# How can we see Evaluation map is a natural transformation.

Anyone can help with it.

Let $F:Set\to Set$ be given by $F(X)=H^S(X)\times S$. Let $\delta :H^S(X)\times S \to X$ be the evaluation map given by $\delta(p,s)=p(s)$, the value of the function p at the point s. How can we prove that $\delta$ is a natural transformation $F \Rightarrow 1_{Set}$.

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what is $H^S(X)$ supposed to mean? – magma Jan 14 '14 at 17:28

Let $H:\mathbf{Set}\to\mathbf{Set}$ denote the covariant hom functor $\text{Hom}(S,-)$, and $G$ the functor sending a map $k:Z\to Z'$ to the map $k\times\text{Id}:Z×S\to Z'×S$. Then your $F$ is just the composition $GH$ and this determines the arrow function of $F$. Now let $f:X\to X'$. The composition $$H(X)×S\xrightarrow\delta X\xrightarrow f X'$$ first maps a pair $(g,s)$ to the element $g(s)$, then to the element $f(g(s))$. On the other hand, the composition $$H(X)×S\xrightarrow{F(f)} H(X')×S\xrightarrow\delta X'$$ first sends the pair $(g,s)$ to the pair $(f\circ g,s)$, then evaluates this to $fg(s)$.

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By the definition. You already have a transformation $\delta\colon F\to I_{\mathbf{Set}}$, so you should only check naturality.

If I understood correctly, $F\colon\mathbf{Set}\to\mathbf{Set}$ is a functor, such that: $$F(X)=X^S\times S;\qquad (F(f))(p,s)=(f\circ p,s).$$ The reason why I understood so is that the "standart" evaluation map in $\mathbf{Set}$ is a map $X^S\times S\to X$.

Let $f\colon X\to Y$ be a mapping, then we have to prove that $f\circ\delta(X)=\delta(Y)\circ F(f)$. Both $f\circ \delta(X)$ and $\delta(Y)\circ F(f)$ are mappings $X^S\times S\to Y$, therefore we should check that they coincide at any point $(p,s)\in X^S\times S$: $$(f\circ \delta(X))(p,s)=f(\delta(X)(p,s))=f(p(s))=(f\circ p)(s)=$$

$$=\delta(Y)(f\circ p,s)=\delta(Y)(F(f)(p,s))=(\delta(Y)\circ F(f))(p,s).$$

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