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For a left-exact reflective subcategory of a Cartesian closed category, an ambient exponential $B^A$ is already fixed by the reflector when $A,B$ are fixed. Indeed maps into $B^A$ transpose to maps from $X\times A$ into $B$, and reflection gives $L(X\times A)\cong LX\times A$. Precomposition with the reflection unit is therefore bijective for maps into $B^A$, characterizing a fixed object.

Every left-exact reflective subcategory of a regular category is regular. The reflector sends a regular-epimorphism--monomorphism factorization to such a factorization among fixed objects. A morphism between fixed objects is a regular epimorphism precisely when the closure of its image under the closure operation induced by a left-exact reflector is the whole codomain; pullback stability follows from pullback stability of images and of closure.

For an idempotent monad, the unit laws give $T\eta =\eta T={\mu }^{-1}$. If $a:TA\to A$ is an algebra, $a{\eta }_A=1$ and naturality giveThus algebras are exactly the objects on which ${\eta }_A$ is invertible, with the unique algebra structure ${\eta }_A^{-1}$. They form a full reflective subcategory with reflector $T$. The assumed preservation of finite limits makes this a left-exact reflective subcategory.

For a subobject $S\hookrightarrow A$, define its closure byusing the unit and the monomorphism $TS\hookrightarrow TA$. This closure operation induced by a left-exact reflector is extensive by unit naturality, stable under pullback and preserves intersections because $T$ preserves finite limits. Applying $T$ to its defining square and using $T\eta$ invertible identifies $T\overline{S}$ with $TS$ over $TA$. Closing a second time therefore gives the same subobject. It is a universal idempotent closure operation, hence defines a local operator $j$ by classifying the closure of $t:1\hookrightarrow \Omega$; closure of any mono is then classified by composing its classifier with $j$.

A mono $s:S\hookrightarrow A$ is $j$-dense exactly when $Ts$ is invertible. One implication follows immediately from the closure square. Conversely if its closure is all of $A$, the unit ${\eta }_A$ factors through $Ts$. Applying $T$ and using the invertible $T{\eta }_A$ makes $Ts$ split epic; since it is also monic, it is invertible. Every fixed object $B$ is consequently a $j$-sheaf, since the adjunction givesand a dense restriction becomes an isomorphism.

Conversely let $B$ be a $j$-sheaf. Its diagonal is closed: the two projections from the closure of the diagonal agree on the dense diagonal, so uniqueness of extension into $B$ makes them agree everywhere. The closure formula identifies this closed diagonal with the kernel pair of ${\eta }_B$, since $T$ takes the diagonal of $B$ to the diagonal of $TB$. Thus ${\eta }_B$ is monic. Also $T{\eta }_B$ is invertible, so this mono is dense. The sheaf property extends $1_B$ across it to a retraction $TB\to B$. Since $TB$ is fixed and hence itself a sheaf, uniqueness across that dense mono also makes the other composite the identity. Therefore ${\eta }_B$ is invertible. We have proved the algebra category is exactly the category of $j$-sheaves.

Its topos structure is concrete. Finite limits are computed in $\mathcal{E}$. For sheaves $A,B$, the ambient exponential $B^A$ is a sheaf: extension across a dense mono reduces, by transposition, to extension into $B$ across its product with $A$, again a dense mono. The classifier is ${\Omega }_j=\mathrm{Eq}(j,1_{\Omega })$ with the induced truth point; it classifies closed subobjects. Colimits are obtained by taking the ambient colimit and applying $T$. These supply an elementary topos, as required.

For the second part, use the square reader monadThe two unit laws select $(a,b)$ again. Each side of associativity selects the entries indexed $(0,0,0)$ and $(1,1,1)$ in an eight-entry array, so the monad laws hold. An algebra is an operation $\ast :A\times A\to A$ satisfyingThese imply associativity, since $(x\ast y)\ast z=(x\ast y)\ast (z\ast z)=x\ast z$ and $x\ast (y\ast z)=(x\ast x)\ast (y\ast z)=x\ast z$. They are the rectangular band laws.

For nonempty $A$, choose $e\in A$, and put $B=\{a\ast e:a\in A\}$, $C=\{e\ast a:a\in A\}$. Define $\varphi (a)=(a\ast e,e\ast a)$ and $\psi (b,c)=b\ast c$. Then $(a\ast e)\ast (e\ast a)=a\ast a=a$, and for $b=x\ast e$, $c=e\ast y$ the laws give $(b\ast c)\ast e=b$ and $e\ast (b\ast c)=c$. Thus $\varphi ,\psi$ are inverse bijections. In these coordinates the operation isConversely any nonempty product with this operation satisfies the algebra laws, proving the requested correspondence with product decompositions. A homomorphism between nonempty products has its first coordinate depending only on the first factor and its second only on the second, as follows by applying it to the displayed operation. Hence it is exactly a pair of factor maps. There is in addition the unique empty algebra, a strict initial object.

This algebra category is Cartesian closed. For nonempty $A=B\times C$ and $D=E\times F$, its exponential is $E^B\times F^C$ with rectangular multiplication: the usual set adjunction in each factor gives the universal property. For the empty cases, $D^0=1$ and $0^A=0$ when $A$ is nonempty; these formulas also give $0^0=1$. They verify the adjunction against empty test objects as well. Thus the algebra category is Cartesian closed.

It is not a topos. If a classifier existed, its global elements would correspond to the two subobjects of the terminal singleton algebra, so its underlying set would have two elements. A two-element rectangular band is either left-zero or right-zero: its two nonempty factors have sizes $(2,1)$ or $(1,2)$. A homomorphism from the four-element band $\{0,1\}\times \{0,1\}$ to a left-zero classifier depends only on the first coordinate, so the preimage of truth cannot be a single point; for a right-zero classifier the same failure occurs in the second coordinate. Yet a single point is a subalgebra and its inclusion is monic. Pullbacks of algebra maps are computed on underlying sets, so this subobject could not be classified. Therefore there is no subobject classifier and the algebra category is not a topos, despite its cartesian closure.

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.