Naturality of bijection given by Yoneda Lemma I'm reading through Wikipedia's proof of the Yoneda Lemma (https://en.wikipedia.org/wiki/Yoneda_lemma), and am having trouble understanding what naturality means in this context.
The articles proves that $Nat(h^A, F) \cong F(A)$. It then goes on to state that if both both sides are seen as functors from $Set^C \times C$ to $Set$, then the bijection provided by the lemma is natural. Other proofs say something like "The bijection is natural in both $F$ and $A$".
My problem is that I don't see what a morphism from $Set^C \times C$ should be mapped to by either of the functors. The functor on the right hand side would need to take a morphism from $f:A\rightarrow B$ from $C$ and a natural transformation $\sigma:F \rightarrow B$  and map them to a function from $F(A)$ to $G(B)$. I don't really see a canonical way of doing so, can somebody help me?
 A: The details are reasonably grim but if you stare at them hard enough you'll see that the morphisms are the only things they could possibly be.
The actions on objects of the two functors $\mathsf{Set}^{\mathcal{C}} \times \mathcal{C} \to \mathsf{Set}$, respectively, are
$$(F,A) \mapsto \mathsf{Nat}(y^A, F) \quad \text{and} \quad (F,A) \mapsto F(A)$$
The actions of the two functors on morphisms are given as follows. Fix $\alpha : F \to G$ and $f : A \to B$.
The first functor sends $(\alpha, f)$ to the function
$$\theta_{\alpha,f} : \mathsf{Nat}(y^A, F) \to \mathsf{Nat}(y^B, G)$$
defined as follows. Given $\eta : y^A \to F$ and $C \in \mathsf{ob}(\mathcal{C})$, we have $\eta_C : \mathsf{Set}(A,C) \to F(C)$. Define
$$\theta_{\alpha,f}(\eta)_C : \mathsf{Set}(B,C) \to G(C)$$
by
$$\theta_{\alpha,f}(\eta)_C(B \xrightarrow{g} C) = \alpha_C(\eta_C(g \circ f))$$
This yields a natural transformation $\theta_{\alpha,f}(\eta) : y^B \to G$, and hence a function $\theta_{\alpha,f} : \mathsf{Nat}(y^A,F) \to \mathsf{Nat}(y^B, G)$.
The second functor sends $(\alpha, f)$ to the function $G(f) \circ \alpha_B : F(A) \to G(B)$:
$$F(A) \xrightarrow{\alpha_A} G(A) \xrightarrow{G(f)} G(B)$$
Equivalently, by naturality of $\alpha$, it sends $(\alpha, f)$ to the function $\alpha_A \circ F(f)$.
A: A morphism $\left(F, A\right) \to \left(G, B\right)$ in $\operatorname{Set}^C \times C$ has the form $\left(\alpha, p\right)$, where $\alpha : F \to G$ is a morphism in $\operatorname{Set}^C$ (that is, a natural transformation $F \Rightarrow G$) and where $p : A \to B$ is a morphism in $C$.
For every morphism $p : A \to B$ in $C$, there is a natural homomorphism $h^p : h^B \to h^A$ of functors from $C$ to $\operatorname{Set}$. It is defined as the natural homomorphism whose $D$-th component (for $D \in C$) is the map $h^B\left(D\right) \to h^A\left(D\right)$ which sends every morphism $q : B \to D$ to the morphism $q \circ p : A \to D$. If you don't already know this construction, do check that it actually defines a natural homomorphism; this is an instructive exercise.
In order to turn $\operatorname{Set}^C \times C \to \operatorname{Set}, \ \left(F, A\right) \mapsto \operatorname{Nat}\left(h^A, F\right)$ into a functor, we need to explain how a morphism $\left(\alpha, p\right) : \left(F, A\right) \to \left(G, B\right)$ in $\operatorname{Set}^C \times C$ gives rise to a map from $\operatorname{Nat}\left(h^A, F\right)$ to $\operatorname{Nat}\left(h^B, G\right)$. Namely, the latter map sends every natural homomorphism $\zeta \in \operatorname{Nat}\left(h^A, F\right)$ to the composition $h^B \overset{h^p}{\to} h^A \overset{\zeta}{\to} F \overset{\alpha}{\to} G$, where $h^p : h^B \to h^A$ is defined as above. It is easy to see that this actually gives a functor (i.e., composition and identities are preserved).
In order to turn $\operatorname{Set}^C \times C \to \operatorname{Set}, \ \left(F, A\right) \mapsto F\left(A\right)$ into a functor, we need to explain how a morphism $\left(\alpha, p\right) : \left(F, A\right) \to \left(G, B\right)$ in $\operatorname{Set}^C \times C$ gives rise to a map from $F\left(A\right)$ to $G\left(B\right)$. Namely, the latter map can be defined either as the composition $F\left(A\right) \overset{F\left(p\right)}{\to} F\left(B\right) \overset{\alpha_B}{\to} G\left(B\right)$, or as the composition $F\left(A\right) \overset{\alpha_A}{\to} G\left(A\right) \overset{G\left(p\right)}{\to} G\left(B\right)$. Both definitions yield the same map, because $\alpha$ is natural. Again, it is easy to check that this actually gives a functor.
A: We can actually exhibit the isomorphism $\phi $ in 
$\tag1Nat(h^A, F) \cong F(A)$. 
For any natural transformation $\nu :h^{A}\to F$, define $\phi :Nat(h^A, F)\to F(A)$ by $\phi (\nu)=\nu _{A}(1_{A})$. 
Note that $\nu _{A}$ is simply the $A$ component of $\nu $. i.e. $\nu _{A}$ is a morphism, in fact a $set$ map: $\hom(A,A)\to FA$ so that $\phi $ sends $\nu $ to the element $\nu _{A}(1_{A})\in FA$. It turns out that with this definition, $\phi $ is bijective, hence an isomporphism of sets. 
And now we may observe that for each $A\in C$ and $F\in Set^{C}$ we get such a $\phi $ which we should then write $\phi _{A,F}$ for clarity. 
To say that $\phi $ is natural in $A$ and $F$ is to say that $\phi $ is a natural transformation of the functors $$\tag2\mathcal F:Set^{C}\times C\to Set$$ defined on objects by 
$\mathcal F(F,A)=Nat(h^A, F)$ 
and on arrows by
$\mathcal F(\tau :F\to G,f:A\to B)=\psi :Nat(h^A, F)\to Nat(h^B, G)$, 
where $\psi $ is defined by $(\psi (\nu ))_{c}=\tau _{c}\circ \nu _{c}\circ f^*$
and 
$$\tag3\mathcal G=eval:Set^{C}\times C\to Set$$ 
$\mathcal G$ is clearly a functor and you can check that $\mathcal F$ is also.
The claim is then that $\phi _{A,F}$ are components of the natural isomorphism $\phi:\mathcal F\cong \mathcal G$.
The proof is somewhat lengthy but routine. Awodey has a clear demostration of this in his book Category Theory (Oxford Logic Guides) 2nd Edition.
A: Since others have given detailed answers, at appropriate math.se length, I won't try to do that. But as Clive says, the details can seem reasonably grim at first sight. So I'll just add that if you want a more slow-motion version, I have a bash at making the full-dress Yoneda Lemma, including the naturality claims, as clear as I can in these draft Notes on Basic Category Theory, esp Ch. 11.  
