Results on subcategories

Proof. If $X \in \C$, then $$\Hom(X,R(\Omega)) \cong \Hom(U(X),\Omega) \cong \Sub(U(X)) \cong \Sub(X).$$ The same proof works for regular subobjects. $\square$

Proof. If $(f_i : X_i \to Y_i)$ is a filtered diagram of monomorphisms in $\C$, it induces a filtered diagram $(U(f_i) : U(X_i) \to U(Y_i))$ of monomorphisms in $\D$. Hence, its colimit $\colim_i U(f_i) : \colim_i U(X_i) \to \colim_i U(Y_i)$ is a monomorphism in $\D$. This morphism is isomorphic to $U(\colim_i f_i) : U(\colim_i X_i) \to U(\colim_i Y_i)$. Since $U(\colim_i f_i)$ is a monomorphism in $\D$ and $U$ is faithful, it follows that $\colim_i f_i$ is a monomorphism in $\C$. $\square$

Proof. It is well known (and easy to prove) that the colimit of a diagram $(X_j)$ in $\C$ is given by $L(\colim_j U(X_j))$, provided that the colimit in $\D$ exists. In particular, $\C$ has filtered colimits. By assumption, it also has finite limits, and $U$ preserves them since it is a right adjoint. Now let $X : \I \times \J \to \C$ be a diagram, where $\I$ is finite and $\J$ is filtered. We compute:

$$\begin{align*} \colim_j {\lim}_i X(i,j) & \cong L(\colim_j U({\lim}_i X(i,j))) \\ & \cong L(\colim_j {\lim}_i U(X(i,j))) \\ & \cong L({\lim}_i \colim_j U(X(i,j))) \\ & \cong {\lim}_i L(\colim_j U(X(i,j))) \\ & \cong {\lim}_i \colim_j X(i,j) \end{align*}$$

It is routine to check that the composite is the natural morphism. $\square$

Proof. Let $X$ be an object of $\C$ and $Y : \I \to \C$ a filtered diagram. Then we have the canonical comparison map $c : \colim_{i\in\I} (X \times Y_i) \to X \times \colim_{i\in\I} Y_i$. By the assumptions, $Uc$ is equivalent to the comparison map $\colim_{i\in\I} (UX \times UY_i) \to UX \times \colim_{i\in\I} UY_i$, which is an isomorphism. Since $U$ is fully faithful and therefore conservative, we conclude that $c$ is an isomorphism. $\square$

Proof. For any objects $X, Y, Z$ of $\C$ we have natural isomorphisms

$$\begin{align*} \Hom_\C(Z\times X, Y) & \cong \Hom_\D(UZ \times UX, UY) \\ & \cong \Hom_\D(UZ, [UX,UY]) \\ & \cong \Hom_\C\bigl(Z, R([UX,UY])\bigr). \end{align*}$$ $\square$

Proof. The forgetful functor $\C / P \to \C$ is fully faithful; it has right adjoint ${-} \times P$; and it preserves binary products (in fact all inhabited limits). Hence, Lemma 5 applies. $\square$

Proof. Since $\C$ has finite limits and coequalizers of kernel pairs, it remains to prove that regular epimorphisms are stable under pullbacks in $\C$. Assume first that $U$ preserves and reflects regular epimorphisms. If $X \to Y$ is a regular epimorphism and $Z \to Y$ is any morphism in $\C$, then $U(X) \to U(Y)$ is a regular epimorphism, and therefore also $U(X) \times_{U(Y)} U(Z) \to U(Z)$ is a regular epimorphism. Since $U$ preserves pullbacks, this identifies with $U(X \times_Y Z) \to U(Z)$, the image under $U$ of $X \times_Y Z \to Z$. Since $U$ reflects regular epimorphisms, it follows that $X \times_Y Z \to Z$ is a regular epimorphism, finishing the proof.

Now assume that $U$ is conservative and preserves coequalizers. Then it clearly preserves regular epimorphisms. Conversely, suppose that $f : X \to Y$ is a morphism in $\C$ such that $U(f)$ is a regular epimorphism. Then in $\C$ we have the diagram $$X \times_Y X \rightrightarrows X \xrightarrow{p} \im(f) \xrightarrow{i} Y$$ where $X \times_Y X \rightrightarrows X$ is the kernel pair of $f$, and $\im(f)$ is the coequalizer. By the assumptions on $U$, its image is equivalent to the diagram in $\D$: $$U(X) \times_{U(Y)} U(X) \rightrightarrows U(X) \xrightarrow{U(p)} U(\im(f)) \xrightarrow{U(i)} U(Y)$$ where $U(X) \times_{U(Y)} U(X)$ is the kernel pair of $U(f)$, and $U(\im(f))$ is the coequalizer. Since $U(f)$ is a regular epimorphism and $\D$ is regular, we must have that $U(i)$ is an isomorphism. Since $U$ is conservative, $i$ is an isomorphism as well, so $f$ is a regular epimorphism. $\square$

Proof. Suppose we have a congruence $E \rightrightarrows X$ in $\C$. We first observe that its image $U(E) \rightrightarrows U(X)$ is a congruence in $\D$, and therefore effective. It is jointly monomorphic by assumption on $U$; the reflexivity and symmetry morphisms for $E$ are easily seen to transform under $U$ into reflexivity and symmetry morphisms for $U(E)$; and similarly, since $U$ preserves pullbacks, the transitivity morphism for $E$ transforms under $U$ into a transitivity morphism for $U(E)$.

