Proposition 8.7. For any locally small category $\mathbb{C}$, the functor category $\mathbb{Sets}^{\mathbb{C}^{op}}$ is complete. Moreover, for every object $C \in \mathbb{C}$, the evaluation functor
$$ev_C:\mathbb{Sets}^{\mathbb{C}^{op}} \to \mathbb{Sets}$$
preserves all limits.Proof. Suppose we have $J$ small and $F:J \to \mathbb{Sets}^{\mathbb{C}^{op}}$. The limit of $F$, if it exists, is an object in $\mathbb{Sets}^{\mathbb{C}^{op}}$, hence is a functor,
$$(\lim_{j \in J}F_j):\mathbb{C}^{op} \to \mathbb{Sets}$$
By the Yoneda lemma, if we had such a functor, then for each object $C \in \mathbb{C}$ we would have a natural isomorphism,
$$(\lim F_j)(C) \cong \mathrm{Hom}(yC,F_j)$$
But then it would be the case that
$$ \mathrm{Hom}(yC, \lim F_j) \cong \lim \mathrm{Hom}(yC, F_j) \cong \lim F_j(C) $$
in $\mathbb{Sets}$
where the first isomorphism is because representable functors preserve limits, and the second is Yoneda again. Thus, we are led to define the limit $\lim_{j \in J}F_j$ to be
$$(\lim_{j \in J}F_j)(C)=\lim_{j \in J}(F_jC) \tag{8.4}$$
that is, the pointwise limit of the functors $F_j$. The reader can easily work out how $\lim F_j$ acts on $\mathbb{C}-arrows$, and what the universal cone is, and our hypothetical argument then shows that it is indeed a limit in $\mathbb{Sets}^{\mathbb{C}^{op}}$.Finally, the preservation of limits by evaluation functors is stated by (8.4).
I'm having some trouble typing the limit signs, so I ignored the $\leftarrow$s below the $\lim$s because I don't know how to stack two lines of subscripts below the limit signs.
I'm confused with 'our hypothetical argument then shows that it is indeed a limit'. Why does this follow from the hypothetical argument? Why isn't there a general cone in $\mathbb{Sets}^{\mathbb{C}^{op}}$ of which the universal mapping property of the limit should be verified?
Thanks!
Best Answer
This is certainly missing from the proof, but here's a justification.
Observation: due to Yoneda, it suffices to check limits in a functor category on the representables.
Why? Suppose $F\to F_j$ is a cone which satisfies the universal property for maps out of representables $\newcommand\C{\mathcal{C}}\C(-,A)$. Suppose $X\to F_j$ is a cone. Let $a\in X(A)$. By Yoneda, this gives a map $\C(-,A)\to X$, and thus a cone $\C(-,A)\to X\to F_j$. By assumption, there is a unique map $\C(-,A)\to F$ compatible with this cone, corresponding to some element $\tau_A(a)\in F(A)$. Define a map $\tau_A : X(A)\to F(A)$ by $a\mapsto \tau_A(a)$.
Then you can check that the $\tau_A$ are natural as $A$ varies, since if $f : A\to B$, $\tau_B(a|_f)$ corresponds to the map $\C(-,B)\xrightarrow{f^*} \C(-,A)\to X\to F$, which is also what $f^*\tau_A(a)$ corresponds to. Thus we have a map of functors $\tau : X\to F$.
It's clear by construction that $X\xrightarrow{\tau} F\to F_j$ gives our original cone back, and any map $X\to F$ compatible with the cones must send $a\in X(A)$ to $\tau_A(a)$ by the uniqueness property of maps into $F$ on representables.