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Use the explicit construction of small limits from products and equalizers from part (i), here with finite indexing sets. If $F$ preserves finite products in a category and equalizers, then $F(P)$ and $F(Q)$, with their image projections, are the products of the objects $FD(j)$ and of the target objects indexed by arrows of $J$.

The image maps $F(a),F(b)$ have componentsAlso $F(e):F(L)\to F(P)$ is their equalizer. Therefore the same universal property proof makes $F(L)$, with projections $F(p_j)$, a categorical limit of $FD$. Any other chosen limit of $D$ is uniquely isomorphic to this construction, and the functor carries that isomorphism to an isomorphism. Consequently $F$ preserves every finite limit. Preservation of finite products includes the empty product, so the terminal object is also preserved.

Let $D:I\to \mathcal{C}$ be a finite diagram. Form the finite categorical productsThere are two morphisms $s,t:P\rightrightarrows Q$: their components at $u:i\to j$ are $D(u){\pi }_i$ and ${\pi }_j$, respectively. Let $e:L\to P$ be their equaliser. Then ${\ell }_i={\pi }_{i}e$ satisfy $D(u){\ell }_i={\ell }_j$, so they constitute a cone over a diagram.

Any other cone $x_i:X\to D(i)$ uniquely determines $x:X\to P$. Its cone equations say precisely that $sx=tx$. The equaliser therefore gives a unique $h:X\to L$ with $eh=x$, and thus ${\ell }_{i}h=x_i$. Conversely these component equations force $eh=x$ by the categorical product universal property, so there is exactly one such mediator. This proves that $(L,{\ell }_i)$ is a categorical limit of $D$.

The empty categorical products are included: if $I$ is empty, $P$ and $Q$ are terminal and the construction gives the empty categorical limit. Thus finite categorical products and equalisers give every finite categorical limit, as a special case of the construction of small limits from products and equalizers.

A diagram in a category of shape $J$ is a functor $D:J\to \mathcal{C}$. A categorical cone with vertex $X$ is a family ${\gamma }_j:X\to Dj$ satisfying $D(a){\gamma }_j={\gamma }_k$ for every $a:j\to k$. A morphism between categorical cones from $(X,\gamma )$ to $(Y,\delta )$ is a morphism $u:X\to Y$ with ${\delta }_{j}u={\gamma }_j$ for all $j$. A categorical limit is a terminal object in this category of cones: each cone has a unique such morphism to the limiting cone.

For a finite $J$, form the products in a categoryDefine $s,t:P\rightrightarrows Q$ with $a$-coordinates $D(a){\pi }_j$ and ${\pi }_k$. The equalizer $e:L\to P$ imposes exactly the cone equations. Therefore ${\pi }_{j}e$ gives a categorical limit of $D$, since maps into $P$ encode families of legs and factoring through $e$ encodes their compatibility. The empty product in a category is the terminal object, covering the empty diagram. This is the construction of small limits from products and equalizers, restricted to finite shapes.

Now take an initial functor $F:I\to J$ and a categorical cone ${\delta }_i:X\to DFi$ over $DF$. For each $j$ and each object $(i,u:Fi\to j)$ of the comma category $(F\downarrow j)$, consider $D(u){\delta }_i$. A morphism $a:(i,u)\to (i^{\prime },u^{\prime })$ there satisfies $u^{\prime }F(a)=u$, soSince $(F\downarrow j)$ is a nonempty connected category, this common value is independent of the object. Define it to be ${\gamma }_j$. This does not require choosing representatives: the value is uniquely determined.

For $b:j\to k$, replacing $(i,u)$ by $(i,bu)$ proves $D(b){\gamma }_j={\gamma }_k$. Taking $(i,1_{Fi})$ proves ${\gamma }_{Fi}={\delta }_i$. Conversely, extending a restricted cone recovers its original legs, since ${\gamma }_j=D(u){\gamma }_{Fi}$. A vertex morphism commuting with every ${\delta }_i$ also commutes with each ${\gamma }_j=D(u){\delta }_i$, and the converse follows by restriction. Thus extension and restriction are strictly inverse functors, not merely an equivalence of categories. This proves cone restriction along an initial functor.

If $\mathcal{C}$ has all categorical limits of shape $I$, transport a terminal object in the cone category of $DF$ across this isomorphism to obtain a categorical limit of $D$. Uniqueness of the induced comparison, and its compatibility with natural transformations of diagrams, givesFor the converse use the representable test for initial functors. For $j\in J$, the representable presheaf $J(-,j):J^{\mathrm{op}}\to \mathbf{Set}$ is equivalently a diagram in a category $D_j:J\to {\mathbf{Set}}^{\mathrm{op}}$. Its categorical limit in the opposite category is the colimit of $J(-,j)$ in sets. Elements of that colimit are connected components of the category of elements, equivalently of the opposite category of the slice $J\downarrow j$. This slice has terminal object $(j,1_j)$, so that colimit is a singleton.

