I am new in category and I am reading Awodey's Category Theory. In the proof of Proposition 8.10 (See the picture of the proof here1, here2
and here3), after identifying $$x\in P(C)$$and
$$x:yC\rightarrow P$$
the author claims the diagram below commutes. But I can not see why.
The author says it comes from naturality in $C$ of Yoneda lemma(see the picture below), but I can not see it.
I am sorry I do not know how to put all the pictures in one link, so there are three parts of them. The proposition is also on the page 196-198 of the book.
Here $y$ is the Yoneda embedding $\mathbf{C} \rightarrow \mathbf{Sets}^{\mathbf{C}^{op}} $ satisfying
$$yC=\operatorname{Hom}_{\mathbf{C}}(-,C):\mathbf{C}^{op}\rightarrow \mathbf{Sets}$$
and taking $f:C\rightarrow D$ to the natural transformation
$$yf=\operatorname{Hom}(-,f):\operatorname{Hom}_{\mathbf{C}}(-,C)\rightarrow\operatorname{Hom}_{\mathbf{C}}(-,D)$$
Thanks for any possible help!
Ps. The commutative diagram of Yoneda lemma in this book is as below. Details can be found on the page 189 of the book
Best Answer
The author claims in your second picture (See here2), that the triangle commutes due to the naturality of $C$, so you take the diagram explaining what this means in point (3) and apply it to your situation. We therefore have the following commuting diagram and we know $P(h)(x)=x'$ from equation (8.5) in your first picture (See here1):
This implies $\operatorname{Hom}(yh,P)(x)=x'$ for the corresponding natural transformations $x\colon yC\Rightarrow P$ and $x'\colon yC'\Rightarrow P$ using the Yoneda lemma (the horizontal arrows). Since $\operatorname{Hom}(yh,P)\colon\operatorname{Hom}(yC,P)\rightarrow\operatorname{Hom}(yC',P)$ is given by postcomposition with $yh\colon yC'\rightarrow yC$, we get the desired result: $$yh\circ x=x'.$$