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For a subobject $m:A^{\prime }\hookrightarrow A$ and a left-exact reflector $L$ with unit $\eta$, define $c_A(A^{\prime })$ by the pullback in a categoryThis operation is monotone, inflationary, idempotent, and stable under pullback. If $A$ is fixed by $L$, then $A^{\prime }$ is fixed by $L$ exactly when $A^{\prime }$ is closed.

A regular category has finite limits, every morphism factors through its image of a morphism in a regular category as a regular epimorphism followed by a monomorphism, and regular epimorphisms are stable under every pullback in a category. A cover is a strong epimorphism. Every regular epimorphism is strong: if $e$ is the coequalizer of $r,s$ and a square has $e$ on the left and a monomorphism $m$ on the right, monicity shows that the upper arrow coequalizes $r,s$. It therefore factors through $e$, and the epimorphism property of $e$ shows that this factor is the required diagonal. Conversely, factor a strong epimorphism $f$ as $f=me$ with $e$ regular epic and $m$ monic. The lifting property gives a two-sided inverse to $m$, so $m$ is an isomorphism and $f$ is regular epic. Thus regular epimorphisms and covers coincide.

Let $L:\mathcal{C}\to \mathcal{D}$ be the left-exact reflector and let $m:A^{\prime }\hookrightarrow A$. Since $L$ preserves finite limits, $Lm$ is monic. Define $c_A(A^{\prime })$ as the pullbackNaturality of the unit supplies a map $A^{\prime }\to c_A(A^{\prime })$ over $A$, proving $A^{\prime }\le c_A(A^{\prime })$. A factorization $A^{\prime }\le A^{\prime \prime }$ induces $L{A}^{\prime }\le L{A}^{\prime \prime }$ and therefore $c_A(A^{\prime })\le c_A(A^{\prime \prime })$, so $c_A$ is order-preserving.

Apply $L$ to the defining pullback. Left exactness and the fact that $L{\eta }_A:LA\to LLA$ is an isomorphism identify $L(c_A(A^{\prime }))$ with $L{A}^{\prime }$. Pulling back once more therefore givesFor a map $f:B\to A$, left exactness identifies $L(f^{\ast }{A}^{\prime })$ with $(Lf{)}^{\ast }(L{A}^{\prime })$. Pasting the two pullback squares then yieldsso this closure operation induced by a left-exact reflector commutes with pullback.

Assume $A$ lies in $\mathcal{D}$, so ${\eta }_A$ is an isomorphism. If $A^{\prime }$ also lies in $\mathcal{D}$, its unit is an isomorphism and the defining square gives $c_A(A^{\prime })\cong A^{\prime }$. Conversely, if $A^{\prime }$ is closed, that square expresses $A^{\prime }$ as a finite limit of $A$, $L{A}^{\prime }$, and $LA$, all fixed by $L$. Fixed objects of a left-exact reflective subcategory are closed under finite limits, so $A^{\prime }$ belongs to $\mathcal{D}$.

Finally suppose $\mathcal{C}$ is regular. The fixed objects have finite limits. For $f:A\to B$ in $\mathcal{D}$, factor it in $\mathcal{C}$ asApplying $L$ gives $A\stackrel{Le}{\to }LI\stackrel{Lm}{\to }B$. The map $Le$ is regular epic because a left adjoint preserves the coequalizer presenting $e$, and $Lm$ is monic because $L$ is left exact. Thus $\mathcal{D}$ has image factorizations. Their image subobject is the closure $c_B(I)$. A map in $\mathcal{D}$ is regular epic exactly when this closure is all of its codomain. Images in $\mathcal{C}$ commute with pullback, and the closure operation also commutes with pullback, so this condition is pullback-stable. Hence $\mathcal{D}$ is regular, as stated by the left-exact reflective subcategory of a regular category theorem.

Because $F:\mathbf{Set}\to \mathbf{Set}$ preserves finite limits and colimits, $F(1)$ is a singleton and $F(\varnothing )=\varnothing$. For $x\in X$, let $\bar{x}:1\to X$ select $x$ and definewhere $\ast$ is the unique element of $F(1)$. This is natural in $X$. Any natural transformation $1_{\mathbf{Set}}\to F$ has the unique possible component at $1$, and naturality along every $\bar{x}$ forces the displayed value, so $\alpha$ is unique.

If $x\ne y$, the equalizer of $\bar{x},\bar{y}:1\rightrightarrows X$ is empty. Since $F$ preserves this equalizer, $F(\bar{x})(\ast )\ne F(\bar{y})(\ast )$. Hence every ${\alpha }_X$ is injective, so $\alpha$ is pointwise monic.

Finite-colimit preservation makes ${\alpha }_n:n\to F(n)$ bijective for every finite set. Use the stated characterization of $\mathbb{N}$ by the coproduct diagramand the coequalizer of $s,1_{\mathbb{N}}:\mathbb{N}\rightrightarrows \mathbb{N}$. Applying $F$ preserves both diagrams. Naturality and uniqueness in this characterization identify ${\alpha }_{\mathbb{N}}:\mathbb{N}\to F(\mathbb{N})$ as an isomorphism.

For a countable family $(A_n)$, let $p:{\coprod }_n{A}_n\to \mathbb{N}$ record the summand. Each squareis a pullback. Applying $F$ and using $F(1)\cong 1$ and $F(\mathbb{N})\cong \mathbb{N}$ shows that $F(A_n)$ is exactly the fiber of $F(p)$ over $n$. Those fibers partition $F({\coprod }_n{A}_n)$, so the canonical mapis bijective. Thus $F$ preserves countable coproducts.