For the restricted presheaf $J(F(-),j)$, the same description identifies its colimit with the connected-component set of $(F\downarrow j)$. Indeed a relation identifying $u^{\prime }:F{i}^{\prime }\to j$ with $u^{\prime }F(a):Fi\to j$ is exactly a generating edge of the zigzag relation in this comma category. The assumed isomorphism of limit functors forces this set to be a singleton. Therefore $(F\downarrow j)$ is nonempty and connected for every $j$, proving

A regular category has finite categorical limits, coequalizers of kernel pairs, and regular epimorphism–monomorphism image factorizations whose regular epimorphisms are stable under every pullback in a category. Equivalently, finite limits, pullback-stable regular epimorphism–monomorphism factorizations suffice. In particular, every strong epimorphism in a regular category is regular.

For the weaker hypotheses here, interpret an image factorization as $f=me$ with $m:I\hookrightarrow B$ the least subobject through which $f$ factors, and $e:A\to I$ a strong epimorphism. With pullback in a category constructions, the least-subobject characterization itself implies that $e$ is strong: in a lifting square against a monomorphism, pull that monomorphism back to $I$; since $e$ factors through this pullback, image minimality forces its mono into $I$ to be invertible, giving the required diagonal. Thus the two descriptions of images agree under the stated hypotheses.

For $a:A^{\prime }\hookrightarrow A$, define ${\exists }_f(A^{\prime })$ as the image factorization subobject of $fa:A^{\prime }\to B$. For $b:B^{\prime }\hookrightarrow B$, the pullback in a category universal property and image minimality giveHenceThis is an adjunction between the posets of subobjects.

For Frobenius reciprocity for subobjects, factor $fa=me$ through its image $m:I\hookrightarrow B$, and put $P=I{\times }_B{B}^{\prime }$. Its mono into $B$ represents ${\exists }_f(A^{\prime })\cap B^{\prime }$. The pullback in a category of $e$ along the monomorphism $P\to I$ has domain canonically $A^{\prime }{\times }_B{B}^{\prime }$, which represents $A^{\prime }\cap f^{\ast }(B^{\prime })$. If strong epimorphisms are stable under pullback along monomorphisms, that pulled-back arrow is strong, so its composite with $P\hookrightarrow B$ is an image factorization. This givesEquality here is equality of subobjects, hence an isomorphism of their representatives.

Conversely, let $e:X\to Y$ be a strong epimorphism and $b:B^{\prime }\hookrightarrow Y$ be monic. Its image is all of $Y$: in any image factorization of $e$, the lifting property makes the mono invertible. Apply Frobenius reciprocity for subobjects to $f=e$ and $A^{\prime }=X$. The induced morphism $p:X{\times }_Y{B}^{\prime }\to B^{\prime }$ has image all of $B^{\prime }$, because the image of $bp$ in $Y$ is $B^{\prime }$. Its image factorization therefore has invertible mono, so $p$ is a strong epimorphism. This proves the converse.

We now use the glued ring categories counterexample to regularity. All rings and ring homomorphisms are unital; the zero ring, where $0=1$, is allowed. It is the terminal object of $\mathbf{Rng}$, and there is no unital ring homomorphism from it to a nonzero ring. Denote the common zero ring by $Z$, and the newly adjoined strict initial object by $\perp$. Nonzero objects in different copies have no morphisms between them. There is one morphism from every object to $Z$ and from $\perp$ to every object, and no morphism into $\perp$ except its identity.

Finite products in a category and equalizers can be described explicitly. The terminal object is $Z$. The product in a category of two nonzero objects in one copy is their ordinary ring product; the product in a category of nonzero objects in different copies is $\perp$, since only $\perp$ can map to both. A product with $Z$ is the other factor, and one with $\perp$ is $\perp$. For parallel morphisms between nonzero objects in the same copy, the ordinary ring equalizer works; it is nonzero because its subring contains distinct $0$ and $1$. All other parallel pairs are equal, and their equalizer is the identity of the domain. Thus the construction of small limits from products and equalizers gives all finite categorical limits.

Within either ring copy, monomorphisms are precisely injective ring homomorphisms: maps from $\mathbb{Z}[x]$ detect unequal elements. The only monos from outside a copy are the maps $\perp \to X$, and the only subobjects of $Z$ are $\perp$ and $Z$. Indeed a nonzero ring $R$ is not subterminal, since $\mathbb{Z}[x]\to R$ can send $x$ to $0$ or to $1$.

