Then there exists an isomorphism-reflecting regular functor $I:C\to C^{\prime }$, where $C^{\prime }$ is also small and almost totally-supported, such that for every well-supported $A\in \mathrm{ob}C$ there exists a morphism $1\to IA$ in $C^{\prime }$ not factoring through $I(m)$ for any proper subobject $m:A^{\prime }\rightarrowtail A$ in $C$.