Hom functor preserving limits - "natural isomorphism"

Question

$\DeclareMathOperator\Hom{Hom}$ I was reading the proof of proposition 2.1.

There is a natural isomorphism $$\Hom(Y, \lim_{\leftarrow} X_i) \cong \lim_{\leftarrow} \Hom(Y,X_i)$$

Are we regarding both sides as functors: $C^{op} \rightarrow Set$? I don't think this is proved but is this what we are actually supposed to show (?)

Answer 1

Yes; in statement 1, the intention seems to be to regard $Y$ as the only variable and show both sides are naturally isomorphic as functors $C^\mathrm{op} \to \mathbf{Set}$.

Another way to organize this fact is that you have the yoneda embedding

$$ \mathbf{y} : C \to \mathbf{Set}^{C^{\mathrm{op}}}$$

There is a a cone from $\mathbf{y}(\lim_i X_i)$ to the $\mathbf{y}(X_i)$, and the universal property of limits gives a map

$$ \mathbf{y}(\lim_i X_i) \to \lim_i \mathbf{y}(X_i)$$

The claim is that this is an isomorphism in $\mathbf{Set}^{C^{\mathrm{op}}}$.

(This equivalence is not immediate; there is another key fact that if $D$ has $J$-shaped limits, then any functor category $D^C$ also has $J$-shaped limits and they are computed pointwise)

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Like most notions in category theory, the notion of a limit cone is only defined up to unique isomorphism; there are lots of limits of the diagram $\mathbf{y}(X_i)$, but any two limit cones are isomorphic in a unique way.

The above isomorphism extends to an isomorphism of cones, so even if it's not a strict equality with whichever limit you've chosen for the right hand side, it's still enough to show the left hand side is also a limit.

Answer 2

I think your conjecture is quite right. In the proof, the first naturality in $Y$ is quite clear, whereas for the second one has to think a bit longer.

But when all inverse limits exist, then we can do the following:

There is the category over a diagram. Namely, if $\mathcal J$ is the index category of the diagram, then it's just the functor category $\mathcal J \to \mathcal C$. We then should have naturality in both variables: The $Y$ variable, which is in $\mathcal C$, and the $(X_i)$ variable, which is in that functor category.