I am reading Dr. Shulman's paper on the variants of epis and monos in a category. I have been unable to verify one statement given as an exercise in the section on adjoints:
Given an adjoint $F\dashv G\colon \mathcal C\to \mathcal D$, $G$ is faithful and conservative (i.e., reflects isos) iff for each $Y\in \mathcal C$, $\varepsilon_Y\colon FGY\to Y$ is extremal epi.
To spell this out:
Supposing for each $Y\in \mathcal C$, $\varepsilon_Y$ is extremal epi, if we are given an iso of the form $Gf\colon GY\to GY'$, we want to show $f$ is iso.
Conversely, supposing $G$ is faithful and conservative, given a factorization $\varepsilon_Y=i\circ p $with $i$ mono, we need to show $i$ is iso.
For the "if" direction, consider the diagram below: $$\require{AMScd} \begin{CD} F G Y @>{\epsilon_Y}>> Y \\ @V{F G f}VV @VV{f}V \\ F G Y^\prime @>>{\epsilon_{Y^\prime}}> Y^\prime \end{CD}$$ Since $G$ is known to be faithful (because the counit components are epimorphisms), it reflects monomorphisms, so $f : Y \to Y'$ is a monomorphism. But $F G f : F G Y \to F G Y'$ is an isomorphism and $\epsilon_{Y'} : F G Y' \to Y'$ is an extremal epimorphism, so $f : Y \to Y'$ must be an isomorphism.
On the other hand, for the "only if" direction, observe that $G \epsilon_Y : G F G Y \to G Y$ is a split epimorphism, and $G$ preserves monomorphisms, so $G i$ must be an isomorphism; but $G$ is also conservative, so $i$ is an isomorphism.