Showing that the Yoneda Map is Natural

Question

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.

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

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

3. What is the morphism function of $E$? That is, what does $E(F:K \rightarrow K', f:r \rightarrow r')$ map to?

Answer 1

$\newcommand{\Nat}{\operatorname{Nat}}$

1. 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. $$ D(f,s)\colon D(r',s)\longrightarrow D(r,s),\quad g\longmapsto g\circ f\quad \text{ (for all objects $s$ in $D$).} $$

2. 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.

3. 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.