Now let $X/E$ be the quotient of $E \rightrightarrows X$, i.e. its coequalizer. We will show that the canonical morphism $$E \to X \times_{X/E} X$$ is an isomorphism. Since $U$ is fully faithful and preserves pullbacks, it suffices to prove that the canonical morphism $$U(E) \to U(X) \times_{U(X/E)} U(X)$$ is an isomorphism, where $U(X/E)$ is the coequalizer of $U(E) \rightrightarrows U(X)$ since $U$ preserves coequalizers. Since the congruence $U(E) \rightrightarrows U(X)$ is effective by the first part of the proof, this follows from this lemma.

Finally, to prove the parenthetical remark, assume that $\C$ has binary products and that $U$ preserves them. Then $U$ preserves jointly monomorphic parallel pairs, because it preserves monomorphisms (as it preserves pullbacks), and a pair $A \rightrightarrows B$ is jointly monomorphic if and only if the induced morphism $A \to B \times B$ is a monomorphism. $\square$

Proof. If $S$ is a generating collection,

$$\begin{align*} \prod_{G\in S} \Hom(L(G),-) & \cong \prod_{G\in S} \Hom(G,U(-)) \\ & \cong \left( \prod_{G\in S} \Hom(G,-) \right) \circ U \end{align*}$$

is a composition of faithful functors, hence faithful. $\square$

Proof. Under the given assumptions, we can factor $\C \to (\Set^+)^S$, $X \mapsto (\Hom_\C(G, X))_{G\in S}$, as being isomorphic to the composition of $U : \C \to \D$ followed by $Y \mapsto (\Hom_\D(UG, Y))_{G\in S}$, using the assumption on $U$ to identify $\Hom_\D(UG, UX)$ with $\Hom_\C(G, X)$ naturally in $X$. In this composition, the first is fully faithful and therefore also conservative; and the second is assumed to be faithful and conservative. Therefore, the composition is also faithful and conservative. $\square$

Proof. This is straightforward. We need to prove that finite coproducts are disjoint and stable under pullbacks in $\C$. If $A,B \in \C$, the coproduct inclusion $A \to A + B$ is a monomorphism: Since $U$ is faithful, it suffices to prove that its image under $U$ is a monomorphism. Since $U$ preserves finite coproducts, the image identifies with the coproduct inclusion $U(A) \to U(A) + U(B)$, which is a monomorphism since $\D$ is extensive. Moreover, the unique morphism $0 \to A \times_{A + B} B$ is an isomorphism: Since $U$ is conservative, it suffices to prove that its image under $U$ is an isomorphism. Since $U$ preserves finite coproducts and pullbacks along coproduct inclusions, the image identifies with the unique morphism $0 \to U(A) \times_{U(A) + U(B)} U(B)$, which is an isomorphism since $\D$ is extensive. This proves that finite coproducts are disjoint in $\C$. To prove that they are stable under pullbacks, let $T \to A + B$ be any morphism in $\C$, and consider the pullbacks $T_A \coloneqq T \times_{A + B} A$ and $T_B \coloneqq T \times_{A + B} B$. We need to show that the canonical morphism $T_A + T_B \to T$ is an isomorphism. Since $U$ is conservative, it suffices to prove that its image under $U$ is an isomorphism. Since $U$ preserves finite coproducts and pullbacks along coproduct inclusions, the image identifies with the canonical morphism $U(T)_{U(A)} + U(T)_{U(B)} \to U(T)$ induced by the morphism $U(T) \to U(A) + U(B)$ in $\D$, which is an isomorphism since $\D$ is extensive. $\square$

Proof. Let $p_1,p_2 : E \rightrightarrows X$ be the kernel pair of $f$ in $\C$. Then $U(p_1),U(p_2)$ is the kernel pair of $U(f)$ in $\D$. Since $U(f)$ is a regular epimorphism, it is the coequalizer of $U(p_1),U(p_2)$ by this lemma. Since $U$ is fully faithful, it reflects colimits, so $f$ is the coequalizer of $p_1,p_2$ in $\C$, i.e. a regular epimorphism.

For the second statement, assume that $\C$ has kernel pairs and that $U$ maps epimorphisms to regular epimorphisms. By the first statement, every epimorphism of $\C$ is regular, i.e. $\C$ is epi-regular. $\square$