Clarify about adjoint functors Consider the adjunction $F\dashv G:\mathbf D\to \mathbf C$, with unity $\eta$ and counity $\varepsilon$. Is it true that $\varepsilon _{FGc}=FG\varepsilon_c $ in general? If not,  under what condition is ture? (For example it seems to me that Mac Lane uses this argument in the proof of  the existence of comparison functor, theorem VI$.3.1$)
 A: No, this is not true in general. For example, take $F$ to be the free group functor $\mathsf{Set}\to \mathsf{Grp}$ and $G$ the forgetful functor $\mathsf{Grp}\to \mathsf{Set}$. For a group $x$, $\varepsilon_x\colon FGx\to x$ is the "evaluation" map which maps a reduced word in the elements of $x$ (an element of the free group on the underlying set of $x$) to the element of $x$ it evaluates to.
Now $\varepsilon_{FGx}$ and $FG\varepsilon_x$ are both maps $FGFGx\to FGx$, and $FGFGx$ consists of "reduced words in reduced words in elements of $x$". But the map $\varepsilon_{FGx}$ evaluates the outer words, while $FG\varepsilon_x$ evaluates the inner words. For example, if $w$ and $w'$ are elements of $FGx$, then $w\cdot w'$ is an element of $FGFGx$. We have $FG\varepsilon_{x}(w\cdot w') = \varepsilon_x(w)\cdot \varepsilon_x(w')$ (this is a word in $FGx$ consisting of a product of two elements of $x$), but $\varepsilon_{FGx}(w\cdot w')$ is the reduction of the word $ww'$ obtained by concatenating the words $w$ and $w'$, which could be an arbitrarily complicated word in $FGx$.

The point you're missing in Mac Lane's argument is that while it's not true that $FG\varepsilon_x = \varepsilon_{FGx}$, these two arrows are coequalizd by $\varepsilon_x$:  $\varepsilon_x\circ (FG\varepsilon_x) = \varepsilon_x\circ \varepsilon_{FGx}$.
Recall that $\varepsilon$ is a natural transformation $FG\to 1_D$. This means that for any arrow $f\colon x\to y$ in $D$, we have a commutative square: $\require{AMScd}$
\begin{CD}
FGx @>{FGf}>> FGy\\
@V{\varepsilon_x}VV @VV{\varepsilon_y}V\\
x @>{f}>> y
\end{CD}
Now set $x = FGy$ and $f = \varepsilon_y$. Then the naturality square above becomes: \begin{CD}
FGFGy @>{FG\varepsilon_y}>> FGy\\
@V{\varepsilon_{FGy}}VV @VV{\varepsilon_y}V\\
FGy @>{\varepsilon_y}>> y
\end{CD}
Finally, apply $G$ to the above square:
\begin{CD}
GFGFGy @>{GFG\varepsilon_y}>> GFGy\\
@V{G\varepsilon_{FGy}}VV @VV{G\varepsilon_y}V\\
GFGy @>{G\varepsilon_y}>> Gy
\end{CD}
This is exactly the square that expresses the first axiom of algebras.
