$\DeclareMathOperator{\sets}{\textbf{Sets}}$ $\DeclareMathOperator{\nat}{Nat}$ $\DeclareMathOperator{\hom}{Hom}$ I am trying to understand the Yoneda lemma from wikipedia and I am stuck at one point.
Let $C$ be a category and $F$ be a functor $F:C\to \sets$. Then for each object $A$ in $C$, there is a bijection $$\nat(\hom(A, -), F)\cong F(A)$$
But then the wiki article states that "Moreover, this isomorphism in natural in $A$ and $F$ when both sides are regarded as functors from $\sets^C\times C$ to $\sets$". (Here $\sets^C$ denotes the category of functors from $C$ to $\sets$.)
I am unable to see how to view $\nat(\hom(A, -), F)$ and $F(A)$ as functors $\sets^C\times C\to \sets$. Can somebody
Can somebody please spell this out a little bit. Thank you.
First, let's fix some notation. For convenience let's denote Hom$(A,-)$ by just $A^\wedge$. For clarity, if $\mathcal{D}$ is any category and $X,Y \in \mathcal{D}$ then denote the Hom-set from $X$ to $Y$ by $\mathcal{D}(X,Y)$.
With these conventions $$\mathrm{Nat}(\mathrm{Hom}(A,−),F) \cong F(A)$$ becomes $$\mathrm{Set}^\mathcal{C}(A^\wedge,F) \cong F(A).$$
An object of $\mathrm{Set}^\mathcal{C}$ is a functor $\mathcal{C} \to \mathrm{Set}$ and morphisms are natural transformations.
If we want to view this isomorphism as a natural transformation then we should first specify its source and target functors.
Its source is the composition $$ \mathcal{C} \times\mathrm{Set}^\mathcal{C} \overset{Y \times 1} \longrightarrow (\mathrm{Set}^\mathcal{C})^{op} \times \mathrm{Set}^\mathcal{C} \overset{\mathrm{Hom}} \longrightarrow \mathrm{Set}, \;\;\;\;\;\; (A,F) \mapsto \mathrm{Set}^\mathcal{C}(A^\wedge,F)$$ where $Y$ is the (contravariant) Yoneda-embedding $A \mapsto A^\wedge$. The target functor is just the evaluation functor $(F,A) \mapsto F(A)$. On morphisms (= pairs of morphisms in $\mathcal{C}$ and morphisms in $\mathrm{Set}^\mathcal{C}$) they are defined accordingly.
These two functors themselves both have source $\mathcal{C}\times \mathrm{Set}^\mathcal{C}$ and target Set.
tl;dr It should be $\mathcal{C}\times \mathrm{Set}^\mathcal{C}$ instead of $\mathrm{Set}^\mathcal{C} \times \mathcal{C}$.