Do adjoint functors "preserve" commutative diagrams? Let $(L,R)$  be a pair of adjoint functor. 
How to show that the commutativity of the left diagram induces the commutativity of the right one?


Let $(\eta,\epsilon):F \dashv G:\mathfrak{C} \to \mathfrak{D}$ be an adjoint pair of functors and assume you have the commuting diagram
$$
\array{
 A & \xrightarrow{f} & GX \\
 h \downarrow & & \downarrow G(k) \\
 B & \xrightarrow{g} & GY 
}
$$
in $\mathfrak{C}$. Then applying the functor $F$ to the diagram gives the commuting diagram
$$
\array{
 FA & \xrightarrow{F(f)} & F(GX) \\
 F(h) \downarrow & & \downarrow F(G(k)) \\
 FB & \xrightarrow{F(g)} & F(GY) 
}
$$
in $\mathfrak{D}$. Using the counit allows us to produce the commuting diagram
$$
\array{
 FA & \xrightarrow{F(f)} & F(GX) & \xrightarrow{\epsilon_{X}} & X \\
 F(h) \downarrow & & \downarrow F(G(k)) & & \downarrow k\\
 FB & \xrightarrow{F(g)} & F(GY)  & \xrightarrow{\epsilon_Y} & Y
}
$$
which contracts to the commuting diagram
$$
\array{
 FA & \xrightarrow{\epsilon_X \circ F(f)} & X \\
 F(h) \downarrow & & \downarrow k \\
 FB & \xrightarrow{\epsilon_Y \circ F(g)} & Y
}
$$
in $\mathfrak{D}$. There is some work to check that the middle rectangle actually commutes, but it all follows from the triangle identities of adjunction and the couniversal property of $\epsilon$.
