From pg. 83 of Categories for the Working Mathematician:
Here, (6) is the equation:
$$ \theta f = Gf \circ \eta_x : x \rightarrow Ga $$
The author's proof is:
Problem: I don't understand the red underlined claims (which are presented in the text without justification).
To start off with the notation from the book, let $k : a \rightarrow a'$ and $h: x' \rightarrow x$. Then the author is saying that the naturality of $\eta$:
implies
while the fact that $G$ is a functor somehow implies
But why?
To understand in what sense $\theta$ is natural in $a$ and $x$, you must first understand in what sense $A(Fx,a)$ and $X(x,Ga)$ are functorial. In fact, $\theta$ is really a natural transformation between the functors $A(F\_,\_)$ and $X(\_,G\_)$ ; these are functors $X^{op}\times A\to \mathbf{Set}$. An arrow in $(x,a)\to (x',a')$ in $X^{op}\times A$ is simply given by an arrow $\xi:x'\to x$ in $X$ and an arrow $\alpha:a\to a'$ in $A$. For such an arrow, the defintion of the first functor is $$f\in A(Fx,a)\mapsto \alpha\circ f\circ F(\xi)\in A(Fx',a')$$and the defintion of the second one is $$g\in X(x,Ga)\mapsto G(\alpha)\circ g\circ \xi \in X(x',Ga').$$ Now the naturality of $\theta$ means that $$\theta_{x',a'}\circ A(F(\xi),\alpha)=X(\xi,G(\alpha))\circ \theta_{x,a},$$where both sides denotes arrows $A(Fx,a)\to X(x',Ga')$ in $\mathbf{Set}$ (i.e. functions).
So all you have to do is check that the two functions agree for every arrow $f:Fx\to a$, which is equivalent to the equality $$\theta_{x',a'}(\alpha\circ f\circ F(\xi))=G(\alpha)\circ \theta_{x,a}(f)\circ \xi.$$ Now with the definition of $\theta$ given above, we find \begin{align}\theta_{x',a'}(\alpha\circ f\circ F(\xi)) & = G(\alpha\circ f\circ F(\xi))\circ \eta_{x'} \\ & = G(\alpha)\circ G(f)\circ GF(\xi)\circ \eta_{x'}\\ & = G(\alpha) \circ G(f)\circ \eta_x\circ \xi \\ & = G(\alpha)\circ \theta_{x,a}(f)\circ\xi,\end{align} where we have used the functoriality of $G$ in the second line and the naturality of $\eta$ for the third one.
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