= Solution
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 $a:A'\hookrightarrow A$, define $\exists_f(A')$ as the <image factorization> subobject of $fa:A'\to B$. For $b:B'\hookrightarrow B$, the <pullback in a category> universal property and image minimality give
$$
\exists_f(A')\le B'\iff fa\text{ factors through }b
\iff A'\le f^*(B').
$$
Hence
$$
\boxed{\exists_f\dashv f^*.}
$$
This 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 $P=I\times_BB'$. Its mono into $B$ represents $\exists_f(A')\cap B'$. The <pullback in a category> of $e$ along the <monomorphism> $P\to I$ has domain canonically $A'\times_BB'$, which represents $A'\cap f^*(B')$. 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 gives
$$
\boxed{\exists_f(A'\cap f^*B')=\exists_f(A')\cap B'.}
$$
Equality here is equality of <subobjects>, hence an isomorphism of their representatives.
Conversely, let $e:X\to Y$ be a <strong epimorphism> and $b:B'\hookrightarrow Y$ 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 $A'=X$. The induced <morphism> $p:X\times_YB'\to B'$ has image all of $B'$, because the image of $bp$ in $Y$ is $B'$. 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 $\bot$. Nonzero objects in different copies have no <morphisms> between them. There is one <morphism> from every object to $Z$ and from $\bot$ to every object, and no <morphism> into $\bot$ 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 $\bot$, since only $\bot$ can map to both. A product with $Z$ is the other factor, and one with $\bot$ is $\bot$. 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 $\bot\to X$, and the only <subobjects> of $Z$ are $\bot$ 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 $\bot\to Z$ would require a nonexistent map from a noninitial domain to $\bot$. Maps with domain $\bot$ 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 $\bot\to X$ with $X\ne\bot$ is monic but noninvertible, and therefore cannot be strong. This classifies the <strong epimorphisms> as the identities of $\bot$ 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 $\bot\to X$ gives $1_\bot$, 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 is
$$
\bot\longrightarrow\mathbb Z[x]_{\mathcal E}.
$$
It is not even an <epimorphism>, since the distinct evaluation <ring homomorphisms> $x\mapsto0$ and $x\mapsto1$ to $\mathbb Z_{\mathcal E}$ agree after composing with $\bot\to\mathbb Z[x]_{\mathcal E}$. Since $e$ is strong, it would be a <regular epimorphism> in a <regular category>, whose pullbacks must be regular, in particular epic. Therefore \b[the glued category has finite limits and images and satisfies Frobenius reciprocity, but is not regular].
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