Showing that the Yoneda Map is Natural From Categories for the Working Mathematician pg. 61:


I understand everything about the Yoneda Lemma except this naturalness component. In particular, I'm having trouble wrapping my mind around the functors $N$ and $E$.
Questions.


*

*What is $D(f,-)$? I understand what $D(o,-)$ for $o$ an object of $D$. But now $o$ is a morphism.

*How is $Nat(D(f,-), F)$ an arrow from $Nat(D(r,-), K)$ to $Nat(D(r',-),K')$?

*What is the morphism function of $E$? That is, what does $E(F:K \rightarrow K', f:r \rightarrow r')$ map to?
 A: $\newcommand{\Nat}{\operatorname{Nat}}$


*

*If $f\colon r\rightarrow r'$ is an arrow, then $D(f,-)\colon D(r',-) \rightarrow D(r,-)$ is the natural transformation given by precomposing $f$, i. e.
$$
D(f,s)\colon D(r',s)\longrightarrow D(r,s),\quad g\longmapsto g\circ f\quad \text{ (for all objects $s$ in $D$).}
$$

*Given an arrow $f\colon r\rightarrow r'$ and a natural map $F\colon K\rightarrow K'$, then $\Nat(D(f,-),F)\colon \Nat(D(r,-),K) \rightarrow \Nat(D(r',-),K')$ is simply given by
$$
\alpha\longmapsto F\circ \alpha\circ D(f,-)
$$
for any $\alpha\in \Nat(D(r,-),K)$, noticing that both $F$ and $D(f,-)$ are natural maps as is the composite of such.

*If again $f\colon r\rightarrow r'$ is an arrow and $F\colon K\rightarrow K'$ is a natural map, then $E(\langle F,f\rangle)\colon Kr\rightarrow K'r'$ is given by the composite
$$
Kr\xrightarrow{F_r} K'r \xrightarrow{K'(f)} K'r'.
$$ 
Notice that this is the same as defining $E(\langle F,f\rangle)$ by
$$
Kr \xrightarrow{K(f)} Kr' \xrightarrow {F_{r'}} K'r'
$$
since $F$ is natural.
