# Why is $\text{Hom}_{\text{Set}}(S, \text{lim} F) \simeq \text{lim } \text{Hom}_{\text{Set}}(S, F(\cdot))$?

Every category theory reference says that the isomorphism in the title is a triviality. How so?

$$\lim F \equiv \text{Hom}_{\hat{C}}(\text{pt}, F) \in \text{Set}$$

so

$$\lim \text{Hom}_{\text{Set}}(S, F(\cdot)) = \text{Hom}_{\hat{C}}(\text{pt}, \text{Hom}_{\text{Set}}(S, F(\cdot))) = ?$$

I am aware of this definition: $$\lim F \simeq \left\lbrace (x_d)_{d \in D} \in \prod_{d \in D} F(d) | \forall (d_i \stackrel{\alpha}{\to} d_j) \in D : F(\alpha)(x_{d_j}) = x_{d_i} \right\rbrace$$

• If you unpack the definitions this should be clear. You can define $\lim F$ quite explicitly as a subset of $\prod F_i$. – Pedro Tamaroff Dec 2 '17 at 18:25
• @PedroTamaroff I know that formula. So you're saying it's a consequence of it. – BananaCats Category Theory App Dec 2 '17 at 18:26

Here's another approach. If you know that limits commute and that products are limits, then you simply have to note that $\text{Hom}_{\text{Set}}(S, X) \simeq \prod_{s\in S}X$, then the result is just an example of commutation of limits.
A map from $S$ to $\lim F$ is uniquely determined by a choice of maps $f_c:S\to F(c)$ for each object $c$ of $C$ such that for any morphism $g:c\to d$ in $C$, $F(g)f_c=f_d$. This is the exact same thing as a natural tranformation from the constant functor pt to $\text{Hom}_{\text{Set}}(S, F(\cdot))$, with $f_c$ being the component of the natural transformation on the object $c$.
If $F : J \to C$ is a functor, there is a canonical function of sets $$\hom(S,\lim F) \xrightarrow{\qquad} \lim \hom(S,F)$$ which is induced by $\pi_{i,*} : \hom(S, \lim F)\to \hom(S,Fi)$ (if $\pi_i$ is the projection from the limit onto $Fi$). Now, the universal property of the limit means that this map is invertible: try to translate the two conditions and see for yourself that they are equivalent!
• This is my favorite one. So we just let $Y = \text{Hom}_{\text{Set}}(S, \lim F)$ and $G(d) = \text{Hom}_{\text{Set}}(S, F(d))$. Then $\pi_d : Y \to G(d)$ satisfies ? How do you get/use the second index under $\pi$? – BananaCats Category Theory App Dec 2 '17 at 22:48