Ordinary ring-image factorizations remain image factorizations in the glued category. Their surjective parts remain strong epimorphisms: a lifting square into a nonzero ring stays in one copy; a square into $Z$ with mono $\perp \to Z$ would require a nonexistent map from a noninitial domain to $\perp$. Maps with domain $\perp$ are already monic and have identity strong part. Conversely, a strong epimorphism within a ring copy has surjective ring image, since its mono image part must be invertible. A map $\perp \to X$ with $X\ne \perp$ is monic but noninvertible, and therefore cannot be strong. This classifies the strong epimorphisms as the identities of $\perp$ and the surjective homomorphisms in the copies, including maps to $Z$.

Pullback along a monomorphism within a copy preserves those surjections, using the given regularity of $\mathbf{Rng}$. Pullback along $\perp \to X$ gives $1_{\perp }$, and these exhaust the additional monos, including those into $Z$. Thus strong epimorphisms are stable under pullback along monos, so Frobenius reciprocity for subobjects holds.

However, let $e:{\mathbb{Z}}_{\mathcal{D}}\to Z$ be the surjective ring homomorphism in the first copy, and pull it back along $\mathbb{Z}[x{]}_{\mathcal{E}}\to Z$ in the other copy. The resulting arrow isIt is not even an epimorphism, since the distinct evaluation ring homomorphisms $x\mapsto 0$ and $x\mapsto 1$ to ${\mathbb{Z}}_{\mathcal{E}}$ agree after composing with $\perp \to \mathbb{Z}[x{]}_{\mathcal{E}}$. Since $e$ is strong, it would be a regular epimorphism in a regular category, whose pullbacks must be regular, in particular epic. Therefore the glued category has finite limits and images and satisfies Frobenius reciprocity, but is not regular.

A diagram in a category is a functor $D:\mathcal{J}\to \mathcal{C}$. A cone over a diagram with vertex $A$ is a familysuch that $D(u){\gamma }_i={\gamma }_j$ for every $u:i\to j$. A categorical limit is a terminal cone: for every cone $(A,\gamma )$ there is a unique map $A\to \lim D$ commuting with all legs.

Suppose $\mathcal{C}$ has small products and equalizers. For a small diagram $D$, formThere are two maps $r,s:P\rightrightarrows Q$. In the coordinate indexed by $u:i\to j$, letThe equalizer $E\to P$ imposes exactly the cone equations. Maps $A\to E$ are therefore naturally the same as cones from $A$ to $D$, so $E\cong \lim D$. This is the construction of small limits from products and equalizers.

Let $F:\mathcal{I}\to \mathcal{J}$ be initial, so every $(F\downarrow j)$ is nonempty and connected. Restriction sends a cone $(A,{\gamma }_j)$ over $D$ to $(A,{\gamma }_{Fi})$ over $DF$. Conversely, given a cone $(A,{\delta }_i)$ over $DF$, choose an objectand defineA morphism in the comma category shows that this expression is unchanged along one edge, and connectedness makes it independent of the chosen object. The cone equations follow by choosing $(i,vu)$ for an arrow $v:j\to j^{\prime }$. This construction is inverse to restriction and acts identically on vertex maps, proving the cone restriction along an initial functor isomorphism.

Terminal objects in the two cone categories therefore correspond. Whenever the $\mathcal{I}$-shaped limit exists,naturally in $D$. Equivalently, the triangle formed by precompositionand the two limit functors commutes up to natural isomorphism.

For the converse, suppose this commutation holds for $\mathcal{C}={\mathbf{Set}}^{\mathrm{op}}$. Passing to opposite categories says that restriction along $F^{\mathrm{op}}$ preserves all set-valued colimits. Fix $j\in \mathcal{J}$ and take the representable functorIts colimit is a singleton: the category of its elements has the initial object $(j,1_j)$. The restricted colimit iswhose elements are precisely the connected components of $(F\downarrow j)$. By the assumed comparison this set is also a singleton. Thus $(F\downarrow j)$ is nonempty and connected for every $j$, so $F$ is initial.

A categorical limit of $D:\mathcal{J}\to \mathcal{C}$ is a terminal cone $(L,(p_j))$: every cone $(X,(x_j))$ has a unique map $X\to L$ commuting with all legs.

Assume all small products and equalizers exist. PutDefine $s,t:P\rightrightarrows Q$ so that the $u:i\to j$ components areA map $X\to P$ equalizes $s,t$ exactly when its components form a cone over $D$. Therefore $\mathrm{eq}(s,t)$ represents cones and is the limit. This is the construction of small limits from products and equalizers.