isomorphism Theorem in the regular category $\mathrm{Mod}(\mathbb{T},\mathcal{C})$$\operatorname{Mod}(\mathbb T,\mathcal C)$ and use the pointwise constructions from the proposition. $◻$$\square$

Specialization to Set-valued models of a Lawvere theory.

Let $\mathbb{T}$$\mathbb T$ be a Lawvere theory with generating object $t$$t$, and let $\alpha :A\to B$$\alpha:A\to B$ be a homomorphism between finite-product-preserving functors $\mathbb{T}\to \mathbf{Set}$$\mathbb T\to\mathbf{Set}$. Write

$$|A|:=A(1),|B|:=B(1),f:={\alpha }_1.$$

Since $\mathbf{Set}$$\mathbf{Set}$ is regular, the preceding corollary applies. On underlying sets, the kernel pair is the equivalence relation

$$a{\sim }_f{a}^{\prime }⟺f(a)=f(a^{\prime }).$$

This relation is a $\mathbb{T}$$\mathbb T$-congruence. Indeed, for every operation $\omega :t^n\to t$$\omega:t^n\to t$, preservation of $\omega$$\omega$ by $f$$f$ implies

$$a_i{\sim }_f{a}_i^{\prime }\text{for all }i⟹{\omega }_A(a_1,\dots ,a_n){\sim }_f{\omega }_A(a_1^{\prime },\dots ,a_n^{\prime }).$$

Consequently, the quotient carries well-defined operations

$${\omega }_{A/{\sim }_f}([a_1],\dots ,[a_n])=[{\omega }_A(a_1,\dots ,a_n)].$$

Likewise, $f(|A|)\subseteq |B|$$f(|A|)\subseteq|B|$ is closed under all operations, including nullary operations, and therefore carries the induced $\mathbb{T}$$\mathbb T$-model structure.

The canonical bijection

$$\overline{f}:|A|/{\sim }_f⟶f(|A|),[a]⟼f(a),$$

is thus an isomorphism of $\mathbb{T}$$\mathbb T$-models:

$$\boxed{{\displaystyle A/{\sim }_f\cong \mathrm{im}(\alpha ).}}$$

This is the usual First Isomorphism Theorem for universal algebra. For groups and modules, the congruence ${\sim }_f$$\sim_f$ is determined by the ordinary kernel, giving the familiar formula $A/\ker (f)\cong \mathrm{im}(f)$$A/\ker(f)\cong\operatorname{im}(f)$.

Why finite products are essential. The proposition does not extend to arbitrary finite-limit sketches. In the proof, the crucial step is that a finite product of regular epimorphisms is regular epic.

For a pullback specification, the analogous argument would require a comparison

$$F(s_1){\times }_{F(s_0)}F(s_2)⟶I(s_1){\times }_{I(s_0)}I(s_2)$$

to be regular epic. Componentwise regular epimorphisms do not guarantee this. Compatibility in the target need not lift to compatibility in the source.

The category $\mathbf{Cat}$$\mathbf{Cat}$ gives a concrete counterexample. Small categories are models of a finite-limit sketch, but $\mathbf{Cat}$$\mathbf{Cat}$ is not regular.

Let

$$A=(0\to 1)⨿(1^{\prime }\to 2),B=(0\to 1\to 2),$$

where $B$$B$ includes the composite arrow $0\to 2$$0\to2$. The functor $q:A\to B$$q:A\to B$ that identifies $1$$1$ and $1^{\prime }$$1'$ is the coequalizer of the two functors $\mathbf{1}\rightrightarrows A$$\mathbf1\rightrightarrows A$ selecting these objects. Thus $q$$q$ is a regular epimorphism.

Let $B^{\prime }\subseteq B$$B'\subseteq B$ be the full subcategory on $\{0,2\}$$\{0,2\}$. Its pullback along $q$$q$ is the discrete category on these two objects:

$$\begin{array}{ccc}\mathrm{Disc}\{0,2\}& ⟶& A\\ p\downarrow & & \downarrow q\\ B^{\prime }& \hookrightarrow & B.\end{array}$$

The functor $p$$p$ is monic but not an isomorphism, so it cannot be a regular epimorphism. Hence regular epimorphisms in $\mathbf{Cat}$$\mathbf{Cat}$ are not pullback-stable.

The same example explains the failure of pointwise images: the image of $q$$q$ contains the two generating arrows of $B$$B$ but not their composite, and therefore does not form a category.