Now choose $x\in F(A)\setminus {\alpha }_A(A)$ and defineIt is upward closed. Since $A=A^{\prime }\bigsqcup (A\setminus A^{\prime })$ and $F$ preserves binary coproducts, $x$ lies in exactly one of the two summands, so exactly one of $A^{\prime }$ and its complement lies in $U$. Pullback preservation gives closure under finite intersections.

For countable completeness, take $A_n\in U$ and put $B={\bigcap }_n{A}_n$. If $B\notin U$, then $A\setminus B\in U$. Partition $A$ into $B$ and the setswhich record the first failed membership. Since $F$ preserves countable coproducts, exactly one cell of this partition lies in $U$. It cannot be $B$, so some $C_n\in U$. But $C_n\subseteq A\setminus A_n$, forcing $A\setminus A_n\in U$, contrary to $A_n\in U$. Hence $B\in U$.

Finally no finite $K$ belongs to $U$. Indeed, $F(K)\cong K$ through ${\alpha }_K$, so if $x$ came from $F(K)$ then naturality would put $x$ in the image of ${\alpha }_A$. Thus $U$ is a countably complete ultrafilter and is nonprincipal.

Conversely, let $U$ be such an ultrafilter on $A$ and define the ultrapower endofunctor of setsIt preserves the terminal object and products: the mapis bijective because $U$ is closed under finite intersections. It preserves equalizers because an equality holding for an equivalence class holds on a $U$-large set, and the representative can be changed off that set to land in the equalizer. Hence it preserves all finite limits.

For $f:A\to {\coprod }_n{B}_n$, countable completeness implies that one index fiberbelongs to $U$; otherwise the countable intersection of all complementary fibers would be empty and belong to $U$. Thus every class $[f]$ lies in one and only one $F(B_n)$, proving preservation of countable coproducts.

The class of the identity map $A\to A$ is not represented by a constant map, since every equality set $\{a:a=a_0\}$ is a singleton and is not in $U$. Therefore ${\alpha }_A:A\to F(A)$ is not surjective. Since $\alpha$ is the unique natural transformation from the identity functor to $F$, a natural isomorphism $1_{\mathbf{Set}}\cong F$ would have to equal $\alpha$, which is impossible.

Let $\mathcal{E}$ be an elementary topos with subobject classifier $\top :1\to \Omega$. A local operator is a map $j:\Omega \to \Omega$ which is inflationary, idempotent, preserves truth, and preserves binary meets:If $m:A^{\prime }\hookrightarrow A$ is classified by ${\chi }_m:A\to \Omega$, its closure operation of a local operator is classified by $j{\chi }_m$. The mono is j-dense monomorphism when this closure is all of $A$, and $j$-closed when it equals its closure. An object $X$ is a j-sheaf when every map $A^{\prime }\to X$ along a $j$-dense mono $A^{\prime }\hookrightarrow A$ extends uniquely to $A\to X$.

The closed-subobject classifier is the equalizerThus maps to ${\Omega }_j$ classify precisely the $j$-closed subobjects. Idempotence factors $j$ as

To prove that ${\Omega }_j$ is a $j$-sheaf, let $m:B\hookrightarrow A$ be dense and let $f:B\to {\Omega }_j$ classify a closed subobject $C\hookrightarrow B$. Take the closure in $A$ of the composite $C\hookrightarrow A$. Pullback stability of closure givesso the classifier $A\to {\Omega }_j$ of ${\overline{C}}^{,A}$ extends $f$. If two closed subobjects of $A$ restrict to the same subobject of dense $B$, the equalizer of their classifiers is a closed subobject containing $B$; it is both closed and dense and hence equals $A$. The extension is therefore unique.

Letbe the sheaf reflector for a local operator. The subobject classifier in the sheaf topos is ${\Omega }_j$. We prove the four assertions through the cycle

The canonical map comparing the reflected ambient classifier with the sheaf classifier isConsequently $L$ preserves the subobject classifier exactly when $L(q)$ is an isomorphism. This proves $(i)⟺(ii)$.

Since $qi=1_{{\Omega }_j}$, one hasIf $L(q)$ is an isomorphism then $L(i)$ is its inverse. Conversely, if $L(i)$ is an isomorphism, the same equation makes $L(q)$ its inverse. A monomorphism is sent to an isomorphism by sheafification exactly when it is $j$-dense, so $(ii)⟺(iii)$.

Assume $(iii)$ and let $m:A^{\prime }\hookrightarrow A$ have characteristic map $\chi :A\to \Omega$. Form the pullbackBecause dense monos are pullback-stable, $d:A^{\prime \prime }\hookrightarrow A$ is $j$-dense. The original $A^{\prime }$ factors through $A^{\prime \prime }$, and its characteristic map inside $A^{\prime \prime }$ is the top horizontal map followed by $i$, which is fixed by $j$. Hence $A^{\prime }\hookrightarrow A^{\prime \prime }$ is $j$-closed. This proves $(iv)$.

Finally assume $(iv)$ and apply it to $i:{\Omega }_j\hookrightarrow \Omega$:where $c$ is closed and $d$ is dense. Since $d$ is monic and $i=dc$, the map $c$ is the pullback of $d$ along $i$. It is therefore dense as well as closed, and hence is an isomorphism. Thus $i$ is, up to an isomorphism, the dense mono $d$, proving $(iii)$. All four conditions are equivalent, as summarized by the subobject-classifier preservation criterion for a sheaf reflector.