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$\DeclareMathOperator{\sets}{\textbf{Sets}}$ $\DeclareMathOperator{\nat}{Nat}$ $\DeclareMathOperator{\hom}{Hom}$ 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.

user557
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caffeinemachine
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    On objects the first functor is defined by $(F,A) \mapsto \text{Nat}(\text{Hom}(A,-),F)$ while the second is defined by $(F,A) \mapsto F(A)$. On morphisms the first functor is defined by $((\alpha,f):(F,A)\to (F',A')) \mapsto (\theta \mapsto (\alpha \cdot \theta \cdot \text{Hom}(f,-)))$ while the second is defined by $((\alpha,f):(F,A)\to (F',A')) \mapsto (x \mapsto \alpha_{A'}(F(f)(x)))$ – Nex Sep 13 '16 at 19:42
  • JIC... if we considered the case where $f : X \to Y$ denotes a function between sets and $x$ denotes an element of $X$, can you see how the expression $f(x)$ can be regarded as (the evaluation of) a function $Y^X \times X \to Y$? –  Sep 15 '16 at 06:51
  • I had this same question, thank you! – Andrius Kulikauskas Apr 14 '20 at 13:22
  • Thank you, @Nex , I found your answer most helpful. If I understood correctly, the morphism $(\alpha , f): (F,A)\rightarrow (F',A')$ gets mapped to the set function which takes $x \in F(A)$ to $\alpha_{A'}(F(f)(x)) \in F'(A')$. And that set function is based on the natural transformation $\alpha$ and could also be written as $x \mapsto F'(f)(\alpha_A(x))$. – Andrius Kulikauskas Apr 15 '20 at 10:23

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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 $\mathrm{Set}^\mathcal{C} \times \mathcal{C}$.

Nicolas Cage
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