Which diagrams are reflected by adjunctions? Functors $F\colon C\to D$ preserve diagrams: if a diagram $A$ commutes in $C$, then $FA$ commutes in $D$ (where $FA$ is the diagram in $D$ that can be obtained from $A$ by applying to every object and morphism in that diagram the functor $F$). However, in general, the converse is not true: functors don't necessarily reflect diagrams.
What is true is that equivalences of categories both preserve and reflect diagrams, i.e., one can without problems translate back and forth between the categories $C$ and $D$.
I wonder: Given a pair of adjoint functors (say $F\dashv U$, where $F\colon C\to D$ and $U\colon D\to C$) between $C$ and $D$, how much can we translate back and forth between $C$ and $D$?
Through a previous thread I came to the following conclusion: a diagram
\begin{array}{cc}
\,\,\,\,\,\,\,A \\
\\
i \downarrow & \,\,\,\,\,\,\,\searrow \,{f} \\
\\
A' & \xrightarrow{r}  & UB
\end{array}
in $C$ commutes if and only if the diagram
\begin{array}{cc}
\,\,\,\,\,\,\,FA \\
\\
Fi \downarrow & \,\,\,\,\,\,\,\searrow \,{\alpha(f)} \\
\\
FA' & \xrightarrow{\alpha(r)}  & B
\end{array}
commutes in $D$, where $\alpha$ denotes the natural bijection $\hom(-, UB)\cong \hom(F-,B)$.
I admit this is not exactly "reflection of diagrams" in the above sense, since only on one arrow we apply $F$ while on the other two we apply $\alpha$.
Question: Can this observation be generalized? Which kind of diagrams are preserved and reflected by adjunctions and in which sense?
 A: Instead of talking about reflecting diagrams, let's talk about reflecting equality of arrows. After all, to say a diagram commutes is to say that any two paths through it (i.e., two arrows from one object in the diagram to another) are equal. To say that a functor $F:C\to D$ reflects equality of arrows is to say that each map $F\colon \mathrm{Hom}_C(A,B)\to \mathrm{Hom}_D(F(A),F(B))$ is injective. That is, I think you're asking when adjoint functors are faithful.
One answer is that the left adjoint $F$ is faithful when the unit $\eta\colon \text{Id}_C\to UF$ is a monomorphism, and the right adjoint $G$ is faithful when the counit $\varepsilon\colon FU\to \text{Id}_D$ is an epimorphism. And these conditions are frequently satisfied!
Given arrows $f,g\colon A\to B$ in $C$, such that $F(f) = F(g)$, we also have $GF(f) = GF(g)$, and we get a naturality square: $$\require{AMScd} 
\begin{CD}
A @>{f,g}>> B\\
@V{\eta_A}VV @VV{\eta_B}V \\
GF(A) @>{GF(f) = GF(g)}>> GF(B)
\end{CD}$$
Now $\eta_B\circ f = GF(f)\circ \eta_A = GF(g) \circ \eta_B = \eta_B\circ g$, so if $\eta_B$ is monic, we get $f = g$.
The argument for the counits is dual, using the naturality square: $$\begin{CD}
FG(A) @>{FG(f)=FG(g)}>> FG(B)\\
@V{\varepsilon_A}VV @VV{\varepsilon_B}V \\
A @>{f,g}>> B
\end{CD}$$
In the general case, we still get that if $F(f) = F(g)$, then $f$ and $g$ are coequalized by $\eta_B$, and if $G(f) = G(g)$, then $f$ and $g$ are equalized by $\varepsilon_A$.
