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Let $\beta : I^{op} \to \text{Set}$ be a functor and $X \in \text{Set}$.

I want to show that there is a natural isomorphism

$$ \text{Hom}_{\text{Set}}(X, \lim_{\longleftarrow} \beta) \xrightarrow{\sim} \lim_{\longleftarrow} \text{Hom}_{\text{Set}}(X, \beta) $$

Firstly, what do they mean by natural here, with respect to what functoriality? It is given that the functor $\beta'$ in the limit on the right is $\beta'(i) = \text{Hom}_{\text{Set}}(X, \beta(i))$.

Secondly, how do we establish the mapping? I know that if $a : X \to \lim\limits_{\longleftarrow} \beta$ in the category $\text{Set}$, then for $x \in X$, $a(x)$ is a natural map in $\text{Hom}_{I^{\wedge}}(\text{pt}_{I^{\wedge}}, \beta)$ such that for all $f : j \to i$ in $I$ we have $\beta(f)\circ a(x)_i = a(x)_j$.

The right side is the same thing except $\text{Hom}_{Set}(X, \beta(f)) \circ a(x)_i =a(x)_j$ .

So they appear to be the same sort of. Does $\text{Hom}_{\text{Set}}(X, g) = g$? If not, how do we show isomorphism?

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The term 'natural isomorphism' really just means natural isomorphism: the two expressions on either side of the $\overset{\sim}{\longrightarrow}$ symbol are both functors $I^{\mathrm{op}} \to \mathbf{Set}$. With that said, the isomorphism is also natural in $X$, but I imagine they're not asking you to prove that.

Now $\mathrm{Hom}_{\mathbf{Set}}(X, {-})$ is a functor $\mathbf{Set} \to \mathbf{Set}$. The action of this functor on functions is given by postcomposition. Explicitly, for a function $g : Y \to Z$, the function $$\mathrm{Hom}_{\mathbf{Set}}(X,g) : \mathrm{Hom}_{\mathbf{Set}}(X, Y) \to \mathrm{Hom}_{\mathbf{Set}}(X,Z)$$ is defined by $\mathrm{Hom}_{\mathbf{Set}}(X,g)(h) = g \circ h$ for all $h : X \to Y$.

Use this to (even more carefully) write down the action of both functors on objects of $I$. The fact that the left and right sides are (almost) 'the same thing' is what gives you the natural isomorphism!

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    $\begingroup$ It does not mean both sides are functors $I^{op}\to\mathbf{Set}$. It almost certainly means that the expression is natural in $X$ as this witnesses the limit as a representation of a functor. (It's possible it's additionally viewing it as natural in $\beta$, but that requires all limits of shape $I$ to exist and is more of a nice to note property.) You probably meant to write $\mathbf{Set}^{op}\to\mathbf{Set}$, though I suspect that's not very clarifying to the OP without indicating how it is such a functor. $\endgroup$ – Derek Elkins Jun 11 '18 at 20:30

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