If the counits of an adjunction are epimorpihsms then the right adjoint reflects monomorphisms? Given an adjunction $F \dashv G$, I need to show that if the counits $\varepsilon_Y: FG(Y) \rightarrow Y$ are epimorphisms then $G$ reflects monomorphisms. I am completely stuck on what properties of epis and monos I should use, so any pointer in the right direction would be great. Thanks.
 A: By the Yoneda Lemma, we have, for any locally small category $\mathcal C$, a bijection of sets
$$
\text{Nat}(\mathcal C(a,-),\mathcal C(b,-)) = \mathcal C(b,a)
$$
That means a natural transformation between the two representable functors is uniquely determined by an arrow $f:b\to a$. Indeed, if $f:b\to a$, then the natural transformation $f^*$ sends $g:a\to c$ to $gf:b\to c$. This implies that $f$ is epic if and only if each component of $f^*:\mathcal C(a,-)\Rightarrow\mathcal C(b,-)$ is an injection.
Given an adjunction $(F,G):\mathcal X\to\mathcal A$, we have a sequence of natural transformations
$$
\mathcal A(a,-)\Rightarrow\mathcal  X(Ga,G-)\Rightarrow\mathcal  A(FGa,-)
$$
The first sends an arrow $g:a\to b$ to the arrow $Gg$, and the second one sends that arrow to its adjunct $(Gg)^\sharp$. By the Yoneda Lemma, the composite transformation $\sigma$ is given by the arrow $\sigma_a(1_a)=(G1_a)^\sharp=1_{Ga}^\sharp=\varepsilon_a$, and it sends $g:a\to b$ to the arrow $g\epsilon_a$, thus $\sigma=\varepsilon_a^*$.
By the above remark, $\varepsilon_a^*$ is always injective iff this counit is an epimorphism. If all counits are epimorphisms, then every map $\mathcal A(a,b)\to \mathcal X(Ga,Gb)$ is injective, which means that $G$ is faithful.
Now you can just use the fact that faithful functors reflect monomorphisms (and epimorphisms).
On the other hand, if $G$ reflects epimorphisms, then each $ε_a$ must be an epimorphism. To see this, recall that for every adjunction $(F,g,\eta,ε)$, we have identities $Gε_a\circ\eta_{Ga}=1_{Ga}$ and $ε_{Fx}\circ F\eta_x=1_{Fx}$, so the image of every counit is a retraction, thus an epi.
