Examples of coregular categories

Question

I'm interested in some examples of coregular categories (and a sketch of a proof, that they are in fact coregular). It would be nice to apply things I know about regular categories to some coregular categories too.

Obviously, I'm not interested in things like: the dual of some topos, the dual of some algebraic category, an Abelian category; unless they go by some specific name.

Remark: This answer lists some examples, but without proof. Also it is unclear to me whether $\mathsf{Grp}$ is coregular or not, which was stated somewhat like an exercise there.

Answer 1

The category of sets is coregular
Every category which is monadic over the category of sets, $\mathbf{Set}$, is regular. In particular this means that $\mathbf{Set}$ is coregular (since the power set functor $P : \mathbf{Set}^{\text{op}} \to \mathbf{Set}$ is monadic.)

The category of topological spaces is coregular
Let $\mathbf{Top}$ be the category of topological spaces. Trivially it has all finite limits and colimits. Therefore, it remains to show that regular monomorphisms are pushout stable. To prove this we will use the following facts and definitions:
(a) monomorphisms in $\mathbf{Top}$ are injective continuous maps;
(b) a morphism $f: A\to B$ is a regular monomorphisms in $\mathbf{Top}$ if and only if its underlying map is injective and the open sets in $A$ are exactly the inverse images of open sets from $B$;
(c) injective maps of sets are up to isomorphism coproduct inclusions;
(d) In what follows the coproduct of sets $S+T$ will be defined as $(S\times \{0\}) \cup (T\times \{1\})$. Note that this means that if $U \subseteq S$ and $V\subseteq T$, then $U+V \subseteq S+T$;
(e) the pushout of topological spaces is computed as in $\mathbf{Set}$ and has the coarsest topology such that the pushout inclusions are continuous. (i.e. a subset of the underlying set of the pushout is open if and only if its inverse image along both pushout inclusions is open).

Now suppose the $f:A\to B$ is a regular monomorphism, and $g : A \to C$ is morphism. Without loss of generality, by (c), we may assume that the underlying set of $B$ is $A + X$ and that the underlying map of $f$ is the first coproduct inclusion. Now forming the pushout $$\require{AMScd}\begin{CD} A @>{f}>> B\\ @V{g}VV @VV{i_2}V \\ C @>>{i_1}> C+_A B \end{CD}$$ we may choose $C+_A B$ to have underlying set $C+X$, $i_1$ to have underlying map the first coproduct inclusion, and $i_2$ to have underlying map $g+1$. Now we need to show that if $U$ is open in $C$ then there exists $V$ open in $C+_A B$ such that $U=i_1^{-1}(V)$. So suppose $U$ is open in $C$. Since $g$ is continuous $g^{-1}(U)$ is open in $A$. Since $f$ is a regular monomorphism there is some $W$, necessarily of the form $g^{-1}(U) +S$ (where $S \subseteq X$) by (d), open in $B$ such that $f^{-1}(W)=g^{-1}(U)$. Now let $V = U + S$. We need to show that $V$ is open in $C+_A B$ and that $i_1^{-1}(V)=U$. By the definition of the topology of the pushout (as mentioned in (e)), to show $V$ is open is to show that $i_1^{-1}(V)$ and $i_2^{-1}(V)$ are open. However $i_1^{-1}(V)=U$ and $i_2^{-1}(V)=W$ are open. This completes the proof.

Every abelian category is coregular
Trivially this is the case since abelian categories are regular and the axioms of an abelian category are self dual.

The category of groups in not coregular
In what follows all groups will be written multiplicatively and their identity elements will be written with $1$. Let $X$ be a group with at least two elements and let $Y$ be the group with two elements $\{1,a\}$. Now let $k : X+X \to X+Y$ be the morphism defined by $k i_1(x) = (x)$ and $k i_2(x) = (a,x,a)$ where $i_1$ and $i_2$ are the first and second coproduct inclusions. I will leave it as an exercise to show that $k$ is a normal monomorphism and hence certainly regular. Now let $g : X+X\to X$ be the morphism defined by $gi_1(x)=x$, $gi_2(x)=1$ and form the pushout $$\require{AMScd}\begin{CD} X+X @>{k}>> X+Y\\ @V{g}VV @VV{j_2}V \\ X @>>{j_1}> Q. \end{CD}$$ Since for any $x$ in $X$ $j_2(a,x^{-1},a)=j_2(k(i_2(x^{-1})))=j_1(g(i_2(x^{-1})))=j_1(1)=1$ it follows that $j_2(a)=j_2(a,x^{-1},a)j_2(a,x)=j_2(a)j_2(x)$ and hence $j_2(x)=1$. This means that $j_1(x)=j_1(g(i_1(x))=j_2(k(i_1(x))=j_2(x)=1$ and hence $j_1$ is a zero morphism and hence certainly not even monomorphism.

Answer 2

$\mathsf{Grp}$ is not coregular:

Let $K$ be a group with $N$ one of its nontrivial subgroup (that is, $N\unlhd K$ and $N$ is not $K$ or $1$), $G$ a simple group containing $K$. Let $f:K\to G$ be the canonical embedding and $g:K\to K/N$ be the canonical projection.

For any $X$, $\varphi:G\to X$ and $\psi:K/N\to X$ such that $\varphi f=\psi g$, since $\varphi(N)=\varphi f(N)=\psi g(N)=1$, $\varphi$ is not injective. But $G$ is simple, so we must have $\varphi$ is trivial, which forces that $\psi g$ ($=\varphi f$) is trivial. This implies that $\psi$ is trivial for $g$ is an epimorphism. In conclusion, the pushout of $f$ and $g$ is $1$.