Colimits in full subcategory (of all monics) of arrow category

Question

Consider the category $\mathrm{mon}(C)$ (objects are monics in $C$) as full subcategory of $\mathrm{Arrow}(C)$. We know that $\mathrm{Arrow}(C)$ is finitely complete and cocomplete, assuming that $C$ has limits as well as colimits. Is $\mathrm{mon}(C)$ (finitely) cocomplete?

For example consider coproducts: For objects $m_1 \colon A \rightarrow C$ and $m_2 \colon B \rightarrow D$ in $\mathrm{Arrow}(C)$, the coproduct is given by $[m_1, m_2] \colon A+B \rightarrow C+D$ (using the couniversal property of $A+B$). Does $m_1$, $m_2$ being monic imply $[m_1, m_2]$ being monic?

I think it is not true in general. If it is assumed that $C$ is a topos, does that help in cocompleteness of $\mathrm{mon}(C)$ in any way? (For $C$ a topos, $\mathrm{mon}(C)$ will have exponents, but not subobject classifier).

Answer 1

The full subcategory of monomorphisms is closed under limits inside the category of morphisms. (Easy exercise.)

In general, colimits in the category of monomorphisms are not the same as colimits in the category of morphisms (assuming they even exist). But there is sometimes a relation. Assuming the category in question has image factorisation:

• The inclusion of the subcategory of monomorphisms has a left adjoint.
• If the category has colimits, then so too does the category of monomorphisms.
• To calculate a particular colimit in the category of monomorphisms, first take the colimit in the category of morphisms and then apply the left adjoint to it.

In particular, if your category is a topos, then you are in the above situation.